Eigenvalues and Special Matrices

This chapter introduces fundamental concepts in Linear Algebra, focusing on special types of matrices, eigenvalues, and eigenvectors. A thorough understanding of these topics, including the Cayley-Hamilton Theorem, is crucial for solving advanced problems and is frequently assessed in the CUET PG MA examination.

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Chapter Contents

| # | Topic |
|---|-------|
| 1 | Special Types of Matrices |
| 2 | Eigenvalues and Eigenvectors |
| 3 | Cayley-Hamilton Theorem |

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We begin with Special Types of Matrices.

Part 1: Special Types of Matrices

Matrices are fundamental mathematical objects in linear algebra, extensively used across various scientific and engineering disciplines. For the CUET PG examination, a comprehensive understanding of special matrix types and their properties is critical for solving problems related to systems of linear equations, transformations, and eigenvalue analysis. We explore these specific classifications, focusing on their definitions and practical applications in problem-solving.

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Core Concepts

1. Symmetric Matrices

We define a square matrix $A$ as symmetric if it is equal to its transpose, i.e., $A = A^T$. This implies that the element $a_{ij}$ in the $i$-th row and $j$-th column is equal to $a_{ji}$ for all $i,j$.

Quick Example:
Consider the matrix $A$. We determine if it is symmetric.

Step 1: Given matrix $A$.

Step 2: Compute the transpose $A^T$.

Step 3: Compare $A$ and $A^T$.
Since $A = A^T$, the matrix $A$ is symmetric.

:::question type="MCQ" question="Which of the following matrices is symmetric?" options=["

","","",""] answer="" hint="A matrix $A$ is symmetric if $A = A^T$. Check this condition for each option." solution="For a matrix $A$ to be symmetric, its elements must satisfy $a_{ij} = a_{ji}$.

$$\begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$$

Here $a_{12} = 2$ and $a_{21} = -2$. Since $a_{12} \ne a_{21}$, it is not symmetric.

$$\begin{bmatrix} 1 & 3 \\ 3 & 2 \end{bmatrix}$$

Here $a_{12} = 3$ and $a_{21} = 3$. Since $a_{12} = a_{21}$, it is symmetric.

$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

Here $a_{12} = 1$ and $a_{21} = 1$. Since $a_{12} = a_{21}$, it is symmetric.

$$\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$$

Here $a_{12} = 0$ and $a_{21} = 2$. Since $a_{12} \ne a_{21}$, it is not symmetric.

Both Option 2 and Option 3 are symmetric. In a typical MCQ, only one option is correct. Assuming this is a single correct answer type, we choose the first valid symmetric matrix encountered.
Answer: $\boxed{\begin{bmatrix} 1 & 3 \\ 3 & 2 \end{bmatrix}}$"
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2. Skew-Symmetric Matrices

A square matrix $A$ is defined as skew-symmetric if it is equal to the negative of its transpose, i.e., $A = -A^T$. This implies that $a_{ij} = -a_{ji}$ for all $i,j$, and consequently, the main diagonal elements must be zero ($a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies a_{ii} = 0$).

Quick Example:
Determine if the matrix $A$ is skew-symmetric.

Step 1: Given matrix $A$.

Step 2: Compute $A^T$.

Step 3: Compute $-A^T$.

Step 4: Compare $A$ and $-A^T$.
Since $A = -A^T$, the matrix $A$ is skew-symmetric.

:::question type="MCQ" question="Let $A$ and $B$ be two symmetric matrices of the same order. Which of the following statements is always correct?" options=["$AB$ is symmetric","$(A+B)$ is symmetric","$A^T B = AB^T$","$AB-BA$ is symmetric"] answer="$(A+B)$ is symmetric" hint="Recall the definitions of symmetric and skew-symmetric matrices and the properties of transpose: $(X+Y)^T = X^T+Y^T$ and $(XY)^T = Y^T X^T$." solution="Let $A$ and $B$ be symmetric matrices. Then $A^T = A$ and $B^T = B$.

$AB$ is symmetric: We check $(AB)^T$.

For $AB$ to be symmetric, $AB = (AB)^T$, which means $AB = BA$. This is not always true; matrix multiplication is not generally commutative. Thus, $AB$ is not always symmetric.

$(A+B)$ is symmetric: We check $(A+B)^T$.

Since $(A+B)^T = A+B$, $(A+B)$ is always symmetric.

$A^T B = AB^T$: Since $A$ and $B$ are symmetric, $A^T = A$ and $B^T = B$.

Therefore, $A^T B = AB$ and $AB^T = AB$.
This implies $A^T B = AB^T$ is always true.

$AB-BA$ is symmetric: We check $(AB-BA)^T$.

This shows that $(AB-BA)$ is skew-symmetric, not symmetric.

The question asks for 'always correct'. Both option 2 and option 3 are always correct based on the properties. If this were an MSQ, the answer would include both. For a standard MCQ where only one option can be chosen, we select the most fundamental and direct property.
Answer: $\boxed{(A+B) \text{ is symmetric}}$"
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3. Hermitian Matrices

For complex matrices, the concept of symmetry extends to Hermitian matrices. A square matrix $A$ with complex entries is Hermitian if it is equal to its conjugate transpose (also known as adjoint), denoted by $A^$A∗. That is, $A = A^$. This implies $a_{ij} = \overline{a_{ji}}$ for all $i,j$. Consequently, the diagonal elements of a Hermitian matrix must be real numbers.

Quick Example:
Verify if matrix $A$ is Hermitian.

Step 1: Given matrix $A$.

Step 2: Compute the conjugate $\overline{A}$.

Step 3: Compute the transpose of $\overline{A}$, which is $A^*$.

Step 4: Compare $A$ and $A^*$.
Since $A = A^*$, the matrix $A$ is Hermitian.

:::question type="MCQ" question="Which of the following matrices is Hermitian?" options=["

","","",""] answer="" hint="A matrix $A$ is Hermitian if $A = A^$A=A∗, where $A^$ = \overline{A}^T. Check that $a_{ij} = \overline{a_{ji}}$ and diagonal elements are real." solution="We check each option for the condition $A = A^*$, or equivalently $a_{ij} = \overline{a_{ji}}$. Also, diagonal elements must be real.

$$\begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix}$$

$a_{11}=1$ (real), $a_{22}=1$ (real).
$a_{12} = i$, $a_{21} = i$. We need $a_{12} = \overline{a_{21}}$. Is $i = \overline{i}$? No, $\overline{i} = -i$. So $i \ne -i$. Not Hermitian.

$$\begin{bmatrix} 0 & 1+i \\ 1-i & 0 \end{bmatrix}$$

$a_{11}=0$ (real), $a_{22}=0$ (real).
$a_{12} = 1+i$, $a_{21} = 1-i$. We need $a_{12} = \overline{a_{21}}$.
Is $1+i = \overline{(1-i)}$? Yes, $\overline{(1-i)} = 1+i$. So $1+i = 1+i$. This matrix is Hermitian.

$$\begin{bmatrix} 2 & 1 \\ 1 & 2i \end{bmatrix}$$

$a_{22}=2i$ is not real. Not Hermitian.

$$\begin{bmatrix} i & 1-i \\ 1+i & i \end{bmatrix}$$

$a_{11}=i$ is not real. Not Hermitian.

Answer: $\boxed{\begin{bmatrix} 0 & 1+i \\ 1-i & 0 \end{bmatrix}}$"
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4. Skew-Hermitian Matrices

A square matrix $A$ with complex entries is skew-Hermitian if it is equal to the negative of its conjugate transpose, i.e., $A = -A^*$. This implies $a_{ij} = -\overline{a_{ji}}$ for all $i,j$. Consequently, the diagonal elements of a skew-Hermitian matrix must be purely imaginary or zero.

Quick Example:
Determine if matrix $A$ is skew-Hermitian.

Step 1: Given matrix $A$.

Step 2: Compute the conjugate $\overline{A}$.

Step 3: Compute $A^* = \overline{A}^T$.

Step 4: Compute $-A^*$.

Step 5: Compare $A$ and $-A^*$.
Since $A = -A^*$, the matrix $A$ is skew-Hermitian.

:::question type="MCQ" question="Let $A$ be a $2 \times 2$ matrix such that

, where $z$ is a complex number. Which type of matrix is $A$?" options=["Symmetric","Hermitian","Skew-Symmetric","Skew-Hermitian"] answer="Skew-Hermitian" hint="Compute $A^$A∗ and $-A^$ and compare with $A$." solution="Given

First, find the conjugate $\overline{A}$:

Next, find the conjugate transpose $A^* = \overline{A}^T$:

Now, let's compare $A$ with $A^$A∗ and $-A^$:
For $A$ to be Hermitian, $A = A^*$:

This would imply $z = -z$, which means $z=0$. This is not generally true for any complex number $z$. So $A$ is not generally Hermitian.

For $A$ to be Skew-Hermitian, $A = -A^*$:

Since $A = -A^*$, the matrix $A$ is skew-Hermitian.

We can also check for symmetric/skew-symmetric. For $A$ to be symmetric, $A=A^T$.

So $A=A^T$ implies $z = -\overline{z}$. This means $z$ must be purely imaginary ($z=ki$ for real $k$). Not generally true.
For $A$ to be skew-symmetric, $A=-A^T$.

So $A=-A^T$ implies $z = \overline{z}$ and $-\overline{z} = -z$. This means $z$ must be real. Not generally true.

Answer: $\boxed{\text{Skew-Hermitian}}$"
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5. Orthogonal Matrices

A square matrix $A$ with real entries is orthogonal if its transpose is equal to its inverse, i.e., $A^T = A^{-1}$. This implies that $A A^T = A^T A = I$, where $I$ is the identity matrix. The columns (and rows) of an orthogonal matrix form an orthonormal basis.

Quick Example:
Verify if matrix $A$ is orthogonal.

Step 1: Given matrix $A$.

Step 2: Compute $A^T$.

Step 3: Compute the product $A A^T$.

Step 4: Compare $A A^T$ with $I$.
Since $A A^T = I$, the matrix $A$ is orthogonal.

:::question type="MCQ" question="If

is an orthogonal matrix, then the value of $x$ is:" options=["$1$","$-1$","$2$","$-2$"] answer="$2$" hint="For an orthogonal matrix $A$, the dot product of any two distinct column vectors (or row vectors) is zero, and the dot product of a column vector (or row vector) with itself is one (if normalized). Use the property that columns form an orthonormal basis." solution="Let the matrix be $M = \begin{bmatrix} 1 & -2 & 2 \\ 2 & -1 & -2 \\ x & 2 & 1 \end{bmatrix}$. If $A = \frac{1}{3}M$ is orthogonal, then the rows of $M$ must be orthogonal to each other and have a squared length of $3^2=9$.

Row 1: $R_1 = \begin{bmatrix} 1 & -2 & 2 \end{bmatrix}$
Row 2: $R_2 = \begin{bmatrix} 2 & -1 & -2 \end{bmatrix}$
Row 3: $R_3 = \begin{bmatrix} x & 2 & 1 \end{bmatrix}$

First, check orthogonality of $R_1$ and $R_2$:

This condition is satisfied.

Next, check orthogonality of $R_1$ and $R_3$:

For orthogonality, $R_1 \cdot R_3 = 0$:

Next, check orthogonality of $R_2$ and $R_3$:

For orthogonality, $R_2 \cdot R_3 = 0$:

Finally, check the squared length of $R_3$. For $A$ to be orthogonal, the rows of $M$ must have a squared length of $3^2=9$.

This must be equal to $9$:

All orthogonality conditions ($R_1 \cdot R_3 = 0$ and $R_2 \cdot R_3 = 0$) consistently give $x=2$. The length condition gives $x=\pm 2$. For all conditions to hold, we must have $x=2$.
Answer: $\boxed{2}$"
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6. Unitary Matrices

A square matrix $A$ with complex entries is unitary if its conjugate transpose is equal to its inverse, i.e., $A^$ = A^{-1}A∗=A−1. This implies that $A A^$ = A^* A = I. Unitary matrices are the complex analogue of orthogonal matrices, preserving the inner product in complex vector spaces.

Quick Example:
Verify if matrix $A$ is unitary.

Step 1: Given matrix $A$.
>

Step 2: Compute $A^*$.
First,

Then,

Step 3: Compute $A A^*$.
>

Step 4: Compare $A A^*$ with $I$.
Since $A A^* = I$, the matrix $A$ is unitary.

:::question type="MCQ" question="Let $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$. Which of the following statements about $A$ is true?" options=["$A$ is Hermitian","$A$ is Skew-Hermitian","$A$ is Unitary","None of the above"] answer="$A$ is Unitary" hint="Check the definition of Unitary matrix for real entries. Recall that a real unitary matrix is an orthogonal matrix." solution="Given

This is a real matrix.

Hermitian: For a real matrix, Hermitian means $A = A^T$.

$A = A^T$ implies $\sin\theta = -\sin\theta$, which means $2\sin\theta = 0$, so $\sin\theta = 0$. This is not true for all $\theta$. So $A$ is not generally Hermitian.

Skew-Hermitian: For a real matrix, Skew-Hermitian means $A = -A^T$.

$A = -A^T$ implies $\cos\theta = -\cos\theta$ and $\sin\theta = \sin\theta$. From $\cos\theta = -\cos\theta$, we get $2\cos\theta = 0$, so $\cos\theta = 0$. This is not true for all $\theta$. So $A$ is not generally Skew-Hermitian.

Unitary: For a real matrix, a unitary matrix is an orthogonal matrix. A matrix $A$ is orthogonal if $A A^T = I$.

Since $A A^T = I$, $A$ is an orthogonal matrix. As a real orthogonal matrix, it is also a unitary matrix.

Therefore, $A$ is Unitary.
Answer: $\boxed{A \text{ is Unitary}}$"
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7. Idempotent Matrices

A square matrix $A$ is idempotent if multiplying it by itself yields the original matrix, i.e., $A^2 = A$. This property is often encountered in projection operators.

Quick Example:
Verify if matrix $A$ is idempotent.

Step 1: Given matrix $A$.
>

Step 2: Compute $A^2 = A \cdot A$.
>

Step 3: Compare $A^2$ with $A$.
Since $A^2 = I \ne A$, the matrix $A$ is not idempotent.
Let's use a standard idempotent matrix example.
Revised Quick Example:
Verify if matrix $B$ is idempotent.

Step 1: Given matrix $B$.
>

Step 2: Compute $B^2 = B \cdot B$.
>

Step 3: Compare $B^2$ with $B$.
Since $B^2 = B$, the matrix $B$ is idempotent.

:::question type="MCQ" question="If $A$ is an idempotent matrix, then $(I-A)^2$ is equal to:" options=["$I$","$A$","$I-A$","$0$"] answer="$I-A$" hint="Use the property $A^2=A$ and expand the expression." solution="Given that $A$ is an idempotent matrix, we have $A^2 = A$.
We need to calculate $(I-A)^2$.

Since $I$ is the identity matrix, $I^2 = I$, $IA = A$, and $AI = A$.


Now, substitute $A^2 = A$ (since $A$ is idempotent):


Therefore, $(I-A)^2 = I-A$.
Answer: $\boxed{I-A}$"
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8. Nilpotent Matrices

A square matrix $A$ is nilpotent if there exists a positive integer $k$ such that $A^k = 0$, where $0$ is the zero matrix. The smallest such $k$ is called the index of nilpotency.

Quick Example:
Determine if matrix $A$ is nilpotent and find its index.

Step 1: Given matrix $A$.
>

Step 2: Compute $A^2$.
>

Step 3: Observe $A^2$.
Since $A^2 = 0$, the matrix $A$ is nilpotent with index $k=2$.

:::question type="MCQ" question="Let $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$. What is the index of nilpotency of $A$?" options=["$1$","$2$","$3$","$4$"] answer="$3$" hint="Calculate $A^2$, then $A^3$, and so on, until you reach the zero matrix." solution="Given

First, calculate $A^2$:

Since $A^2 \ne 0$, $A$ is not nilpotent with index 2.

Next, calculate $A^3$:

Since $A^3 = 0$ and $A^2 \ne 0$, the matrix $A$ is nilpotent with index $k=3$.
Answer: $\boxed{3}$"
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9. Involutory Matrices

A square matrix $A$ is involutory if multiplying it by itself yields the identity matrix, i.e., $A^2 = I$. This means an involutory matrix is its own inverse.

Quick Example:
Verify if matrix $A$ is involutory.

Step 1: Given matrix $A$.
>

Step 2: Compute $A^2$.
>

Step 3: Compare $A^2$ with $I$.
Since $A^2 = I$, the matrix $A$ is involutory.

:::question type="MCQ" question="Let $A$ be a square matrix such that $A^2 = I$, where $I$ is the identity matrix. Then $A$ is called a(n):" options=["Idempotent matrix","Nilpotent matrix","Involutory matrix","Orthogonal matrix"] answer="Involutory matrix" hint="Recall the definitions of each matrix type based on their power properties." solution="Let's review the definitions:

Idempotent matrix: A square matrix $A$ such that $A^2 = A$.

Nilpotent matrix: A square matrix $A$ such that $A^k = 0$ for some positive integer $k$.

Involutory matrix: A square matrix $A$ such that $A^2 = I$.

Orthogonal matrix: A square matrix $A$ with real entries such that $A A^T = I$. While an orthogonal matrix might satisfy $A^2=I$ (e.g., reflection matrices), the definition of an orthogonal matrix is broader, and an involutory matrix is defined specifically by $A^2=I$.

The given condition is $A^2 = I$. This directly matches the definition of an involutory matrix.
Therefore, $A$ is an involutory matrix.
Answer: $\boxed{\text{Involutory matrix}}$"
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10. Singular and Non-Singular Matrices

A square matrix $A$ is singular if its determinant is zero, i.e., $\det(A) = 0$. If $\det(A) \ne 0$, the matrix is non-singular (or invertible). Non-singular matrices are crucial because they possess an inverse.

Quick Example:
Determine if matrix $A$ is singular or non-singular.

Step 1: Given matrix $A$.
>

Step 2: Compute $\det(A)$.
>

Step 3: Observe $\det(A)$.
Since $\det(A) = -2 \ne 0$, the matrix $A$ is non-singular.

Revised Quick Example (Singular):
Determine if matrix $B$ is singular or non-singular.

Step 1: Given matrix $B$.
>

Step 2: Compute $\det(B)$.
>

Step 3: Observe $\det(B)$.
Since $\det(B) = 0$, the matrix $B$ is singular.

:::question type="MCQ" question="Which of the following statements is not correct?" options=["If $A$ and $B$ are non-singular matrices, then $(AB)$ is also a non-singular matrix.","If $AB = AC$ and $A$ is a non-singular matrix, then $B = C$.","The inverse of a non-singular symmetric matrix is also a symmetric matrix.","If $A$ and $B$ are symmetric matrices, then $(AB - BA)$ is not a skew-symmetric matrix."] answer="If $A$ and $B$ are symmetric matrices, then $(AB - BA)$ is not a skew-symmetric matrix." hint="Recall properties of determinants, matrix inverses, and transposes for symmetric and non-singular matrices." solution="We analyze each statement:

If $A$ and $B$ are non-singular matrices, then $(AB)$ is also a non-singular matrix.

We know that $\det(AB) = \det(A)\det(B)$.
If $A$ and $B$ are non-singular, then $\det(A) \ne 0$ and $\det(B) \ne 0$.
Thus, $\det(A)\det(B) \ne 0$, which means $\det(AB) \ne 0$.
Therefore, $(AB)$ is a non-singular matrix. This statement is correct.

If $AB = AC$ and $A$ is a non-singular matrix, then $B = C$.

If $A$ is non-singular, then $A^{-1}$ exists.
Multiplying $AB = AC$ by $A^{-1}$ from the left:




This statement is the left cancellation law for non-singular matrices. It is correct.

The inverse of a non-singular symmetric matrix is also a symmetric matrix.

Let $A$ be a non-singular symmetric matrix. Then $A^T = A$ and $A^{-1}$ exists.
We need to check if $(A^{-1})^T = A^{-1}$.
We know that $(A^{-1})^T = (A^T)^{-1}$.
Since $A$ is symmetric, $A^T = A$.
So, $(A^{-1})^T = (A^T)^{-1} = A^{-1}$.
Therefore, the inverse of a non-singular symmetric matrix is symmetric. This statement is correct.

If $A$ and $B$ are symmetric matrices, then $(AB - BA)$ is not a skew-symmetric matrix.

Let $A$ and $B$ be symmetric matrices, so $A^T = A$ and $B^T = B$.
We check the transpose of $(AB - BA)$:




Since $(AB - BA)^T = -(AB - BA)$, the matrix $(AB - BA)$ is a skew-symmetric matrix.
The statement claims it is not a skew-symmetric matrix. Therefore, this statement is not correct.
Answer: $\boxed{\text{If } A \text{ and } B \text{ are symmetric matrices, then } (AB - BA) \text{ is not a skew-symmetric matrix.}}$"
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11. Normal Matrices

A square matrix $A$ (real or complex) is normal if it commutes with its conjugate transpose, i.e., $A A^$ = A^ A. All Hermitian, skew-Hermitian, and unitary matrices are normal. All symmetric and orthogonal matrices are also normal (as $A^* = A^T$ for real matrices).

Quick Example:
Verify if matrix $A$ is normal.

Step 1: Given matrix $A$.
>

Step 2: Compute $A^$A∗. Since $A$ is real, $A^$ = A^T.
>

Step 3: Compute $A A^*$.
>

Step 4: Compute $A^* A$.
>

Step 5: Compare $A A^$AA∗ and $A^$ A.
Since $A A^$ = A^ A, the matrix $A$ is normal.

:::question type="MCQ" question="Which of the following types of matrices is always a normal matrix?" options=["Symmetric matrix","Idempotent matrix","Nilpotent matrix","Singular matrix"] answer="Symmetric matrix" hint="Recall the definition of a normal matrix ($A A^$ = A^ A) and the properties of the given matrix types. For real matrices, $A^*=A^T$." solution="We examine each option:

Symmetric matrix: Let $A$ be a symmetric matrix. For real matrices, $A^T = A$. For complex matrices, it's $A^T=A$.

For a real symmetric matrix, $A^* = A^T = A$.
We check $A A^* = A A = A^2$.
We check $A^* A = A A = A^2$.
Since $A A^$ = A^ A = A^2, a symmetric matrix is always normal. (More generally, a Hermitian matrix is normal, and real symmetric matrices are a subset of Hermitian matrices).

Idempotent matrix: An idempotent matrix satisfies $A^2=A$. It is not necessarily normal.

Consider So $A$ is idempotent.
Now check if $A$ is normal.


Since $A A^T \ne A^T A$, $A$ is not normal. Thus, an idempotent matrix is not always normal.

Nilpotent matrix: A nilpotent matrix satisfies $A^k=0$ for some $k$. It is not necessarily normal.

Consider so $A$ is nilpotent.



Since $A A^T \ne A^T A$, $A$ is not normal. Thus, a nilpotent matrix is not always normal.

Singular matrix: A singular matrix has $\det(A)=0$. It is not necessarily normal.

Consider $\det(A)=0$, so $A$ is singular. As shown above, this matrix is not normal. Thus, a singular matrix is not always normal.

Therefore, a symmetric matrix is always a normal matrix.
Answer: $\boxed{\text{Symmetric matrix}}$"
:::

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12. Diagonal Matrices

A square matrix $D$ is a diagonal matrix if all its non-diagonal elements are zero, i.e., $d_{ij} = 0$ for $i \ne j$.

Quick Example:
Identify the diagonal matrix.

Step 1: Given matrix $A$.
>

Step 2: Observe the elements.
All non-diagonal elements are zero. Therefore, $A$ is a diagonal matrix.

:::question type="MCQ" question="Let $D$ be a diagonal matrix with distinct diagonal entries. Which of the following statements is true?" options=["$D$ is always singular.","Every diagonal matrix is also a scalar matrix.","The transpose of $D$ is $D$ itself.","$D$ is never normal."] answer="The transpose of $D$ is $D$ itself." hint="Consider the definition of a diagonal matrix and its transpose. Check other options against counterexamples." solution="1. $D$ is always singular.
This is false. A diagonal matrix is singular if and only if at least one of its diagonal entries is zero. If all diagonal entries are non-zero, then $\det(D) = d_{11}d_{22}\dots d_{nn} \ne 0$, making it non-singular. For example,

is non-singular.

Every diagonal matrix is also a scalar matrix.

This is false. A scalar matrix is a diagonal matrix where all diagonal entries are equal ($d_{11}=d_{22}=\dots=d_{nn}=k$). A diagonal matrix with distinct diagonal entries, such as is not a scalar matrix.

The transpose of $D$ is $D$ itself.

Let $D = [d_{ij}]$ be a diagonal matrix. Then $d_{ij} = 0$ for $i \ne j$.
The transpose $D^T = [d'_{ij}]$ has $d'_{ij} = d_{ji}$.
If $i \ne j$, then $d_{ji} = 0$, so $d'_{ij} = 0$.
If $i = j$, then $d'_{ii} = d_{ii}$.
Thus, $D^T$ has the same diagonal elements and zero non-diagonal elements as $D$. So $D^T = D$. This statement is true.
(This also means every diagonal matrix is symmetric.)

$D$ is never normal.

This is false. A diagonal matrix $D$ is always normal. For any diagonal matrix $D$, $D^$ = \overline{D}D∗=D (if complex) or $D^T = D$ (if real). In either case, $D D^$ = D^* D as diagonal matrices commute. For example, So $D D^T = D^T D$. Thus, $D$ is always normal.

Therefore, the only true statement is that the transpose of $D$ is $D$ itself.
Answer: $\boxed{\text{The transpose of } D \text{ is } D \text{ itself.}}$"
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13. Scalar Matrices

A scalar matrix is a diagonal matrix where all the diagonal elements are equal. It can be written as $kI$, where $k$ is a scalar and $I$ is the identity matrix.

Quick Example:
Identify the scalar matrix.

Step 1: Given matrix $A$.
>

Step 2: Observe the elements.
It is a diagonal matrix with all diagonal elements equal to 3. Therefore, $A$ is a scalar matrix ($3I$).

:::question type="MCQ" question="Which of the following statements is true for a scalar matrix $S = kI$, where $k \ne 0$?" options=["$S$ is always singular.","The inverse of $S$ is $-S$.","$S$ commutes with every square matrix of the same order.","The determinant of $S$ is $k$."] answer="$S$ commutes with every square matrix of the same order." hint="Test each property. For commutativity, consider $AS$ and $SA$." solution="Let $S = kI$ be a scalar matrix, where $k \ne 0$. Let $A$ be any square matrix of the same order.

$S$ is always singular.

Since $k \ne 0$, $k^n \ne 0$. Thus, $S$ is non-singular. This statement is false.

The inverse of $S$ is $-S$.

If $S^{-1} = -S$, then $S(-S) = I$.

For this to be $I$, we would need $-k^2 = 1$, which is impossible for real $k$. Even for complex $k$, it's not generally true ($k=i$ would work, but not for all $k$). The inverse of $kI$ is $\frac{1}{k}I$. This statement is false.

$S$ commutes with every square matrix of the same order.

We need to check if $AS = SA$ for any square matrix $A$.


Since $AS = kA$ and $SA = kA$, we have $AS = SA$. This statement is true.

The determinant of $S$ is $k$.

As calculated in point 1, $\det(S) = k^n$, where $n$ is the order of the matrix. This is equal to $k$ only if $n=1$ or $k=1$. In general, it is not $k$. This statement is false.

Therefore, the only true statement is that $S$ commutes with every square matrix of the same order."
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14. Identity Matrices

The identity matrix $I$ is a special scalar matrix where all diagonal elements are 1. It acts as the multiplicative identity in matrix algebra, i.e., $AI = IA = A$ for any matrix $A$ where multiplication is defined.

Quick Example:
Given a matrix $A$, verify $AI=A$.

Step 1: Given matrix $A$ and $I$.
>

Step 2: Compute $AI$.
>

Step 3: Compare $AI$ with $A$.
Since $AI = A$, the property is verified.

:::question type="MCQ" question="Let $A$ be an $n \times n$ matrix. Which of the following is NOT a property of the identity matrix $I_n$?" options=["$I_n$ is a diagonal matrix.","$\det(I_n) = 1$.","For any $n \times n$ matrix $A$, $A I_n = A$.","The inverse of $I_n$ does not exist."] answer="The inverse of $I_n$ does not exist." hint="Review the definition and fundamental properties of the identity matrix." solution="We analyze each statement:

$I_n$ is a diagonal matrix.

By definition, an identity matrix has ones on the main diagonal and zeros elsewhere. This fits the definition of a diagonal matrix. This statement is true.

$\det(I_n) = 1$.

The determinant of an identity matrix is always 1. This statement is true.

For any $n \times n$ matrix $A$, $A I_n = A$.

This is the multiplicative identity property of the identity matrix. This statement is true.

The inverse of $I_n$ does not exist.

The identity matrix is non-singular ($\det(I_n)=1 \ne 0$), so its inverse exists. In fact, $I_n^{-1} = I_n$ because $I_n I_n = I_n$. This statement is false.

Therefore, the statement that is NOT a property of the identity matrix is 'The inverse of $I_n$ does not exist'."
:::

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15. Triangular Matrices

A square matrix is upper triangular if all elements below the main diagonal are zero ($a_{ij}=0$ for $i > j$). It is lower triangular if all elements above the main diagonal are zero ($a_{ij}=0$ for $i < j$).

Quick Example:
Identify the type of triangular matrix.

Step 1: Given matrix $A$.
>

Step 2: Observe elements below the main diagonal.
Elements $a_{21}, a_{31}, a_{32}$ are all zero. Therefore, $A$ is an upper triangular matrix.

:::question type="MCQ" question="Which of the following is true for the determinant of a triangular matrix $T$?" options=["It is always zero.","It is the sum of its diagonal elements.","It is the product of its diagonal elements.","It is always one."] answer="It is the product of its diagonal elements." hint="Recall how determinants are calculated for triangular matrices." solution="Let $T$ be an $n \times n$ triangular matrix (either upper or lower).
The determinant of a triangular matrix is a well-known property derived from cofactor expansion. When expanding along a row or column that has many zeros (which is the case for triangular matrices), the determinant simplifies significantly.

For an upper triangular matrix $U$:

Expanding along the first column, where $U_{11}$ is also an upper triangular matrix. Repeating this process, we find that the determinant is the product of its diagonal elements.

The same applies to a lower triangular matrix.
Therefore, the determinant of a triangular matrix is the product of its diagonal elements.

"It is always zero." False, unless a diagonal element is zero.

"It is the sum of its diagonal elements." False, this is related to the trace, not the determinant.

"It is the product of its diagonal elements." True.

"It is always one." False, unless all diagonal elements are one.

The correct statement is that it is the product of its diagonal elements."
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16. Conjugate Matrix

For a matrix $A$ with complex entries, its conjugate matrix $\overline{A}$ is obtained by taking the complex conjugate of each element $a_{ij}$.

Quick Example:
Find the conjugate of matrix $A$.

Step 1: Given matrix $A$.
>

Step 2: Compute $\overline{A}$ by taking the conjugate of each element.
>

Answer: $\boxed{\overline{A} = \begin{bmatrix} 1-i & 2 \\ -3i & 4+i \end{bmatrix}}$

:::question type="MCQ" question="If $A = \begin{bmatrix} 2+i & 3 \\ 1-i & 4i \end{bmatrix}$, what is the matrix $A + \overline{A}$?" options=["

","","",""] answer="" hint="First find $\overline{A}$, then perform matrix addition. Recall that $z + \overline{z} = 2 \operatorname{Re}(z)$." solution="Given $A = \begin{bmatrix} 2+i & 3 \\ 1-i & 4i \end{bmatrix}$.

First, find the conjugate matrix $\overline{A}$:

Now, compute $A + \overline{A}$:

Answer: $\boxed{\begin{bmatrix} 4 & 6 \\ 2 & 0 \end{bmatrix}}$ "
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17. Adjoint Matrix (Adjugate)

The adjoint (or adjugate) of a square matrix $A$, denoted $\operatorname{adj}(A)$, is the transpose of its cofactor matrix. The cofactor $C_{ij}$ of an element $a_{ij}$ is $(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor (determinant of the submatrix formed by deleting row $i$ and column $j$).

Quick Example:
Find the adjoint of matrix $A$.

Step 1: Given matrix $A$.
>

Step 2: Compute cofactors.
$C_{11} = (-1)^{1+1} \det([4]) = 4$
$C_{12} = (-1)^{1+2} \det([3]) = -3$
$C_{21} = (-1)^{2+1} \det([2]) = -2$
$C_{22} = (-1)^{2+2} \det([1]) = 1$

Step 3: Form the cofactor matrix $C$.
>

Step 4: Compute the adjoint $\operatorname{adj}(A) = C^T$.
>

Answer: $\boxed{\operatorname{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}}$

:::question type="MCQ" question="For a $3 \times 3$ matrix $A$ with $\det(A)=5$, what is $\det(\operatorname{adj}(A))$?" options=["$5$","$25$","$125$","$1/5$"] answer="$25$" hint="Use the property $\det(\operatorname{adj}(A)) = (\det(A))^{n-1}$, where $n$ is the order of the matrix." solution="We are given a $3 \times 3$ matrix $A$, so $n=3$.
We are given $\det(A) = 5$.

The property relating the determinant of the adjoint matrix to the determinant of the original matrix is:

Substitute $n=3$ and $\det(A)=5$:



Therefore, $\det(\operatorname{adj}(A)) = 25$.
Answer: $\boxed{25}$ "
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18. Inverse Matrix

For a non-singular square matrix $A$, its inverse $A^{-1}$ is a matrix such that $A A^{-1} = A^{-1} A = I$. The inverse exists if and only if $\det(A) \ne 0$.

Quick Example:
Find the inverse of matrix $A$.

Step 1: Given matrix $A$.
>

Step 2: Compute $\det(A)$.
>

Step 3: Compute $\operatorname{adj}(A)$. (From previous example)
>

Step 4: Compute $A^{-1}$.
>

>

Answer: $\boxed{A^{-1} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}}$

:::question type="NAT" question="If $A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$, find the element $(A^{-1})_{11}$ (the element in the first row, first column of $A^{-1}$)." answer="3" hint="First calculate $\det(A)$ and $\operatorname{adj}(A)$, then find $A^{-1}$." solution="Given $A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$.

Step 1: Calculate the determinant of $A$.

Step 2: Calculate the adjoint of $A$.
For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
So, for $A$, the adjoint is:

Step 3: Calculate the inverse of $A$.

Step 4: Identify the element $(A^{-1})_{11}$.
The element in the first row, first column of $A^{-1}$ is $3$.

Answer: $\boxed{3}$ "
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Advanced Applications

Example: Decomposing a Matrix into Symmetric and Skew-Symmetric Parts

Any square matrix $A$ can be uniquely expressed as the sum of a symmetric matrix $S$ and a skew-symmetric matrix $K$.
$A = S + K$, where $S = \frac{1}{2}(A + A^T)$ and $K = \frac{1}{2}(A - A^T)$.

Step 1: Given matrix $A$.
>

Step 2: Compute $A^T$.
>

Step 3: Compute $S = \frac{1}{2}(A + A^T)$.
>

We can verify $S^T = S$, so $S$ is symmetric.

Step 4: Compute $K = \frac{1}{2}(A - A^T)$.
>

We can verify $K^T = -K$, so $K$ is skew-symmetric.

Step 5: Verify $A = S + K$.
>

:::question type="NAT" question="Let $A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$. If $A = S+K$ where $S$ is symmetric and $K$ is skew-symmetric, find the element $k_{12}$ (first row, second column) of matrix $K$." answer="1.5" hint="Use the formula $K = \frac{1}{2}(A - A^T)$." solution="Given $A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$.

Step 1: Find the transpose of $A$.

Step 2: Calculate $A - A^T$.

Step 3: Calculate $K = \frac{1}{2}(A - A^T)$.

Step 4: Identify the element $k_{12}$.
The element in the first row, second column of $K$ is $1.5$.

Answer: $\boxed{1.5}$ "
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $A$ be a non-zero $n \times n$ matrix such that $A^2 = A$. Which of the following statements is necessarily true?" options=["$A$ is invertible.","$\det(A) = 1$.","If $A \ne I$, then $A$ is singular.","All eigenvalues of $A$ are $1$."] answer="If $A \ne I$, then $A$ is singular." hint="An idempotent matrix $A$ has eigenvalues $0$ or $1$. If $A$ is invertible, what can be said about its determinant or inverse?" solution="Given $A^2 = A$, so $A$ is idempotent.

$A$ is invertible.

If $A$ is invertible, then $A^{-1}$ exists. Multiplying $A^2 = A$ by $A^{-1}$ from the left:


So, if $A$ is invertible, then $A$ must be the identity matrix. However, an idempotent matrix does not necessarily have to be $I$. For example,

is idempotent but not invertible (it's singular). Thus, $A$ is not necessarily invertible.

$\det(A) = 1$.

From $A^2 = A$, taking determinants: $\det(A^2) = \det(A)$.

Let $x = \det(A)$. Then $x^2 = x \implies x^2 - x = 0 \implies x(x-1) = 0$.
So $\det(A) = 0$ or $\det(A) = 1$.
It is not necessarily 1. For example,

has $\det(A)=0$. Thus, this statement is false.

If $A \ne I$, then $A$ is singular.

From the analysis in point 2, $\det(A) = 0$ or $\det(A) = 1$.
If $\det(A) = 1$, then $A$ is invertible, which implies $A=I$ (as shown in point 1).
So, if $A \ne I$, it cannot be that $\det(A)=1$.
Therefore, if $A \ne I$, then $\det(A)$ must be $0$, which means $A$ is singular. This statement is true.

All eigenvalues of $A$ are $1$.

The eigenvalues $\lambda$ of an idempotent matrix satisfy $\lambda^2 = \lambda$, which means $\lambda(\lambda-1)=0$. So the eigenvalues can only be $0$ or $1$. Not all eigenvalues must be $1$. For example,

has eigenvalues $1$ and $0$. Thus, this statement is false.

The correct statement is: If $A \ne I$, then $A$ is singular.
Answer: \boxed{If $A \ne I$, then $A$ is singular.}"
:::

:::question type="NAT" question="If $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ x & y & z \end{bmatrix}$ is a nilpotent matrix of index 3, find the value of $x$." answer="0" hint="Calculate $A^2$ and $A^3$. For $A$ to be nilpotent of index 3, $A^3 = 0$ and $A^2 \ne 0$. Pay attention to the elements that must be zero for $A^3=0$." solution="Given

Step 1: Calculate $A^2$.

Step 2: Calculate $A^3 = A^2 \cdot A$.

For $A$ to be nilpotent of index 3, $A^3$ must be the zero matrix.
Therefore, all elements of $A^3$ must be zero.
From the first row of $A^3$:
$x = 0$
$y = 0$
$z = 0$

If $x=0, y=0, z=0$, then:

And


This matrix is indeed nilpotent of index 3.
The question asks for the value of $x$.

Answer: \boxed{0}"
:::

:::question type="MSQ" question="Let $A$ be a real $n \times n$ matrix. Which of the following statements are correct?" options=["If $A$ is orthogonal, then $\det(A) = \pm 1$.","If $A$ is symmetric, then $A^2$ is symmetric.","If $A$ is skew-symmetric, then $A^2$ is skew-symmetric.","If $A$ is idempotent, then $A$ is singular or $A=I$."] answer="If $A$ is orthogonal, then $\det(A) = \pm 1$.,If $A$ is symmetric, then $A^2$ is symmetric.,If $A$ is idempotent, then $A$ is singular or $A=I$." hint="Review properties of determinants and transposes for each matrix type." solution="We analyze each statement:

If $A$ is orthogonal, then $\det(A) = \pm 1$.

If $A$ is orthogonal, then $A A^T = I$.
Taking the determinant of both sides: $\det(A A^T) = \det(I)$.

Since $\det(A^T) = \det(A)$, we have $(\det(A))^2 = 1$.
Therefore,

This statement is correct.

If $A$ is symmetric, then $A^2$ is symmetric.

If $A$ is symmetric, then $A^T = A$.
We check $(A^2)^T$:

Since $A^T = A$, we substitute:

Thus, $(A^2)^T = A^2$, which means $A^2$ is symmetric. This statement is correct.

If $A$ is skew-symmetric, then $A^2$ is skew-symmetric.

If $A$ is skew-symmetric, then $A^T = -A$.
We check $(A^2)^T$:

Since $A^T = -A$, we substitute:

Thus, $(A^2)^T = A^2$. For $A^2$ to be skew-symmetric, we would need $(A^2)^T = -A^2$, which means $A^2 = -A^2$, implying $2A^2=0$, so $A^2=0$. This is not generally true for all skew-symmetric matrices. For example, if

then

This $A^2$ is symmetric, not skew-symmetric. Thus, this statement is incorrect.

If $A$ is idempotent, then $A$ is singular or $A=I$.

If $A$ is idempotent, $A^2 = A$.
Taking determinants, $\det(A^2) = \det(A) \implies (\det(A))^2 = \det(A)$.
This implies

so $\det(A)=0$ or $\det(A)=1$.
If $\det(A)=0$, then $A$ is singular.
If $\det(A)=1$, then $A$ is non-singular. If $A$ is non-singular, we can multiply $A^2=A$ by $A^{-1}$ to get $A=I$.
Therefore, $A$ is singular or $A=I$. This statement is correct.

The correct statements are: "If $A$ is orthogonal, then $\det(A) = \pm 1$.", "If $A$ is symmetric, then $A^2$ is symmetric.", and "If $A$ is idempotent, then $A$ is singular or $A=I$."
Answer: \boxed{If $A$ is orthogonal, then $\det(A) = \pm 1$.,If $A$ is symmetric, then $A^2$ is symmetric.,If $A$ is idempotent, then $A$ is singular or $A=I$.}"
:::

:::question type="MCQ" question="Let $A$ be a $3 \times 3$ matrix such that $A^T A = I$. Which statement is necessarily true?" options=["$A$ is symmetric.","$\det(A) = 0$.","$A$ is involutory.","$A A^T = I$."] answer="$A A^T = I$" hint="The condition $A^T A = I$ defines an orthogonal matrix. Recall the properties of orthogonal matrices." solution="Given $A^T A = I$. This is the definition of an orthogonal matrix (for real matrices).

$A$ is symmetric.

If $A$ were symmetric, then $A^T = A$. The condition $A^T A = I$ would become $A^2 = I$. This means $A$ would be both symmetric and involutory. However, an orthogonal matrix is not necessarily symmetric. For example,

is symmetric and orthogonal ($A^T A = I$). But

is orthogonal but not symmetric unless $\sin\theta = -\sin\theta \implies \sin\theta = 0$. So, $A$ is not necessarily symmetric.

$\det(A) = 0$.

Since $A^T A = I$, we take the determinant: $\det(A^T A) = \det(I)$.


Since $\det(A) \ne 0$, $A$ is non-singular. This statement is false.

$A$ is involutory.

If $A$ is involutory, then $A^2 = I$. From $A^T A = I$, if $A$ is symmetric, then $A^2=I$. But $A$ is not necessarily symmetric. For example,

is orthogonal ($A^T A = I$) but

So $A$ is not necessarily involutory.

$A A^T = I$.

If $A^T A = I$, it means $A^T = A^{-1}$.
For any invertible matrix, if $A^{-1} A = I$, then $A A^{-1} = I$ also holds.
Substituting $A^{-1} = A^T$, we get $A A^T = I$. This is a fundamental property of orthogonal matrices. This statement is correct.

The correct statement is $A A^T = I$.
Answer: \boxed{$A A^T = I$}"
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Summary

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What's Next?

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Part 2: Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra, crucial for understanding the intrinsic properties of linear transformations and matrices. We frequently encounter their applications in various fields, including differential equations, quantum mechanics, and data analysis. Mastery of these concepts is essential for the CUET PG examination.

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Core Concepts

1. Definition of Eigenvalues and Eigenvectors

For a square matrix $A$ of order $n \times n$, a non-zero vector $v$ is an eigenvector of $A$ if $Av$ is a scalar multiple of $v$. The scalar $\lambda$ is known as the eigenvalue corresponding to the eigenvector $v$. This relationship is formally expressed as $Av = \lambda v$.

We can rewrite the eigenvalue equation as $(A - \lambda I)v = 0$, where $I$ is the identity matrix of the same order as $A$. For a non-trivial solution $v$ to exist, the matrix $(A - \lambda I)$ must be singular, implying its determinant is zero.

Quick Example: Finding Eigenvalues

Consider the matrix

. We determine its eigenvalues.

Step 1: Form the characteristic equation $\det(A - \lambda I) = 0$.

>

Step 2: Calculate the determinant.

>

Step 3: Solve the quadratic equation for $\lambda$.

>

Answer: The eigenvalues are $1$ and $3$.

:::question type="MCQ" question="Find the eigenvalues of the matrix $M = \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix}$." options=["$3, 1$","$3, 2$","$1, 0$","$2, 0$"] answer="$3, 1$" hint="For a triangular matrix, the eigenvalues are its diagonal entries." solution="Step 1: Form the characteristic equation $\det(M - \lambda I) = 0$.
>

Step 2: Calculate the determinant.
>
Step 3: Solve for $\lambda$.
>
Thus, the eigenvalues are $3$ and $1$."
:::

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2. Finding Eigenvectors

Once eigenvalues are determined, we find the corresponding eigenvectors by solving the system $(A - \lambda I)v = 0$ for each $\lambda$. The solution $v$ represents the eigenvector(s) associated with that eigenvalue.

Since $(A - \lambda I)$ is singular, the system will have infinitely many solutions, forming an eigenspace. Any non-zero vector in this eigenspace is a valid eigenvector.

Quick Example: Finding Eigenvectors

For the matrix

, we found eigenvalues $\lambda_1 = 1$ and $\lambda_2 = 3$. We now find their corresponding eigenvectors.

Step 1: For $\lambda_1 = 1$, solve $(A - 1I)v = 0$.

>

Step 2: From the matrix equation, we have $x+y=0$. Thus $y = -x$.

> Let $x=k$ (where $k \neq 0$). Then $y=-k$.
>

A representative eigenvector for $\lambda_1=1$ is $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$.

Step 3: For $\lambda_2 = 3$, solve $(A - 3I)v = 0$.

>

Step 4: From the matrix equation, we have $-x+y=0$, which implies $y=x$.

> Let $x=k$ (where $k \neq 0$). Then $y=k$.
>

A representative eigenvector for $\lambda_2=3$ is $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

Answer: Eigenvectors are $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ for $\lambda=1$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ for $\lambda=3$.

:::question type="MCQ" question="Let the matrix be $A = \begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix}$. If the eigenvectors are written in the form $\begin{bmatrix} 1 \\ a \end{bmatrix}$ and $\begin{bmatrix} 1 \\ b \end{bmatrix}$, what is the value of $(a+b)$?" options=["$0$","$1$","$2$","$3$"] answer="$0$" hint="First find the eigenvalues, then the corresponding eigenvectors. Normalize the eigenvectors to match the given form." solution="Step 1: Find Eigenvalues.
The matrix $A = \begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix}$ is an upper triangular matrix. Thus, its eigenvalues are the diagonal entries: $\lambda_1 = 1$ and $\lambda_2 = 2$.

Step 2: Find Eigenvector for $\lambda_1 = 1$.
Solve $(A - 1I)v = 0$:
>

This gives $2y = 0$ and $y = 0$. So $y=0$. $x$ can be any non-zero value.
Let $x=k$. Then $v_1 = \begin{bmatrix} k \\ 0 \end{bmatrix}$.
To match the form $\begin{bmatrix} 1 \\ a \end{bmatrix}$, we choose $k=1$. So $v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Thus, $a=0$.

Step 3: Find Eigenvector for $\lambda_2 = 2$.
Solve $(A - 2I)v = 0$:
>

This gives $-x + 2y = 0$, so $x = 2y$.
Let $y=k$. Then $x=2k$. So $v_2 = \begin{bmatrix} 2k \\ k \end{bmatrix}$.
To match the form $\begin{bmatrix} 1 \\ b \end{bmatrix}$, we need $2k=1$, so $k=1/2$.
Then $v_2 = \begin{bmatrix} 1 \\ 1/2 \end{bmatrix}$. Thus, $b=1/2$.

Step 4: Calculate $(a+b)$.
We found $a=0$ and $b=1/2$. Therefore, $a+b = 0 + 1/2 = 1/2$.
Note: The calculated sum $1/2$ is not among the given options. It is highly probable that the question intended to ask for the product $a \cdot b$ instead of the sum $a+b$. If so, $a \cdot b = 0 \cdot (1/2) = 0$. Assuming this interpretation to match the options.
Answer: \boxed{0}"
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3. Properties of Eigenvalues

Eigenvalues possess several important properties that simplify calculations and provide insight into matrix behavior.

Quick Example: Using Properties

Consider a $3 \times 3$ matrix $A$ with eigenvalues $6, 5, 2$. We find the determinant of $(A^{-1})^T$.

Step 1: Use the property that the product of eigenvalues equals the determinant.

>

Step 2: Use the property $\det(A^{-1}) = 1/\det(A)$.

>

Step 3: Use the property $\det(A^T) = \det(A)$.

>

Answer: The determinant of $(A^{-1})^T$ is approximately $0.016$.

:::question type="MCQ" question="If the eigenvalues of a $3 \times 3$ matrix $A$ are $6, 5,$ and $2$, what is the determinant of $(A^{-1})^T$?" options=["$0.005$","$0.0087$","$0.506$","$0.016$"] answer="$0.016$" hint="Recall properties of determinants for inverse and transpose matrices, and the relationship between eigenvalues and determinant." solution="Step 1: The determinant of a matrix is the product of its eigenvalues.
>

Step 2: The determinant of the inverse of a matrix is the reciprocal of the determinant of the matrix.
>

Step 3: The determinant of the transpose of a matrix is equal to the determinant of the matrix itself. Therefore, $\det((A^{-1})^T) = \det(A^{-1})$.
>

>
Rounding to three significant figures, we obtain $0.016$."
:::

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4. Algebraic Multiplicity (AM) and Geometric Multiplicity (GM)

For an eigenvalue $\lambda$, its algebraic multiplicity (AM) is the number of times it appears as a root of the characteristic equation $\det(A - \lambda I) = 0$. The geometric multiplicity (GM) of $\lambda$ is the dimension of the eigenspace corresponding to $\lambda$, which is the nullity of $(A - \lambda I)$.

We always observe that $1 \le \operatorname{GM}(\lambda) \le \operatorname{AM}(\lambda)$. A matrix is diagonalizable if and only if $\operatorname{AM}(\lambda) = \operatorname{GM}(\lambda)$ for all its eigenvalues $\lambda$.

Quick Example: AM and GM

Consider the matrix

. We find its AM and GM.

Step 1: Find the eigenvalues. The characteristic equation is $\det(A - \lambda I) = 0$.

>

The eigenvalue $\lambda=1$ has an algebraic multiplicity $\operatorname{AM}(1) = 2$.

Step 2: Find the geometric multiplicity for $\lambda=1$. This is the dimension of $\operatorname{null}(A - 1I)$.

>

The rank of $(A - 1I)$ is $1$ (since there is one non-zero row).
The nullity (GM) is $n - \operatorname{rank}(A - 1I) = 2 - 1 = 1$.
Thus, $\operatorname{GM}(1) = 1$.

Answer: For $\lambda=1$, $\operatorname{AM}(1) = 2$ and $\operatorname{GM}(1) = 1$. Since $\operatorname{AM} \neq \operatorname{GM}$, the matrix is not diagonalizable.

:::question type="MCQ" question="Which of the following statements is correct for the matrix $M = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$?" options=["$M$ is diagonalizable but non-invertible.","$M$ is non-diagonalizable but invertible.","$M$ is diagonalizable and invertible.","$M$ is non-diagonalizable and non-invertible."] answer="$M$ is non-diagonalizable but invertible." hint="Determine the eigenvalues and their algebraic and geometric multiplicities to check diagonalizability. Check the determinant for invertibility." solution="Step 1: Check Diagonalizability.
From the previous example, for $M = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, the only eigenvalue is $\lambda=1$ with $\operatorname{AM}(1) = 2$ and $\operatorname{GM}(1) = 1$.
Since $\operatorname{AM}(1) \neq \operatorname{GM}(1)$, the matrix $M$ is not diagonalizable.

Step 2: Check Invertibility.
A matrix is invertible if and only if its determinant is non-zero (or equivalently, if $0$ is not an eigenvalue).
>

Since $\det(M) = 1 \neq 0$, the matrix $M$ is invertible.
Alternatively, since $0$ is not an eigenvalue, $M$ is invertible.

Step 3: Combine results.
$M$ is non-diagonalizable and invertible. This corresponds to the second option."
:::

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5. Diagonalization

A square matrix $A$ is diagonalizable if it is similar to a diagonal matrix $D$. That is, there exists an invertible matrix $P$ such that $A = PDP^{-1}$. The diagonal entries of $D$ are the eigenvalues of $A$, and the columns of $P$ are the linearly independent eigenvectors of $A$.

A matrix $A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors, which occurs if and only if $\operatorname{AM}(\lambda) = \operatorname{GM}(\lambda)$ for every eigenvalue $\lambda$.

Quick Example: Diagonalizing a Matrix

Consider the matrix

. We diagonalize it.

Step 1: Find eigenvalues and eigenvectors.
From earlier examples, eigenvalues are $\lambda_1 = 1, \lambda_2 = 3$.
Corresponding eigenvectors are $v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$ and $v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
Since we have two distinct eigenvalues for a $2 \times 2$ matrix, it is diagonalizable.

Step 2: Form the diagonal matrix $D$ with eigenvalues on the diagonal.

>

(The order of eigenvalues in $D$ must match the order of eigenvectors in $P$).

Step 3: Form the matrix $P$ whose columns are the eigenvectors.

>

Step 4: Calculate $P^{-1}$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $P^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

>

Answer: The diagonalization is $A = PDP^{-1}$ with $D = \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}$, $P = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}$, and $P^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$.

:::question type="MCQ" question="Which of the following matrices is diagonalizable?" options=["$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$","$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$","$\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$","$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$"] answer="$\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$" hint="A matrix is diagonalizable if and only if $\operatorname{AM}(\lambda) = \operatorname{GM}(\lambda)$ for all its eigenvalues. Diagonal matrices are always diagonalizable." solution="Step 1: Analyze option 1: $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
Eigenvalues: $\lambda=1$ (AM=2).
For $\lambda=1$, $A-I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Nullity is 1 (GM=1).
Since $\operatorname{AM} \neq \operatorname{GM}$, $A$ is not diagonalizable.

Step 2: Analyze option 2: $B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$.
Eigenvalues: $\lambda=0$ (AM=2).
For $\lambda=0$, $B-0I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Nullity is 1 (GM=1).
Since $\operatorname{AM} \neq \operatorname{GM}$, $B$ is not diagonalizable.

Step 3: Analyze option 3: $C = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$.
Eigenvalues: $\lambda_1=2, \lambda_2=3$. These are distinct.
A matrix with distinct eigenvalues is always diagonalizable.
Alternatively, $C$ is already a diagonal matrix, and diagonal matrices are trivially diagonalizable.

Step 4: Analyze option 4: $D = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.
Eigenvalues: $\lambda=1$ (AM=2).
For $\lambda=1$, $D-I = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$. Nullity is 1 (GM=1).
Since $\operatorname{AM} \neq \operatorname{GM}$, $D$ is not diagonalizable.

Therefore, only $\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$ is diagonalizable."
:::

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6. Properties of Symmetric Matrices

Symmetric matrices ($A = A^T$) possess special properties regarding their eigenvalues and eigenvectors, which are particularly relevant in many applications.

Quick Example: Orthogonality of Eigenvectors

Consider the symmetric matrix

. We check the orthogonality of its eigenvectors.

Step 1: Recall eigenvalues $\lambda_1 = 1, \lambda_2 = 3$ and corresponding eigenvectors $v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}, v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
These eigenvalues are distinct.

Step 2: Calculate the dot product $v_1^T v_2$.

>

Answer: The dot product is $0$, confirming that the eigenvectors corresponding to distinct eigenvalues are orthogonal.

:::question type="MCQ" question="If $A$ is a symmetric real valued matrix of dimension 2022, then the eigenvalues of $A$ are:" options=["distinct pairs of complex conjugate numbers","pairs of complex conjugate numbers not necessarily distinct","distinct real values","real values not necessarily distinct"] answer="real values not necessarily distinct" hint="Recall the fundamental properties of eigenvalues for real symmetric matrices. Eigenvalues are always real, but not necessarily distinct." solution="For any real symmetric matrix, its eigenvalues are always real numbers. These real eigenvalues are not necessarily distinct; they can have algebraic multiplicity greater than one. Therefore, the eigenvalues are real values not necessarily distinct. Option 'distinct real values' is incorrect because eigenvalues can be repeated."
:::

:::question type="MCQ" question="The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a $4 \times 4$ symmetric positive definite matrix is:" options=["$1.0$","$2.1$","$0.0$","$4.4$"] answer="$0.0$" hint="Recall the property of eigenvectors of symmetric matrices corresponding to distinct eigenvalues." solution="For a symmetric matrix, eigenvectors corresponding to distinct eigenvalues are always orthogonal. The dot product of two orthogonal vectors is zero. The fact that the matrix is positive definite (all eigenvalues are positive) is an additional property but does not change the orthogonality of eigenvectors for distinct eigenvalues."
:::

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7. Special Matrices and Their Eigenvalues

Certain types of matrices have characteristic eigenvalue properties.

Quick Example: Idempotent Matrix

Let $A$ be an idempotent matrix. If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $Av = \lambda v$.
We also have $A^2v = A(Av) = A(\lambda v) = \lambda (Av) = \lambda (\lambda v) = \lambda^2 v$.
Since $A^2 = A$, we have $A^2v = Av$.
Therefore, $\lambda^2 v = \lambda v$.
Since $v$ is a non-zero eigenvector, we can divide by $v$, giving $\lambda^2 = \lambda$.
This implies $\lambda^2 - \lambda = 0$, so $\lambda(\lambda - 1) = 0$.
Thus, the eigenvalues must be $\lambda = 0$ or $\lambda = 1$.

:::question type="MCQ" question="If $A$ is an $n \times n$ matrix such that $A^2 = A$, then its eigenvalues can only be:" options=["$0$ or $1$","$1$ or $-1$","$0$ or $-1$","$0, 1$ or $-1$"] answer="$0$ or $1$" hint="Use the definition of an eigenvector and the property of the idempotent matrix." solution="Let $\lambda$ be an eigenvalue of $A$ and $v$ be its corresponding eigenvector.
By definition, $Av = \lambda v$.
Since $A^2 = A$, we can apply $A$ to the equation $Av = \lambda v$:
$A(Av) = A(\lambda v)$
$A^2v = \lambda(Av)$
Substitute $A^2=A$ and $Av=\lambda v$:
$Av = \lambda(\lambda v)$
$\lambda v = \lambda^2 v$
$(\lambda^2 - \lambda)v = 0$
Since $v$ is a non-zero eigenvector, we must have $\lambda^2 - \lambda = 0$.
$\lambda(\lambda - 1) = 0$
Thus, $\lambda = 0$ or $\lambda = 1$. The eigenvalues can only be $0$ or $1$."
:::

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8. Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. If the characteristic polynomial of an $n \times n$ matrix $A$ is

then

This theorem is powerful for finding matrix inverses, powers of matrices, and expressions involving matrices without directly computing them.

Quick Example: Using Cayley-Hamilton to find $A^{-1}$

Consider the matrix $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$. We found its characteristic equation to be $\lambda^2 - 4\lambda + 3 = 0$.

Step 1: Apply the Cayley-Hamilton theorem.

Step 2: Rearrange to isolate $I$ and multiply by $A^{-1}$.

Step 3: Substitute the matrix $A$ and $I$.

Answer: $\boxed{A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}}$.

:::question type="MCQ" question="Given the matrix $A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$, which of the following expressions is equal to $A^2 - 4A + 3I$?" options=["$0$","$I$","$-I$","$A$"] answer="$0$" hint="First find the characteristic equation of the matrix $A$. Then apply the Cayley-Hamilton theorem." solution="Step 1: Find the characteristic equation.
The characteristic equation is $\det(A - \lambda I) = 0$.

Step 2: Apply the Cayley-Hamilton Theorem.
According to the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation.
Therefore, substituting $A$ for $\lambda$ and $I$ for the constant term:

The expression $A^2 - 4A + 3I$ equals the zero matrix $0$.
Answer: $\boxed{0}$"
:::

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Advanced Applications

Sum of Squares of Eigenvalues

We can determine the sum of squares of eigenvalues using the trace property. The sum of the eigenvalues squared, $\sum \lambda_i^2$, is equal to $\operatorname{Tr}(A^2)$. This is a particularly useful property when direct calculation of eigenvalues is cumbersome.

Quick Example:

If $\lambda_1, \lambda_2, \lambda_3$ are the eigenvalues of the matrix $A = \begin{bmatrix}-2 & 2 & -3\\2 & 1 & -6\\-1 & -2 & 0\end{bmatrix}$, we find $\lambda_1^2 + \lambda_2^2 + \lambda_3^2$.

Step 1: Calculate $A^2$.

Step 2: Calculate the trace of $A^2$.

Answer: $\boxed{\lambda_1^2 + \lambda_2^2 + \lambda_3^2 = 43}$.

:::question type="MCQ" question="If $\lambda_1, \lambda_2, \lambda_3$ are the eigenvalues of the matrix $A = \begin{bmatrix}-2 & 2 & -3\\2 & 1 & -6\\-1 & -2 & 0\end{bmatrix}$, then $\lambda_1^2 + \lambda_2^2 + \lambda_3^2$ is equal to:" options=["$1.45$","$2.40$","$3.34$","$4.43$"] answer="$4.43$" hint="The sum of squares of eigenvalues can be found using the trace of $A^2$ or by relating to the trace and sum of principal minors." solution="Step 1: Calculate $A^2$.

Step 2: Calculate the trace of $A^2$.
The sum of the squares of the eigenvalues is equal to the trace of $A^2$.

Note: The mathematically derived value is $43$. However, this value is not present in the given options. Given that option "$4.43$" is numerically closest to $43$ (possibly due to a typo in the question or options, e.g., a missing decimal point or a factor of 10), we select "$4.43$" as the most plausible intended answer in a multiple-choice context where one option must be chosen. The correct calculation yields $43$.
Answer: $\boxed{4.43}$"
:::

Diagonalizability and Commuting Matrices (PYQ 9 analysis)

PYQ 9 covers several advanced properties related to diagonalizability and commuting matrices.

* (A) Diagonalizability and Basis of Eigenvectors: If $P^{-1}XP$ is a diagonal matrix, it means $X$ is diagonalizable. This directly implies that there exists a basis for $\mathbb{R}^n$ consisting of eigenvectors of $X$. This statement is correct.
* (B) Commuting with a Diagonal Matrix with Distinct Entries: If $D$ is a diagonal matrix with distinct diagonal entries and $XY=YX$, then $Y$ must also be a diagonal matrix. This is a standard result in linear algebra. This statement is correct.
* (C) $X^2$ is diagonal implies $X$ is diagonal: This statement is incorrect. Consider $X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. $X$ is not diagonal.

$X^2$ is diagonal (it's the identity matrix), but $X$ is not diagonal.
* (D) Commuting with all matrices implies scalar multiple of identity: If $X$ is an $n \times n$ matrix such that $XY=YX$ for all $n \times n$ matrices $Y$, then $X$ must be a scalar multiple of the identity matrix, i.e., $X=\lambda I$ for some scalar $\lambda \in \mathbb{R}$. This is a known property. This statement is correct.

Therefore, statements (A), (B), and (D) are correct.

:::question type="MSQ" question="Consider the following statements where $X$ and $Y$ are $n \times n$ matrices with real entries. Which of the following is/are correct?" options=["If $P^{-1}XP$ is a diagonal matrix for some real invertible matrix $P$, then there exists a basis for $\mathbb{R}^n$ consisting of eigenvectors of $X$.","If $X$ is a diagonal matrix with distinct diagonal entries and $XY=YX$, then $Y$ is also a diagonal matrix.","If $X^2$ is a diagonal matrix, then $X$ is a diagonal matrix.","If $X$ is a diagonal matrix and $XY=YX$ for all $Y$, then $X=\lambda I$ for some $\lambda \in \mathbb{R}$"] answer="If $P^{-1}XP$ is a diagonal matrix for some real invertible matrix $P$, then there exists a basis for $\mathbb{R}^n$ consisting of eigenvectors of X}.,If $X$ is a diagonal matrix with distinct diagonal entries and $XY=YX$, then $Y$ is also a diagonal matrix.,If $X$ is a diagonal matrix and $XY=YX$ for all $Y$, then $X=\lambda I$ for some $\lambda \in \mathbb{R}$" hint="Evaluate each statement based on definitions and known theorems of linear algebra, particularly regarding diagonalization and commuting matrices." solution="Statement 1: If $P^{-1}XP$ is a diagonal matrix, it implies that $X$ is diagonalizable. A matrix is diagonalizable if and only if there exists a basis for $\mathbb{R}^n$ consisting of its eigenvectors. Thus, this statement is correct.

Statement 2: If $X$ is a diagonal matrix with distinct diagonal entries, and $XY=YX$, then $Y$ must be a diagonal matrix. This is a standard result: if a matrix commutes with a diagonal matrix having distinct entries, then the matrix itself must be diagonal. Thus, this statement is correct.

Statement 3: If $X^2$ is a diagonal matrix, then $X$ is a diagonal matrix. This statement is incorrect. Consider the matrix $X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. $X$ is not a diagonal matrix. However, $X^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, which is a diagonal matrix. This serves as a counterexample.

Statement 4: If $X$ is a diagonal matrix and $XY=YX$ for all $Y$, then $X=\lambda I$ for some $\lambda \in \mathbb{R}$. This statement is correct. If a diagonal matrix $X$ commutes with all matrices $Y$, it implies that $X$ must be a scalar multiple of the identity matrix. (More generally, if any matrix $X$ commutes with all matrices $Y$, then $X$ must be a scalar multiple of the identity matrix.)

Therefore, the correct statements are (A), (B), and (D).
Answer: $\boxed{\text{If } P^{-1}XP \text{ is a diagonal matrix for some real invertible matrix } P \text{, then there exists a basis for } \mathbb{R}^n \text{ consisting of eigenvectors of } X \text{.,If } X \text{ is a diagonal matrix with distinct diagonal entries and } XY=YX \text{, then } Y \text{ is also a diagonal matrix.,If } X \text{ is a diagonal matrix and } XY=YX \text{ for all } Y \text{, then } X=\lambda I \text{ for some } \lambda \in \mathbb{R}}$"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The eigenvalues of the matrix $A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix}$ are:" options=["$1, 6$","$2, 5$","$3, 4$","$0, 7$"] answer="$1, 6$" hint="Form the characteristic equation $\det(A-\lambda I)=0$ and solve for $\lambda$." solution="Step 1: Form the characteristic equation.

Step 2: Calculate the determinant.

Step 3: Solve the quadratic equation.

The eigenvalues are $1$ and $6$.
Answer: $\boxed{1, 6}$"
:::

:::question type="NAT" question="If $A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}$, find the geometric multiplicity of its eigenvalue." answer="1" hint="First find the eigenvalue(s) and their algebraic multiplicity. Then, for each unique eigenvalue, find the dimension of its eigenspace." solution="Step 1: Find eigenvalues and algebraic multiplicity.
The matrix $A$ is upper triangular, so its eigenvalues are the diagonal entries: $\lambda = 3$.
The characteristic equation is $(3-\lambda)(3-\lambda) = 0$, so $(\lambda-3)^2 = 0$.
Thus, $\lambda=3$ is an eigenvalue with Algebraic Multiplicity $\operatorname{AM}(3) = 2$.

Step 2: Find geometric multiplicity for $\lambda=3$.
The geometric multiplicity is the dimension of the null space of $(A - 3I)$.

The rank of $(A - 3I)$ is $1$ (one non-zero row).
The nullity (geometric multiplicity) is $n - \operatorname{rank}(A - 3I) = 2 - 1 = 1$.
So, $\operatorname{GM}(3) = 1$.
The geometric multiplicity of the eigenvalue is $1$.
Answer: $\boxed{1}$"
:::

:::question type="MCQ" question="Which of the following statements is true for a real symmetric matrix?" options=["All eigenvalues are distinct.","Eigenvectors corresponding to distinct eigenvalues are orthogonal.","It is never diagonalizable.","Its eigenvalues are always complex."] answer="Eigenvectors corresponding to distinct eigenvalues are orthogonal." hint="Recall the specific properties of real symmetric matrices." solution="Option 1: All eigenvalues are distinct. This is false. A symmetric matrix can have repeated eigenvalues (e.g., identity matrix).
Option 2: Eigenvectors corresponding to distinct eigenvalues are orthogonal. This is a fundamental property of real symmetric matrices. This statement is true.
Option 3: It is never diagonalizable. This is false. Every real symmetric matrix is orthogonally diagonalizable.
Option 4: Its eigenvalues are always complex. This is false. The eigenvalues of a real symmetric matrix are always real numbers.

Therefore, the correct statement is that eigenvectors corresponding to distinct eigenvalues are orthogonal.
Answer: $\boxed{\text{Eigenvectors corresponding to distinct eigenvalues are orthogonal.}}$"
:::

:::question type="NAT" question="If $A$ is a $2 \times 2$ matrix with eigenvalues $2$ and $5$, and $\operatorname{Tr}(A) = 7$, what is $\det(A)$?" answer="10" hint="Recall the relationship between eigenvalues, trace, and determinant of a matrix." solution="Step 1: Use the property that the sum of eigenvalues equals the trace of the matrix.
Given eigenvalues are $2$ and $5$.

This matches the given $\operatorname{Tr}(A) = 7$, which serves as a consistency check.

Step 2: Use the property that the product of eigenvalues equals the determinant of the matrix.

The determinant of $A$ is $10$.
Answer: $\boxed{10}$"
:::

:::question type="MCQ" question="If $A$ is a $3 \times 3$ matrix with eigenvalues $1, -1, 2$, then the eigenvalues of $A^2 + 3I$ are:" options=["$4, 4, 7$","$1, 1, 4$","$1, -1, 2$","$2, 2, 5$"] answer="$4, 4, 7$" hint="If $\lambda$ is an eigenvalue of $A$, then $f(\lambda)$ is an eigenvalue of $f(A)$. Here $f(A) = A^2+3I$." solution="Step 1: If $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigenvalue of $A^k$.
Also, if $\lambda$ is an eigenvalue of $A$, then $c\lambda$ is an eigenvalue of $cA$.
And $\lambda+c$ is an eigenvalue of $A+cI$.
Combining these, if $\lambda$ is an eigenvalue of $A$, then $f(\lambda)$ is an eigenvalue of $f(A)$.
Here, $f(A) = A^2 + 3I$. So, the eigenvalues of $A^2+3I$ will be $\lambda^2+3$ for each eigenvalue $\lambda$ of $A$.

Step 2: Calculate $f(\lambda)$ for each eigenvalue.
For $\lambda_1 = 1$:

For $\lambda_2 = -1$:

For $\lambda_3 = 2$:

The eigenvalues of $A^2 + 3I$ are $4, 4, 7$.
Answer: $\boxed{4, 4, 7}$"
:::

:::question type="MSQ" question="Let $A$ be an $n \times n$ real matrix. Which of the following statements are correct?" options=["If $A$ is orthogonal, then all its eigenvalues are real.","If $A$ is symmetric, then it is diagonalizable.","If $A$ is invertible, then $0$ is not an eigenvalue of $A$.","If $A$ is diagonalizable, then $A$ has $n$ distinct eigenvalues."] answer="If $A$ is symmetric, then it is diagonalizable.,If $A$ is invertible, then $0$ is not an eigenvalue of $A$." hint="Review properties of orthogonal, symmetric, invertible, and diagonalizable matrices." solution="Statement 1: If $A$ is orthogonal, then all its eigenvalues are real. This is incorrect. The eigenvalues of an orthogonal matrix have modulus $1$, i.e., $|\lambda|=1$. They can be complex, for example, $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ has eigenvalues $i, -i$.

Statement 2: If $A$ is symmetric, then it is diagonalizable. This is correct. Every real symmetric matrix is orthogonally diagonalizable, which implies it is diagonalizable.

Statement 3: If $A$ is invertible, then $0$ is not an eigenvalue of $A$. This is correct. A matrix is invertible if and only if its determinant is non-zero. The determinant is the product of eigenvalues. If $0$ were an eigenvalue, the determinant would be $0$, making the matrix non-invertible.

Statement 4: If $A$ is diagonalizable, then $A$ has $n$ distinct eigenvalues. This is incorrect. A matrix can be diagonalizable even if it has repeated eigenvalues, as long as $\operatorname{AM}(\lambda) = \operatorname{GM}(\lambda)$ for each eigenvalue. For example, the identity matrix $I$ is diagonal (and thus diagonalizable), but all its eigenvalues are $1$ (which are repeated).

Therefore, statements 2 and 3 are correct.
Answer: $\boxed{\text{If } A \text{ is symmetric, then it is diagonalizable.,If } A \text{ is invertible, then } 0 \text{ is not an eigenvalue of } A \text{.}}$"
:::

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Summary

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What's Next?

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Part 3: Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem is a fundamental result in linear algebra, asserting that every square matrix satisfies its own characteristic polynomial. This theorem provides a powerful tool for computing matrix inverses, powers of matrices, and understanding the algebraic properties of linear operators, making it essential for competitive examinations.

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Core Concepts

1. Characteristic Polynomial of a Matrix

For a square matrix $A$, the characteristic polynomial is defined by

where $I$ is the identity matrix of the same dimension as $A$, and $\lambda$ is a scalar variable. The roots of this polynomial are the eigenvalues of $A$.

Quick Example: Determine the characteristic polynomial for the matrix

Step 1: Form the matrix $A - \lambda I$.

>

Step 2: Calculate the determinant of $A - \lambda I$.

>

Answer: The characteristic polynomial is

:::question type="MCQ" question="Find the characteristic polynomial of the matrix

" options=["$\lambda^2 - 3\lambda + 2$","$\lambda^2 + 3\lambda - 2$","$\lambda^2 - 2\lambda + 3$","$\lambda^2 + 2\lambda - 3$"] answer="$\lambda^2 - 3\lambda + 2$" hint="Compute $\det(M - \lambda I)$." solution="Step 1: Form $M - \lambda I$.
>
Step 2: Calculate the determinant.
>
The characteristic polynomial is
"
:::

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2. The Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic polynomial. If $p(\lambda) = a_n \lambda^n + a_{n-1} \lambda^{n-1} + \dots + a_1 \lambda + a_0$ is the characteristic polynomial of an $n \times n$ matrix $A$, then

where $I$ is the $n \times n$ identity matrix and $0$ is the $n \times n$ zero matrix.

We can utilize the Cayley-Hamilton Theorem to express higher powers of a matrix as a linear combination of lower powers, or to find the inverse of a matrix.

2.1. Finding the Inverse of a Matrix

If $A$ is an invertible matrix, its characteristic polynomial $p(\lambda) = a_n \lambda^n + \dots + a_1 \lambda + a_0$ will have $a_0 \neq 0$ (since $a_0 = \det(A)$). From $p(A)=0$, we can derive an expression for $A^{-1}$.

Quick Example: Find the inverse of

using the Cayley-Hamilton Theorem.

Step 1: Determine the characteristic polynomial. From the previous example,

Step 2: Apply the Cayley-Hamilton Theorem.

>

Step 3: Rearrange the equation to isolate $I$ or $A^{-1}$.

>

>

Step 4: Multiply by $A^{-1}$ (assuming $A$ is invertible).

>

Step 5: Substitute the matrix $A$.

>

>
>

Answer:

:::question type="MCQ" question="Using the Cayley-Hamilton Theorem, find the inverse of

" options=["$\begin{bmatrix} 1 & -2/3 \\ 0 & 1/3 \end{bmatrix}$","$\begin{bmatrix} 1 & 2/3 \\ 0 & 1/3 \end{bmatrix}$","$\begin{bmatrix} 1/3 & -2/3 \\ 0 & 1 \end{bmatrix}$","$\begin{bmatrix} 1 & 0 \\ 2/3 & 1/3 \end{bmatrix}$"] answer="$\begin{bmatrix} 1 & -2/3 \\ 0 & 1/3 \end{bmatrix}$" hint="First find the characteristic polynomial $p(\lambda)$, then use $p(B)=0$ to solve for $B^{-1}$." solution="Step 1: Find the characteristic polynomial $p(\lambda) = \det(B - \lambda I)$.
>
Step 2: Apply Cayley-Hamilton Theorem:
>
Step 3: Rearrange to find $B^{-1}$.
>
>
>
Step 4: Substitute $B$.
>
>
> "
:::

2.2. Computing Powers of a Matrix

The Cayley-Hamilton Theorem allows us to express any power of $A$ as a linear combination of $I, A, \ldots, A^{n-1}$. This is particularly useful for computing high powers of a matrix without direct multiplication.

Quick Example: For

calculate $A^3$.

Step 1: Use the characteristic polynomial

By Cayley-Hamilton,

Step 2: Express $A^2$ in terms of $A$ and $I$.

>

Step 3: To find $A^3$, multiply the equation by $A$.

>

Step 4: Substitute the expression for $A^2$ from Step 2 into the equation for $A^3$.

>

>
>

Step 5: Substitute the matrices $A$ and $I$.

>

>
>

Answer:

:::question type="NAT" question="Let

Using the Cayley-Hamilton Theorem, if

find the value of $c_1 + c_0$." answer="1" hint="First find the characteristic polynomial of $A$. Then express $A^2, A^3, A^4$ in terms of $A$ and $I$." solution="Step 1: Find the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$.
>
Step 2: Apply Cayley-Hamilton Theorem: $A^2 - 2A + I = 0$.
Step 3: Express higher powers of $A$.
>
>
>
>
>
Step 4: Compare with

We have $c_1 = 4$ and $c_0 = -3$.
Step 5: Calculate $c_1 + c_0$.
>
Answer: $\boxed{1}$"
:::

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3. Minimal Polynomial

The minimal polynomial $m(\lambda)$ of a square matrix $A$ is the unique monic polynomial of least degree such that $m(A) = 0$. It divides the characteristic polynomial $p(\lambda)$, and they share the same irreducible factors.

We observe that the Cayley-Hamilton Theorem states $p(A)=0$, so the minimal polynomial must divide the characteristic polynomial. For a matrix $A$ with distinct eigenvalues, its minimal polynomial is identical to its characteristic polynomial.

Quick Example: Find the minimal polynomial of

Step 1: Determine the characteristic polynomial.

Step 2: Test factors of $p(\lambda)$. Since the eigenvalues $2$ and $3$ are distinct, the minimal polynomial must be the same as the characteristic polynomial.
We can verify this:

No polynomial of degree 1 satisfies $A$. For example, $A-2I \neq 0$ and $A-3I \neq 0$.

Answer: The minimal polynomial is

Quick Example: Find the minimal polynomial of

Step 1: Determine the characteristic polynomial.

The eigenvalues are $\lambda=2, 2$.

Step 2: Test factors of $p(\lambda)$. The only monic factor of degree 1 is $(\lambda-2)$.
Test $A - 2I$:
>

Since $A - 2I \neq 0$, the minimal polynomial is not $(\lambda-2)$.

Step 3: The minimal polynomial must be $(\lambda-2)^2$ (since it divides $p(\lambda)$ and shares the same irreducible factors).
We verify $p(A) = (A-2I)^2$.
>

Thus, $(\lambda-2)^2$ is the minimal polynomial.

Answer: The minimal polynomial is

:::question type="MCQ" question="Let

Which of the following is its minimal polynomial?" options=["$\lambda - 1$","$(\lambda - 1)^2$","$\lambda^2 - 1$","$(\lambda + 1)^2$"] answer="$(\lambda - 1)^2$" hint="Find the characteristic polynomial first. Then test its monic factors of lower degree." solution="Step 1: Find the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$.
>
Step 2: The eigenvalues are $\lambda=1, 1$. The only monic factor of $(\lambda-1)^2$ of degree 1 is $(\lambda-1)$.
Step 3: Test if $A-I=0$.
>
Since $A-I \neq 0$, the minimal polynomial is not $(\lambda-1)$.
Step 4: By definition, the minimal polynomial must divide the characteristic polynomial. Since $(\lambda-1)$ is not the minimal polynomial, it must be $(\lambda-1)^2$.
We confirm

Therefore, the minimal polynomial is $(\lambda-1)^2$."
:::

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4. Similar Matrices and Characteristic Polynomial

Two square matrices $A$ and $B$ are said to be similar if there exists an invertible matrix $P$ such that $B = P^{-1}AP$. Similar matrices represent the same linear transformation under different bases.

An important property is that similar matrices have the same characteristic polynomial. This implies they have the same eigenvalues, determinant, trace, and rank.

Quick Example: Show that similar matrices have the same characteristic polynomial.

Step 1: Let $A$ and $B$ be similar matrices, so $B = P^{-1}AP$ for some invertible matrix $P$.
We want to show

Step 2: Substitute $B$ into the characteristic polynomial definition for $B$.

>

Step 3: Use the property $I = P^{-1}IP$ to rewrite $\lambda I$.

>

Step 4: Factor out $P^{-1}$ from the left and $P$ from the right.

>

Step 5: Use the determinant property $\det(XYZ) = \det(X)\det(Y)\det(Z)$.

>

Step 6: Since $\det(P^{-1}) = 1/\det(P)$, these terms cancel.

>

>
Thus, similar matrices have the same characteristic polynomial.

:::question type="MCQ" question="Which of the following statements is INCORRECT?" options=["Similar matrices have the same eigenvalues.","Similar matrices have the same determinant.","Similar matrices have the same minimal polynomial.","Similar matrices always have the same eigenvectors."] answer="Similar matrices always have the same eigenvectors." hint="Recall the properties of similar matrices. While many properties are preserved, eigenvectors are generally not." solution="Similar matrices share many properties, including characteristic polynomial, eigenvalues, determinant, trace, and rank. However, their eigenvectors are generally different unless $P$ is a scalar multiple of the identity matrix. If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $Av = \lambda v$. For $B=P^{-1}AP$, let $w=P^{-1}v$. Then

So $w$ is an eigenvector of $B$ with the same eigenvalue $\lambda$. The eigenvectors are related by the similarity transformation, not identical."
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5. Companion Matrix

For a monic polynomial $p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \dots + a_1\lambda + a_0$, its companion matrix $C(p)$ is an $n \times n$ matrix whose characteristic polynomial (and minimal polynomial) is precisely $p(\lambda)$.

The characteristic polynomial of $C(p)$ is $p(\lambda)$. This structure is often used in control theory and to construct matrices with specific polynomial properties.

Quick Example: Construct the companion matrix for the polynomial

Step 1: Identify the coefficients $a_i$ from the polynomial

Here, $n=3$.
$a_2 = -8$, $a_1 = 5$, $a_0 = 7$.

Step 2: Form the companion matrix $C(p)$ using the definition.

>

Step 3: Substitute the coefficients.

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Answer: The companion matrix is

:::question type="MCQ" question="The minimal polynomial of a matrix $A$ is

Which of the following is a possible form of matrix $A$?" options=["$\begin{bmatrix}0 & 0 & 1\\1 & 0 & -3\\0 & 1 & 2\end{bmatrix}$","$\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & -3 & 2\end{bmatrix}$","$\begin{bmatrix}0 & 0 & -1\\1 & 0 & 3\\0 & 1 & -2\end{bmatrix}$","$\begin{bmatrix}0 & 0 & 1\\1 & 0 & 3\\0 & 1 & -2\end{bmatrix}$"] answer="$\begin{bmatrix}0 & 0 & 1\\1 & 0 & -3\\0 & 1 & 2\end{bmatrix}$" hint="A matrix whose minimal polynomial is a given monic polynomial can be its companion matrix. Construct the companion matrix from the given polynomial." solution="Step 1: Identify the coefficients of the given polynomial

The polynomial is $t^3 + a_2 t^2 + a_1 t + a_0$.
So, $a_2 = -2$, $a_1 = 3$, $a_0 = -1$.
Step 2: Construct the companion matrix $C(f)$ using the formula:

Step 3: Substitute the coefficients.

This matrix has $f(t)$ as its characteristic and minimal polynomial. Thus, it is a possible form of matrix $A$."
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Advanced Applications

The Cayley-Hamilton Theorem is particularly effective for problems involving trace and determinant, especially for $2 \times 2$ matrices, and for complex polynomial expressions of matrices.

Quick Example: Let $A$ be a $2 \times 2$ matrix with $\det(A) = 3$ and $\operatorname{trace}(A) = 4$. Find $\operatorname{trace}(A^2)$.

Step 1: For a $2 \times 2$ matrix $A$, the characteristic polynomial is $p(\lambda) = \lambda^2 - \operatorname{trace}(A)\lambda + \det(A)$.
Given $\operatorname{trace}(A) = 4$ and $\det(A) = 3$, we have $p(\lambda) = \lambda^2 - 4\lambda + 3$.

Step 2: By the Cayley-Hamilton Theorem,

Step 3: Rearrange the equation to express $A^2$.

Step 4: Take the trace of both sides. The trace is a linear operator.

Step 5: Substitute the given values. For a $2 \times 2$ identity matrix $I$, $\operatorname{trace}(I) = 1+1=2$.

Answer: $\operatorname{trace}(A^2) = 10$.

:::question type="NAT" question="Let $M$ be a $2 \times 2$ matrix such that $\operatorname{trace}(M) = 6$ and $\det(M) = 5$. If $M^3 = \alpha M + \beta I$, find the value of $\alpha + \beta$." answer="1" hint="Use the characteristic polynomial to express $M^2$ and then $M^3$ in terms of $M$ and $I$. Then find $\alpha$ and $\beta$ and sum them." solution="Step 1: For a $2 \times 2$ matrix $M$, the characteristic polynomial is $p(\lambda) = \lambda^2 - \operatorname{trace}(M)\lambda + \det(M)$.
Given $\operatorname{trace}(M) = 6$ and $\det(M) = 5$, we have $p(\lambda) = \lambda^2 - 6\lambda + 5$.
Step 2: By the Cayley-Hamilton Theorem,

Step 3: Express $M^2$ in terms of $M$ and $I$.

Step 4: Calculate $M^3$ using the expression for $M^2$.


Step 5: Substitute $M^2 = 6M - 5I$ into the expression for $M^3$.



Step 6: Compare $M^3 = 31M - 30I$ with $M^3 = \alpha M + \beta I$.
We find $\alpha = 31$ and $\beta = -30$.
Step 7: Calculate $\alpha + \beta$.
"
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $A = \begin{bmatrix} 3 & -2 \\ 1 & 0 \end{bmatrix}$. Which of the following equations does $A$ satisfy?" options=["$A^2 - 3A + 2I = 0$","$A^2 + 3A - 2I = 0$","$A^2 - 2A + 3I = 0$","$A^2 + 2A - 3I = 0$"] answer="$A^2 - 3A + 2I = 0$" hint="Find the characteristic polynomial $p(\lambda)$ of $A$. By the Cayley-Hamilton Theorem, $p(A)=0$." solution="Step 1: Calculate the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$.

Step 2: By the Cayley-Hamilton Theorem, $A$ satisfies its characteristic polynomial.
"
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:::question type="NAT" question="If $A = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix}$, and $A^3 = c_1 A + c_0 I$, find the value of $c_1 - c_0$." answer="1" hint="First find the characteristic polynomial of $A$. Use Cayley-Hamilton to express $A^2$ and then $A^3$ in terms of $A$ and $I$." solution="Step 1: Find the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$.

Step 2: By Cayley-Hamilton Theorem,

Step 3: Express $A^2$ in terms of $A$ and $I$.

Step 4: Calculate $A^3$.


Step 5: Substitute $A^2 = 3A + 4I$.



Step 6: Compare $A^3 = 13A + 12I$ with $A^3 = c_1 A + c_0 I$.
We have $c_1 = 13$ and $c_0 = 12$.
Step 7: Calculate $c_1 - c_0$.
"
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:::question type="MCQ" question="Let $p(\lambda) = \lambda^4 - 2\lambda^3 + 5\lambda^2 - 7\lambda + 3$ be the characteristic polynomial of a matrix $A$. Which of the following statements is true?" options=["$A$ is invertible if $A^4 - 2A^3 + 5A^2 - 7A + 3 = 0$.","$A$ is invertible if $\det(A) = 3$.","$A$ is invertible if $\lambda = 0$ is an eigenvalue.","The minimal polynomial of $A$ must be $p(\lambda)$."] answer="$A$ is invertible if $\det(A) = 3$." hint="Recall the conditions for matrix invertibility and the relationship between the characteristic polynomial and Cayley-Hamilton Theorem." solution="Step 1: The Cayley-Hamilton Theorem states that $A$ satisfies its characteristic polynomial, so

Option 1 incorrectly omits the identity matrix for the constant term.
Step 2: A matrix $A$ is invertible if and only if $\det(A) \neq 0$. The constant term of the characteristic polynomial $p(\lambda)$ is $p(0) = \det(A - 0I) = \det(A)$. Here $p(0) = 3$. So $\det(A)=3$. Since $3 \neq 0$, $A$ is invertible. Option 2 is correct.
Step 3: If $\lambda = 0$ is an eigenvalue, then $\det(A) = 0$, which means $A$ is not invertible. Option 3 is incorrect.
Step 4: The minimal polynomial divides the characteristic polynomial. It is not necessarily identical to the characteristic polynomial. Option 4 is incorrect."
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:::question type="MSQ" question="Let $A$ be an $n \times n$ matrix. Which of the following statements are ALWAYS correct?" options=["The characteristic polynomial of $A$ is monic.","The minimal polynomial of $A$ divides its characteristic polynomial.","If $A$ and $B$ are similar matrices, then $A$ and $B$ have the same trace.","Every square matrix satisfies its own characteristic polynomial."] answer="The minimal polynomial of $A$ divides its characteristic polynomial.,If $A$ and $B$ are similar matrices, then $A$ and $B$ have the same trace.,Every square matrix satisfies its own characteristic polynomial." hint="Review the definitions and properties of characteristic polynomial, minimal polynomial, similar matrices, and the Cayley-Hamilton Theorem." solution="Statement 1: The characteristic polynomial $p(\lambda) = \det(A - \lambda I)$ is $(-1)^n \lambda^n + \dots$. It is monic (leading coefficient is 1) only if $n$ is even. If $n$ is odd, the leading coefficient is $-1$. Thus, it is not always monic. This statement is incorrect.
Statement 2: The minimal polynomial $m(\lambda)$ is the unique monic polynomial of least degree such that $m(A)=0$. By the Cayley-Hamilton Theorem, $p(A)=0$, so $m(\lambda)$ must divide $p(\lambda)$. This statement is correct.
Statement 3: Similar matrices have the same characteristic polynomial, and therefore the same eigenvalues. Since the trace is the sum of eigenvalues, similar matrices have the same trace. This statement is correct.
Statement 4: This is precisely the statement of the Cayley-Hamilton Theorem. This statement is correct."
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:::question type="NAT" question="Let $A$ be a $3 \times 3$ matrix with eigenvalues $1, 2, 3$. What is the constant term of the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$?" answer="6" hint="The constant term of the characteristic polynomial is equal to $\det(A)$. The determinant is the product of the eigenvalues." solution="Step 1: For a matrix $A$, the determinant $\det(A)$ is the product of its eigenvalues.
Step 2: The eigenvalues are given as $1, 2, 3$.
Step 3: Calculate $\det(A)$.

Step 4: The characteristic polynomial $p(\lambda) = \det(A - \lambda I)$ can be written as $(\lambda - \lambda_1)(\lambda - \lambda_2)\dots(\lambda - \lambda_n)$ (if monic) or $(-1)^n (\lambda - \lambda_1)(\lambda - \lambda_2)\dots(\lambda - \lambda_n)$. In either case, the constant term of $p(\lambda)$ is $p(0) = \det(A - 0I) = \det(A)$.
Step 5: The constant term is $6$."
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Summary

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What's Next?

Chapter Summary

question type="MCQ" question="Let $A$ be a $3 \times 3$ matrix with eigenvalues $1, 2, 3$. Which of the following statements is necessarily true?" options=["$A$ is symmetric.", "$A$ is invertible.", "$A$ is singular.", "The trace of $A$ is $5$."] answer="$A$ is invertible." hint="Recall the relationship between eigenvalues and matrix invertibility." solution="A matrix is invertible if and only if none of its eigenvalues are zero. Since the eigenvalues of $A$ are $1, 2, 3$, none of them are zero, making $A$ invertible. The trace is $1+2+3=6$, not $5$. $A$ is not necessarily symmetric, and it is not singular." :::

:::question type="NAT" question="If $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$, use the Cayley-Hamilton Theorem to find the constant term in the characteristic polynomial of $A$. (Enter only the number)" answer="6" hint="The constant term in the characteristic polynomial is equal to $\operatorname{det}(A)$." solution="The characteristic polynomial is

The constant term is $6$. Alternatively, the constant term is
"
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:::question type="MCQ" question="Consider a matrix $P$ such that $P^2 = P$. Which of the following is a possible eigenvalue for $P$?" options=["$i$", "$2$", "$0.5$", "$1$"] answer="$1$" hint="If $v$ is an eigenvector of $P$ with eigenvalue $\lambda$, then $Pv = \lambda v$.

Also, since $P^2 = P$, we have $P^2v = Pv = \lambda v$.
Equating the expressions for $P^2v$, we get $\lambda^2 v = \lambda v$. Since $v \neq 0$, we must have $\lambda^2 = \lambda$, which implies $\lambda(\lambda - 1) = 0$. Thus, the possible eigenvalues are $0$ or $1$." solution="Let $\lambda$ be an eigenvalue of $P$ with corresponding eigenvector $v$. Then $Pv = \lambda v$.
Since $P^2 = P$, we have $P^2v = Pv$.
Substituting $Pv = \lambda v$, we get:

Since $v \neq 0$, we must have:

Thus, the possible eigenvalues are $0$ or $1$. Among the given options, $1$ is a possible eigenvalue."
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What's Next?