Linear Transformations and Matrices

This chapter delves into linear transformations and their essential matrix representations, fundamental concepts for understanding mappings between vector spaces. A strong grasp of these topics, including range space, null space, and the Rank-Nullity Theorem, is critical for success in the Algebra section of the CUET PG examination.

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Chapter Contents

| # | Topic |
|---|-------|
| 1 | Linear Transformations |
| 2 | Range Space and Null Space |
| 3 | Rank-Nullity Theorem |
| 4 | Matrix Representation |

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We begin with Linear Transformations.

Part 1: Linear Transformations

Linear transformations are fundamental mappings between vector spaces that preserve the operations of vector addition and scalar multiplication. We explore their properties, matrix representations, and implications for understanding vector space structures, which are critical for various applications in mathematics and related fields.

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Core Concepts

1. Definition of a Linear Transformation

A mapping $T: V \rightarrow W$ between two vector spaces $V$ and $W$ (over the same field $F$) is defined as a linear transformation if it satisfies two conditions:

Additivity: For all $u, v \in V$, $T(u + v) = T(u) + T(v)$.

Homogeneity (Scalar Multiplication): For all $u \in V$ and $c \in F$, $T(cu) = cT(u)$.

These two conditions can be combined into a single condition: for all $u, v \in V$ and $c_1, c_2 \in F$, $T(c_1 u + c_2 v) = c_1 T(u) + c_2 T(v)$.

Quick Example: Determine if $T: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ defined by $T(x, y) = (x + y, x)$ is a linear transformation.

Step 1: Check additivity. Let $u = (x_1, y_1)$ and $v = (x_2, y_2)$.

And

We observe $T(u+v) = T(u) + T(v)$.

Step 2: Check homogeneity. Let $c$ be a scalar.

And

We observe $T(cu) = cT(u)$.

Answer: Since both conditions are satisfied, $T(x, y) = (x + y, x)$ is a linear transformation.

:::question type="MCQ" question="Which of the following functions $T: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ is a linear transformation?" options=["$T(x, y) = (x^2, y)$","$T(x, y) = (x+1, y)$","$T(x, y) = (xy, x)$","$T(x, y) = (2x - y, x + 3y)$"] answer="$T(x, y) = (2x - y, x + 3y)$" hint="Test both additivity and homogeneity. Functions with constant terms, products of variables, or non-linear powers are generally not linear." solution="Let $u=(x_1, y_1)$ and $v=(x_2, y_2)$, and $c$ be a scalar.
For $T(x, y) = (2x - y, x + 3y)$:

Additivity:

$T(u+v) = T(x_1+x_2, y_1+y_2) = (2(x_1+x_2) - (y_1+y_2), (x_1+x_2) + 3(y_1+y_2))$
$= (2x_1-y_1 + 2x_2-y_2, x_1+3y_1 + x_2+3y_2)$
$T(u)+T(v) = (2x_1-y_1, x_1+3y_1) + (2x_2-y_2, x_2+3y_2)$
$= (2x_1-y_1+2x_2-y_2, x_1+3y_1+x_2+3y_2)$.
Additivity holds.

Homogeneity:

$T(cu) = T(cx_1, cy_1) = (2cx_1 - cy_1, cx_1 + 3cy_1) = c(2x_1 - y_1, x_1 + 3y_1)$.
$cT(u) = c(2x_1 - y_1, x_1 + 3y_1)$.
Homogeneity holds.
Thus, $T(x, y) = (2x - y, x + 3y)$ is a linear transformation.

The other options fail one or both conditions:

• $T(x, y) = (x^2, y)$: $T(2(1,0)) = T(2,0) = (4,0)$, but $2T(1,0) = 2(1,0) = (2,0)$. Fails homogeneity.


• $T(x, y) = (x+1, y)$: $T(0,0) = (1,0) \neq (0,0)$. For any linear transformation, $T(0)=0$. Fails this property.


• $T(x, y) = (xy, x)$: $T((1,1)+(1,1)) = T(2,2) = (4,2)$. $T(1,1)+T(1,1) = (1,1)+(1,1) = (2,2)$. Fails additivity.

Answer: \boxed{$T(x, y) = (2x - y, x + 3y)$}"
:::

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2. Properties of Linear Transformations

Several essential properties follow directly from the definition of a linear transformation. These properties are often useful in proofs and problem-solving.

Quick Example: Let $T: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ be a linear transformation such that $T(1,0) = (2,3)$ and $T(0,1) = (-1,1)$. Find $T(3, -2)$.

Step 1: Express $(3, -2)$ as a linear combination of the basis vectors $(1,0)$ and $(0,1)$.

Step 2: Apply the linearity property $T(c_1 v_1 + c_2 v_2) = c_1 T(v_1) + c_2 T(v_2)$.

Answer: $T(3, -2) = (8,7)$.

:::question type="MCQ" question="Let $T: \mathbf{R}^3 \rightarrow \mathbf{R}^2$ be a linear transformation defined by $T(1,0,0) = (1,2)$, $T(0,1,0) = (3,-1)$, and $T(0,0,1) = (0,4)$. What is $T(2, -1, 3)$?" options=["$(-1, 17)$","$(2, 12)$","$(1, 15)$","$(7, 10)$"] answer="$(-1, 17)$" hint="Express the input vector as a linear combination of the standard basis vectors and apply the linearity property." solution="Step 1: Express $(2, -1, 3)$ as a linear combination of the standard basis vectors.

Step 2: Apply the linear transformation $T$.

So, $T(2, -1, 3) = (-1, 17)$.
Answer: \boxed{$(-1, 17)$}"
:::

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3. Kernel and Range of a Linear Transformation

For a linear transformation $T: V \rightarrow W$, two crucial subspaces are defined: the kernel and the range.

Quick Example: Let $T: \mathbf{R}^3 \rightarrow \mathbf{R}^2$ be defined by $T(x, y, z) = (x+y, y-z)$. Find a basis for $\operatorname{ker}(T)$.

Step 1: Set $T(x, y, z) = (0,0)$ to find vectors in the kernel.

This yields a system of linear equations:

Step 2: Solve the system. From the equations, we have $y = -x$ and $y = z$.
So, $z = y = -x$.
The general form of a vector in the kernel is $(x, -x, -x)$.

Step 3: Express the general form as a scalar multiple of a basis vector.

Answer: A basis for $\operatorname{ker}(T)$ is $\{(1, -1, -1)\}$.

:::question type="MCQ" question="Let $T: \mathbf{R}^3 \rightarrow \mathbf{R}^2$ be defined by $T(x, y, z) = (x-y, y-z)$. Which of the following vectors is in $\operatorname{ker}(T)$?" options=["$(1, 1, 0)$","$(1, 0, 1)$","$(0, 1, 1)$","$(1, 1, 1)$"] answer="$(1, 1, 1)$" hint="A vector $v$ is in $\operatorname{ker}(T)$ if $T(v) = 0_W$. Substitute each option into $T(x,y,z)$ and check if it results in $(0,0)$." solution="For a vector $(x,y,z)$ to be in $\operatorname{ker}(T)$, we must have $T(x,y,z) = (x-y, y-z) = (0,0)$.
This implies:

$x-y = 0 \Rightarrow x = y$

$y-z = 0 \Rightarrow y = z$

Thus, for a vector to be in the kernel, all its components must be equal: $x=y=z$.
Checking the options:

• $(1, 1, 0)$: $x=1, y=1, z=0$. $x=y$ but $y \neq z$. Not in $\operatorname{ker}(T)$.


• $(1, 0, 1)$: $x=1, y=0, z=1$. $x \neq y$. Not in $\operatorname{ker}(T)$.


• $(0, 1, 1)$: $x=0, y=1, z=1$. $x \neq y$. Not in $\operatorname{ker}(T)$.


• $(1, 1, 1)$: $x=1, y=1, z=1$. $x=y=z$. This vector is in $\operatorname{ker}(T)$.

Therefore, $(1, 1, 1)$ is in $\operatorname{ker}(T)$.
Answer: \boxed{$(1, 1, 1)$}"
:::

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4. Rank-Nullity Theorem

The Rank-Nullity Theorem establishes a fundamental relationship between the dimensions of the domain, kernel, and range of a linear transformation.

Quick Example: Let $T: \mathbf{R}^4 \rightarrow \mathbf{R}^3$ be a linear transformation. If $\operatorname{rank}(T) = 2$, find $\operatorname{nullity}(T)$.

Step 1: Identify the dimension of the domain $V$.
Here, $V = \mathbf{R}^4$, so $\dim(V) = 4$.

Step 2: Apply the Rank-Nullity Theorem.

Step 3: Solve for $\operatorname{nullity}(T)$.

Answer: $\operatorname{nullity}(T) = 2$.

:::question type="NAT" question="A linear transformation $T: \mathbf{R}^5 \rightarrow \mathbf{R}^3$ has a kernel with dimension 3. What is the dimension of its range?" answer="2" hint="Apply the Rank-Nullity Theorem: $\dim(V) = \operatorname{nullity}(T) + \operatorname{rank}(T)$." solution="Given $T: \mathbf{R}^5 \rightarrow \mathbf{R}^3$.
The dimension of the domain $V$ is $\dim(\mathbf{R}^5) = 5$.
The dimension of the kernel, $\operatorname{nullity}(T)$, is given as 3.
Using the Rank-Nullity Theorem:

The dimension of the range is 2.
Answer: \boxed{2}"
:::

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5. Matrix Representation of Linear Transformations

Any linear transformation $T: V \rightarrow W$ between finite-dimensional vector spaces can be represented by a matrix. If $B = \{v_1, \dots, v_n\}$ is a basis for $V$ and $C = \{w_1, \dots, w_m\}$ is a basis for $W$, then the matrix representation of $T$ with respect to bases $B$ and $C$, denoted $[T]_{C \leftarrow B}$, is an $m \times n$ matrix whose columns are the coordinate vectors of $T(v_j)$ with respect to basis $C$.

Specifically, if $T(v_j) = a_{1j}w_1 + a_{2j}w_2 + \dots + a_{mj}w_m$, then the $j$-th column of $[T]_{C \leftarrow B}$ is

.

Quick Example: Let $T: \mathbf{R}^2 \rightarrow \mathbf{R}^3$ be defined by $T(x, y) = (x+y, x-y, 2x)$. Find the matrix representation of $T$ with respect to the standard bases $B = \{(1,0), (0,1)\}$ for $\mathbf{R}^2$ and $C = \{(1,0,0), (0,1,0), (0,0,1)\}$ for $\mathbf{R}^3$.

Step 1: Apply $T$ to each basis vector of $B$.

>

Step 2: Express these images as coordinate vectors with respect to basis $C$. Since $C$ is the standard basis, the coordinate vectors are the vectors themselves.

>

>

Step 3: Form the matrix $[T]_{C \leftarrow B}$ by using these coordinate vectors as columns.

>

Answer: The matrix representation is

.

:::question type="MCQ" question="Let $T: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ be defined by $T(x,y) = (2x+y, x-3y)$. Find the matrix representation of $T$ with respect to the standard basis $B = \{(1,0), (0,1)\}$ for $\mathbf{R}^2$." options=["

", "", "", ""] answer="" hint="Apply the transformation to each standard basis vector. The images form the columns of the matrix." solution="Step 1: Apply $T$ to the first standard basis vector $(1,0)$.
>
The first column of the matrix will be .

Step 2: Apply $T$ to the second standard basis vector $(0,1)$.
>

The second column of the matrix will be .

Step 3: Form the matrix.
>

Answer: "
:::

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6. Composition of Linear Transformations

If $T_1: U \rightarrow V$ and $T_2: V \rightarrow W$ are linear transformations, their composition $T_2 \circ T_1: U \rightarrow W$ is also a linear transformation. The matrix representation of the composite transformation is the product of their individual matrix representations.

Quick Example: Let $T_1: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ be $T_1(x,y) = (x+y, x)$ and $T_2: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ be $T_2(x,y) = (2x, y-x)$. Find the matrix representation of $T_2 \circ T_1$ with respect to the standard basis.

Step 1: Find the matrix representation for $T_1$.
$T_1(1,0) = (1,1)$, $T_1(0,1) = (1,0)$.
>

Step 2: Find the matrix representation for $T_2$.
$T_2(1,0) = (2, -1)$, $T_2(0,1) = (0,1)$.
>

Step 3: Compute the product $[T_2][T_1]$ to find $[T_2 \circ T_1]$.

>

Answer: The matrix representation of $T_2 \circ T_1$ is

.

:::question type="MCQ" question="Let $T_1: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ be defined by $T_1(x,y) = (x-y, y)$ and $T_2: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ be defined by $T_2(x,y) = (2x, x+y)$. Find the matrix representation of $T_1 \circ T_2$ with respect to the standard basis." options=["

", "", "", ""] answer="" hint="First find the matrix representations for $T_1$ and $T_2$ individually. Then multiply them in the correct order for $T_1 \circ T_2$." solution="Step 1: Find the matrix representation for $T_1$.
$T_1(1,0) = (1-0, 0) = (1,0)$.
$T_1(0,1) = (0-1, 1) = (-1,1)$.
>

Step 2: Find the matrix representation for $T_2$.
$T_2(1,0) = (2(1), 1+0) = (2,1)$.
$T_2(0,1) = (2(0), 0+1) = (0,1)$.
>

Step 3: Compute the product $[T_1][T_2]$ for $T_1 \circ T_2$.
>

Answer: "
:::

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7. Inverse Linear Transformations and Isomorphisms

An invertible linear transformation is a bijection between vector spaces. If such a transformation exists, the vector spaces are said to be isomorphic.

Quick Example: Determine if $T: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ defined by $T(x,y) = (x+y, x-y)$ is an isomorphism.

Step 1: Check if $T$ is one-to-one by finding $\operatorname{ker}(T)$.
Set $T(x,y) = (0,0)$:
>

Adding the two equations yields $2x=0 \Rightarrow x=0$. Substituting $x=0$ into the first equation yields $y=0$.
So, $\operatorname{ker}(T) = \{(0,0)\}$. Since the kernel contains only the zero vector, $T$ is one-to-one.

Step 2: Check if $T$ is onto by comparing dimensions.
Since $T: \mathbf{R}^2 \rightarrow \mathbf{R}^2$, $\dim(V) = 2$ and $\dim(W) = 2$.
From the Rank-Nullity Theorem: $\dim(\mathbf{R}^2) = \operatorname{nullity}(T) + \operatorname{rank}(T)$.
$2 = 0 + \operatorname{rank}(T) \Rightarrow \operatorname{rank}(T) = 2$.
Since $\operatorname{rank}(T) = \dim(\operatorname{Im}(T)) = 2$, and $\dim(W) = 2$, we have $\operatorname{Im}(T) = W$. Thus, $T$ is onto.

Answer: Since $T$ is both one-to-one and onto, it is an isomorphism.

.

:::question type="MCQ" question="Let $T: \mathbf{R}^3 \rightarrow \mathbf{R}^3$ be a linear transformation defined by $T(x,y,z) = (x+y, y+z, x+z)$. Is $T$ an isomorphism?" options=["Yes, because $\operatorname{det}([T]) \neq 0$", "No, because $\operatorname{ker}(T) \neq \{0\}$", "Yes, because $\operatorname{rank}(T) < 3$", "No, because $\operatorname{Im}(T) = \mathbf{R}^2$"] answer="Yes, because $\operatorname{det}([T]) \neq 0$" hint="An isomorphism is both one-to-one and onto. For a transformation from $V$ to $V$, this is equivalent to the kernel being trivial or the determinant of its matrix representation being non-zero." solution="Step 1: Find the matrix representation of $T$ with respect to the standard basis.
$T(1,0,0) = (1,0,1)$
$T(0,1,0) = (1,1,0)$
$T(0,0,1) = (0,1,1)$
>

Step 2: Calculate the determinant of $[T]$.
>

Step 3: Interpret the determinant.
Since $\det([T]) = 2 \neq 0$, the matrix is invertible. An invertible matrix corresponds to an invertible linear transformation. An invertible linear transformation is an isomorphism.
Alternatively, since $\det([T]) \neq 0$, the nullity of the matrix is 0, meaning $\operatorname{ker}(T) = \{0\}$, so $T$ is one-to-one. By Rank-Nullity Theorem, $\operatorname{rank}(T) = \dim(\mathbf{R}^3) - \operatorname{nullity}(T) = 3 - 0 = 3$. Since $\operatorname{rank}(T) = \dim(\mathbf{R}^3)$, $T$ is onto.
Since $T$ is both one-to-one and onto, it is an isomorphism.
Answer:

"
:::

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Advanced Applications

Example: Consider the linear transformation $T: P_2(\mathbf{R}) \rightarrow P_2(\mathbf{R})$ defined by $T(p(x)) = p(x) + p'(x)$, where $P_2(\mathbf{R})$ is the space of polynomials of degree at most 2. Find the matrix representation of $T$ with respect to the standard basis $B = \{1, x, x^2\}$.

Step 1: Apply $T$ to each basis vector in $B$.
For $p(x) = 1$: $T(1) = 1 + 0 = 1$.
For $p(x) = x$: $T(x) = x + 1$.
For $p(x) = x^2$: $T(x^2) = x^2 + 2x$.

Step 2: Express the images as coordinate vectors with respect to basis $B$.

Step 3: Form the matrix $[T]_B$.

>

Answer: The matrix representation is

.

:::question type="NAT" question="Let $T: M_{2 \times 2}(\mathbf{R}) \rightarrow \mathbf{R}$ be a linear transformation defined by $T(A) = \operatorname{tr}(A)$, where $\operatorname{tr}(A)$ is the trace of matrix $A$. If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $T(A) = a+d$. What is the nullity of $T$?" answer="3" hint="First determine the dimension of the domain space $M_{2 \times 2}(\mathbf{R})$. Then find the rank of $T$ by identifying the range, and finally use the Rank-Nullity Theorem." solution="Step 1: Determine the dimension of the domain space.
The domain is $M_{2 \times 2}(\mathbf{R})$, the space of $2 \times 2$ matrices. A basis for this space is

Thus, $\dim(M_{2 \times 2}(\mathbf{R})) = 4$.

Step 2: Determine the range of $T$.
The transformation $T(A) = a+d$ maps a $2 \times 2$ matrix to a real number. The range $\operatorname{Im}(T)$ is a subspace of $\mathbf{R}$.
Can we get any real number $k$ in the range? Yes, for example, $T\left(\begin{bmatrix} k & 0 \\ 0 & 0 \end{bmatrix}\right) = k$.
So, $\operatorname{Im}(T) = \mathbf{R}$.
The dimension of the range, $\operatorname{rank}(T)$, is $\dim(\mathbf{R}) = 1$.

Step 3: Apply the Rank-Nullity Theorem.
>

>
>
>
The nullity of $T$ is 3. Answer: "
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $T: \mathbf{R}^2 \rightarrow \mathbf{R}^3$ be a linear transformation defined by $T(x,y) = (x+2y, y, x-y)$. What is the dimension of the range of $T$?" options=["0","1","2","3"] answer="2" hint="Find the images of the standard basis vectors of $\mathbf{R}^2$. These images span the range. Determine the number of linearly independent vectors among them." solution="Step 1: Find the images of the standard basis vectors of $\mathbf{R}^2$.
$T(1,0) = (1+2(0), 0, 1-0) = (1,0,1)$
$T(0,1) = (0+2(1), 1, 0-1) = (2,1,-1)$

Step 2: The range of $T$ is spanned by these image vectors: $\operatorname{Im}(T) = \operatorname{span}\{(1,0,1), (2,1,-1)\}$.

Step 3: Determine if these vectors are linearly independent.
We check if $c_1(1,0,1) + c_2(2,1,-1) = (0,0,0)$ implies $c_1=c_2=0$.

From the second equation, $c_2 = 0$.
Substituting $c_2=0$ into the first equation, $c_1 + 2(0) = 0 \Rightarrow c_1 = 0$.
Since both $c_1=0$ and $c_2=0$, the vectors $(1,0,1)$ and $(2,1,-1)$ are linearly independent.

Step 4: The dimension of the range is the number of linearly independent vectors in its spanning set.
$\operatorname{rank}(T) = 2$.
Alternatively, $\dim(\text{domain}) = \dim(\mathbf{R}^2) = 2$.
$\operatorname{ker}(T)$ for $T(x,y) = (x+2y, y, x-y) = (0,0,0)$:
$y=0$.
$x+2(0)=0 \Rightarrow x=0$.
$x-y=0 \Rightarrow 0-0=0$.
So $\operatorname{ker}(T) = \{(0,0)\}$, meaning $\operatorname{nullity}(T) = 0$.
By Rank-Nullity Theorem: $\dim(\mathbf{R}^2) = \operatorname{nullity}(T) + \operatorname{rank}(T)$
$2 = 0 + \operatorname{rank}(T) \Rightarrow \operatorname{rank}(T) = 2$.
Answer: $\boxed{2}$"
:::

:::question type="NAT" question="Consider the vector space $V = \{ (x,y,z) \in \mathbf{R}^3 \mid x-y+z=0 \}$. Let $T: V \rightarrow \mathbf{R}^2$ be a linear transformation defined by $T(x,y,z) = (x,z)$. What is the nullity of $T$?" answer="0" hint="First find a basis for $V$ to determine $\dim(V)$. Then find the kernel of $T$ for vectors in $V$." solution="Step 1: Determine the dimension of the domain space $V$.
The condition $x-y+z=0$ defines a plane in $\mathbf{R}^3$. We can express $y$ in terms of $x$ and $z$: $y = x+z$.
A vector in $V$ is of the form $(x, x+z, z)$.

The vectors $(1,1,0)$ and $(0,1,1)$ are linearly independent and span $V$.
Thus, $\dim(V) = 2$.

Step 2: Determine the kernel of $T$.
A vector $(x,y,z) \in V$ is in $\operatorname{ker}(T)$ if $T(x,y,z) = (0,0)$.
Given $T(x,y,z) = (x,z)$, this implies $x=0$ and $z=0$.
Since $(x,y,z)$ must also be in $V$, it must satisfy $x-y+z=0$.
Substituting $x=0$ and $z=0$ into the equation for $V$:
$0 - y + 0 = 0 \Rightarrow y = 0$.
Therefore, the only vector in $\operatorname{ker}(T)$ is $(0,0,0)$.
So, $\operatorname{nullity}(T) = \dim(\operatorname{ker}(T)) = 0$.
Answer: $\boxed{0}$"
:::

:::question type="MSQ" question="Let $T: \mathbf{R}^3 \rightarrow \mathbf{R}^2$ be a linear transformation given by $T(x,y,z) = (x+y, y-z)$. Which of the following statements are true?" options=["$T$ is one-to-one.","The range of $T$ is $\mathbf{R}^2$.","The nullity of $T$ is 1.","The matrix representation of $T$ with respect to standard bases is

"] answer="The range of $T$ is $\mathbf{R}^2$,The nullity of $T$ is 1,The matrix representation of $T$ with respect to standard bases is $\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & -1 \end{bmatrix}$" hint="Determine the kernel and range of $T$. Calculate the matrix representation using standard basis vectors. Use the Rank-Nullity Theorem." solution="Step 1: Find the matrix representation of $T$ with respect to the standard bases.
$T(1,0,0) = (1,0)$
$T(0,1,0) = (1,1)$
$T(0,0,1) = (0,-1)$

So, option D is true.

Step 2: Find the kernel of $T$.
Set $T(x,y,z) = (0,0)$:

From these, $y = -x$ and $z = y$. So $z = -x$.
The vectors in the kernel are of the form $(x, -x, -x) = x(1, -1, -1)$.
A basis for $\operatorname{ker}(T)$ is $\{(1,-1,-1)\}$.
The dimension of the kernel, $\operatorname{nullity}(T)$, is 1. So, option C is true.

Step 3: Check if $T$ is one-to-one.
Since $\operatorname{ker}(T) \neq \{0\}$, $T$ is not one-to-one. So, option A is false.

Step 4: Find the dimension of the range using the Rank-Nullity Theorem.
$\dim(\text{Domain}) = \dim(\mathbf{R}^3) = 3$.
$\operatorname{nullity}(T) = 1$.
$\dim(\text{Domain}) = \operatorname{nullity}(T) + \operatorname{rank}(T)$
$3 = 1 + \operatorname{rank}(T)$
$\operatorname{rank}(T) = 2$.
Since the codomain is $\mathbf{R}^2$ and $\dim(\mathbf{R}^2) = 2$, and $\operatorname{rank}(T) = 2$, the range of $T$ is $\mathbf{R}^2$. So, option B is true.

Therefore, the true statements are B, C, and D.
Answer: $\boxed{\text{B, C, and D}}$"
:::

:::question type="MCQ" question="Let $P_1(\mathbf{R})$ be the vector space of polynomials of degree at most 1. Consider $T: P_1(\mathbf{R}) \rightarrow \mathbf{R}$ defined by $T(p(x)) = \int_0^1 p(x) dx$. Which of the following is true?" options=["$T$ is not a linear transformation.","$\operatorname{ker}(T) = \{0\}$.","$\operatorname{rank}(T) = 1$.","The range of $T$ is $\{0\}$.""] answer="$\operatorname{rank}(T) = 1$." hint="First, verify linearity. Then find the kernel by setting the integral to zero. Use Rank-Nullity." solution="Step 1: Verify linearity.
Let $p(x) = ax+b$ and $q(x) = cx+d$. Let $k$ be a scalar.
$T(p(x)+q(x)) = T((a+c)x + (b+d))$

$T(p(x)) + T(q(x)) = \int_0^1 (ax+b) dx + \int_0^1 (cx+d) dx$

Additivity holds.
$T(kp(x)) = T(k(ax+b))$

$kT(p(x)) = k(\frac{a}{2}+b)$.
Homogeneity holds.
So, $T$ is a linear transformation. Option A is false.

Step 2: Determine the kernel of $T$.
Let $p(x) = ax+b$. We need $T(p(x))=0$.

So, polynomials in $\operatorname{ker}(T)$ are of the form $ax - \frac{a}{2} = a(x - \frac{1}{2})$.
Since $a$ can be any real number, $\operatorname{ker}(T)$ is not $\{0\}$. For example, $x - \frac{1}{2}$ is in $\operatorname{ker}(T)$. Option B is false.

Step 3: Determine the rank of $T$.
The domain is $P_1(\mathbf{R})$, which has a basis $\{1, x\}$, so $\dim(P_1(\mathbf{R})) = 2$.
The nullity is $\dim(\operatorname{ker}(T))$. Since $\operatorname{ker}(T) = \operatorname{span}\{x - \frac{1}{2}\}$, $\operatorname{nullity}(T) = 1$.
Using the Rank-Nullity Theorem:
$\dim(P_1(\mathbf{R})) = \operatorname{nullity}(T) + \operatorname{rank}(T)$
$2 = 1 + \operatorname{rank}(T)$
$\operatorname{rank}(T) = 1$.
Option C is true.

Step 4: Determine the range of $T$.
The codomain is $\mathbf{R}$. Since $\operatorname{rank}(T) = 1$, the range of $T$ is $\mathbf{R}$ (any real number can be obtained). For example, $T(2x) = \int_0^1 2x dx = [x^2]_0^1 = 1$. $T(4x) = 2$. So the range is not $\{0\}$. Option D is false.
Answer: $\boxed{\text{Option C}}$"
:::

:::question type="MSQ" question="Let $T: \mathbf{R}^2 \rightarrow \mathbf{R}^2$ be a linear transformation represented by the matrix

with respect to the standard basis. Which of the following statements are true?" options=["$T$ is invertible.","The kernel of $T$ is $\operatorname{span}\{(1,1)\}$.","The range of $T$ is $\operatorname{span}\{(1,-1)\}$.","The nullity of $T$ is 1.""] answer="The kernel of $T$ is $\operatorname{span}\{(1,1)\}$,The range of $T$ is $\operatorname{span}\{(1,-1)\}$,The nullity of $T$ is 1." hint="Calculate the determinant of the matrix to check invertibility. Find the null space of the matrix to determine the kernel and its dimension. Find the column space of the matrix to determine the range." solution="Step 1: Check if $T$ is invertible.
Calculate the determinant of $A$:

Since $\det(A) = 0$, the matrix $A$ is not invertible, and thus $T$ is not invertible. So, option A is false.

Step 2: Find the kernel of $T$.
We need to find vectors $(x,y)$ such that

This gives the system:

Both equations simplify to $x=y$.
So, vectors in the kernel are of the form $(x,x) = x(1,1)$.
Thus, $\operatorname{ker}(T) = \operatorname{span}\{(1,1)\}$. So, option B is true.

Step 3: Find the nullity of $T$.
Since $\operatorname{ker}(T) = \operatorname{span}\{(1,1)\}$, the dimension of the kernel is 1.
So, $\operatorname{nullity}(T) = 1$. So, option D is true.

Step 4: Find the range of $T$.
The range of $T$ is the column space of $A$. The columns of $A$ are $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$.
These two columns are linearly dependent, as $\begin{bmatrix} -1 \\ 1 \end{bmatrix} = -1 \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
Thus, the column space is spanned by a single vector, e.g., $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
So, $\operatorname{Im}(T) = \operatorname{span}\{(1,-1)\}$. So, option C is true.

Alternatively, using Rank-Nullity Theorem:
$\dim(\mathbf{R}^2) = \operatorname{nullity}(T) + \operatorname{rank}(T)$
$2 = 1 + \operatorname{rank}(T)$
$\operatorname{rank}(T) = 1$.
A basis for the range is a single non-zero vector from the image, e.g., $T(1,0)=(1,-1)$. So, $\operatorname{Im}(T) = \operatorname{span}\{(1,-1)\}$.

Therefore, the true statements are B, C, and D.
Answer: $\boxed{\text{B, C, and D}}$"
:::

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Summary

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What's Next?

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Part 2: Range Space and Null Space

Linear transformations are fundamental to linear algebra, mapping vectors from one vector space to another while preserving vector addition and scalar multiplication. The range space and null space are two critical subspaces associated with any linear transformation, providing insight into its structure and behavior. We examine these concepts for their direct utility in analyzing linear systems and their prevalence in competitive examinations.

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Core Concepts

1. Range Space of a Linear Transformation

The range space, also known as the image, of a linear transformation $T: V \to W$ is the set of all possible output vectors in $W$ that result from applying $T$ to vectors in the domain $V$. It is a subspace of the codomain $W$.

If $T$ is represented by a matrix $A$, then the range space of $T$ is equivalent to the column space of $A$. The dimension of the range space is called the rank of the transformation or matrix.

Quick Example: Determine the range space and rank of the linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(x,y) = (x, x+y, y)$.

Step 1: Find the matrix representation of $T$.
We apply $T$ to the standard basis vectors of $\mathbb{R}^2$:
$T(1,0) = (1, 1+0, 0) = (1,1,0)$
$T(0,1) = (0, 0+1, 1) = (0,1,1)$

The matrix $A$ whose columns are these image vectors is:

Step 2: Find a basis for the column space of $A$.
The columns of $A$ are

and

We check for linear independence.
If $k_1 c_1 + k_2 c_2 = 0$, then:

This implies $k_1=0$ and $k_2=0$, so the columns are linearly independent.

Step 3: State the range space and rank.
The range space $\operatorname{Im}(T)$ is the span of these linearly independent column vectors.

The rank of $T$ is the dimension of its range space, which is the number of linearly independent vectors in its basis.
Answer: $\operatorname{rank}(T) = 2$.

:::question type="MCQ" question="Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation defined by $T(x,y,z) = (x+y, y-z)$. What is the rank of $T$?" options=["$0$","$1$","$2$","$3$"] answer="$2$" hint="Find the matrix representation of $T$ and determine the dimension of its column space." solution="Step 1: Find the matrix representation of $T$.
Apply $T$ to the standard basis vectors of $\mathbb{R}^3$:
$T(1,0,0) = (1+0, 0-0) = (1,0)$
$T(0,1,0) = (0+1, 1-0) = (1,1)$
$T(0,0,1) = (0+0, 0-1) = (0,-1)$

The matrix $A$ is formed by these column vectors:

Step 2: Determine the rank of $A$.
The rank of $A$ is the number of linearly independent columns (or rows). We can perform row operations to find the row echelon form or observe linear independence.
The first two columns

and

are clearly linearly independent. Since the maximum possible rank for a $2 \times 3$ matrix is 2, the rank of $A$ is 2.
Alternatively, the row echelon form is already achieved or easily obtained.

There are two non-zero rows, so $\operatorname{rank}(A)=2$.

Step 3: The rank of $T$ is the rank of its matrix representation.
Answer: $\boxed{2}$"
:::

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2. Null Space of a Linear Transformation

The null space, also known as the kernel, of a linear transformation $T: V \to W$ is the set of all vectors in the domain $V$ that are mapped to the zero vector in the codomain $W$. It is a subspace of the domain $V$.

If $T$ is represented by a matrix $A$, then the null space of $T$ is the solution space of the homogeneous system $Ax=0$. The dimension of the null space is called the nullity of the transformation or matrix.

Quick Example: Determine the null space and nullity of the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $T(x,y,z) = (x+y, y-z)$.

Step 1: Find the matrix representation of $T$.
As derived in the previous example, the matrix $A$ for $T$ is:

Step 2: Find the null space by solving $Ax=0$.
We set $Ax=0$:

This gives the system of equations:

From the second equation, $y=z$.
Substitute $y=z$ into the first equation: $x+z=0$, so $x=-z$.

Step 3: Express the general solution and find a basis for the null space.
Let $z=t$ be a free variable. Then $y=t$ and $x=-t$.
Any vector in the null space is of the form:

The null space $\operatorname{Ker}(T)$ is the span of the vector $\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$.

The nullity of $T$ is the dimension of its null space.
Answer: $\operatorname{nullity}(T) = 1$.

:::question type="MCQ" question="Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation defined by $T(x,y,z) = (x-y+z, 2x-2y+2z, -x+y-z)$. What is the nullity of $T$?" options=["$0$","$1$","$2$","$3$"] answer="$2$" hint="Form the matrix $A$ for $T$ and solve the homogeneous system $Ax=0$ to find the basis for the null space." solution="Step 1: Find the matrix representation of $T$.
Apply $T$ to the standard basis vectors of $\mathbb{R}^3$:
$T(1,0,0) = (1, 2, -1)$
$T(0,1,0) = (-1, -2, 1)$
$T(0,0,1) = (1, 2, -1)$

The matrix $A$ for $T$ is:

Step 2: Find the null space by solving $Ax=0$.
We row-reduce the matrix $A$:

The system $Ax=0$ is equivalent to the single equation:

We have two free variables. Let $y=s$ and $z=t$.
Then $x = y - z = s - t$.

Step 3: Express the general solution and find a basis for the null space.
Any vector in the null space is of the form:

The null space $\operatorname{Ker}(T)$ is spanned by the linearly independent vectors $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$.

The nullity of $T$ is the dimension of this null space.
Answer: $\boxed{2}$."
:::

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3. Rank-Nullity Theorem

The Rank-Nullity Theorem establishes a fundamental relationship between the dimension of the domain, the rank, and the nullity of a linear transformation. This theorem is crucial for understanding the properties of linear maps.

Quick Example: Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be a linear transformation whose range space $\operatorname{Im}(T)$ has dimension 3. What is the nullity of $T$?

Step 1: Identify the known quantities.
The domain $V = \mathbb{R}^4$, so $\dim(V) = 4$.
The rank of $T$ is given as $\operatorname{rank}(T) = \dim(\operatorname{Im}(T)) = 3$.

Step 2: Apply the Rank-Nullity Theorem.

Step 3: Solve for $\operatorname{nullity}(T)$.

Answer: The nullity of $T$ is $1$.

:::question type="NAT" question="A linear transformation $T: \mathbb{R}^5 \to \mathbb{R}^4$ is given by a $4 \times 5$ matrix $A$. If the column space of $A$ has dimension 3, what is the dimension of the solution space to $Ax=0$?" answer="2" hint="Recall that the dimension of the column space is the rank, and the dimension of the solution space to $Ax=0$ is the nullity. Apply the Rank-Nullity Theorem." solution="Step 1: Identify the given information in terms of the Rank-Nullity Theorem.
The domain of the transformation is $\mathbb{R}^5$, so $\dim(V) = 5$.
The dimension of the column space of $A$ is 3. The dimension of the column space is precisely the rank of the matrix $A$, and thus the rank of the transformation $T$. So, $\operatorname{rank}(T) = 3$.
The dimension of the solution space to $Ax=0$ is the nullity of $T$, $\operatorname{nullity}(T)$.

Step 2: Apply the Rank-Nullity Theorem.
The theorem states: $\dim(V) = \operatorname{rank}(T) + \operatorname{nullity}(T)$.
Substitute the known values:

Step 3: Solve for $\operatorname{nullity}(T)$.

Answer: The dimension of the solution space to $Ax=0$ is $\boxed{2}$."
:::

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4. Injectivity and Surjectivity

The null space and range space provide direct criteria for determining whether a linear transformation is injective (one-to-one) or surjective (onto).

Quick Example: Consider $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x,y) = (x+y, x-y)$. Is $T$ injective? Is $T$ surjective?

Step 1: Determine the null space for injectivity.
We solve $T(x,y) = (0,0)$:

Adding the two equations yields $2x=0 \implies x=0$.
Substituting $x=0$ into the first equation yields $y=0$.
Thus, $\operatorname{Ker}(T) = \{(0,0)\}$.

Step 2: Conclude on injectivity.
Since $\operatorname{Ker}(T) = \{(0,0)\}$, $T$ is injective.

Step 3: Determine the range space for surjectivity.
The matrix for $T$ is:

The columns are $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$. These are linearly independent, so $\operatorname{rank}(T) = 2$.
The codomain is $\mathbb{R}^2$, and $\dim(\mathbb{R}^2) = 2$.

Step 4: Conclude on surjectivity.
Since $\operatorname{rank}(T) = \dim(\mathbb{R}^2)$, $T$ is surjective.

Answer: $T$ is both injective and surjective.

:::question type="MSQ" question="Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation defined by $T(x,y,z) = (x+y-z, y+z)$. Select ALL correct statements regarding $T$." options=["$T$ is injective.","$\operatorname{nullity}(T) = 1$.","$T$ is surjective.","$\operatorname{rank}(T) = 2$."] answer="$\operatorname{nullity}(T) = 1$.,$T$ is surjective.,$\operatorname{rank}(T) = 2$." hint="First, find the matrix representation of $T$. Then, determine its rank and nullity using row reduction and the Rank-Nullity Theorem. Finally, evaluate injectivity and surjectivity." solution="Step 1: Find the matrix representation of $T$.
Apply $T$ to the standard basis vectors of $\mathbb{R}^3$:
$T(1,0,0) = (1,0)$
$T(0,1,0) = (1,1)$
$T(0,0,1) = (-1,1)$

The matrix $A$ for $T$ is:

Step 2: Determine the rank of $T$.
The matrix $A$ is already in row echelon form. There are two non-zero rows, so the rank of $A$ is 2.
Thus, $\operatorname{rank}(T) = 2$. (Statement D is correct)

Step 3: Determine the nullity of $T$ using the Rank-Nullity Theorem.
The domain is $\mathbb{R}^3$, so $\dim(\mathbb{R}^3) = 3$.

(Statement B is correct)

Step 4: Evaluate injectivity.
For $T$ to be injective, $\operatorname{nullity}(T)$ must be 0. Since $\operatorname{nullity}(T) = 1$, $T$ is not injective. (Statement A is incorrect)

Step 5: Evaluate surjectivity.
For $T$ to be surjective, $\operatorname{rank}(T)$ must be equal to the dimension of the codomain. The codomain is $\mathbb{R}^2$, and $\dim(\mathbb{R}^2) = 2$.
Since $\operatorname{rank}(T) = 2 = \dim(\mathbb{R}^2)$, $T$ is surjective. (Statement C is correct)

Answer: $\operatorname{nullity}(T) = 1$.,$T$ is surjective.,$\operatorname{rank}(T) = 2$."
:::

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Advanced Applications

We consider a more complex example combining multiple concepts.

Quick Example: Let $A$ be a $3 \times 4$ matrix given by:

Find a basis for the range space of $A$ and a basis for the null space of $A$.

Step 1: Row-reduce $A$ to find its row echelon form.

This is the row echelon form.

Step 2: Find a basis for the range space (column space).
The pivot columns in the row echelon form are the 1st and 3rd columns.
Thus, the corresponding columns from the original matrix $A$ form a basis for the range space.

The rank of $A$ is 2.

Step 3: Find a basis for the null space.
From the row echelon form, the system $Ax=0$ is equivalent to:

From the second equation, $x_3 = x_4$.
Substitute $x_3$ into the first equation: $x_1 + 2x_2 + x_4 = 0 \implies x_1 = -2x_2 - x_4$.
The free variables are $x_2$ and $x_4$. Let $x_2 = s$ and $x_4 = t$.
Then $x_3 = t$ and $x_1 = -2s - t$.

The general solution vector is:

Step 4: State the basis for the null space.

The nullity of $A$ is 2.

Answer: Basis for $\operatorname{Im}(A) = \left\{ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \right\}$, Basis for $\operatorname{Ker}(A) = \left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \\ 1 \end{bmatrix} \right\}$.
We observe $\dim(V) = 4$, $\operatorname{rank}(A) = 2$, $\operatorname{nullity}(A) = 2$. This satisfies the Rank-Nullity Theorem: $4 = 2+2$.

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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 $5 \times 7$ matrix with $\operatorname{rank} 3$. What is the dimension of the null space of $A$?" options=["$2$","$3$","$4$","$5$"] answer="$4$" hint="Apply the Rank-Nullity Theorem. The dimension of the domain is the number of columns of the matrix." solution="Step 1: Identify the dimensions.
The matrix $A$ is $5 \times 7$, meaning it defines a linear transformation from $\mathbb{R}^7$ to $\mathbb{R}^5$.
The dimension of the domain $V$ is the number of columns, so $\dim(V) = 7$.
The rank of $A$ is given as $\operatorname{rank}(A) = 3$.

Step 2: Apply the Rank-Nullity Theorem.
$\dim(V) = \operatorname{rank}(A) + \operatorname{nullity}(A)$

Step 3: Solve for $\operatorname{nullity}(A)$.

Answer: The dimension of the null space of $A$ is $\boxed{4}$."
:::

:::question type="NAT" question="Let $T: \mathbb{P}_2 \to \mathbb{R}^2$ be a linear transformation defined by $T(p(x)) = (p(0), p(1))$. What is the $\operatorname{nullity}$ of $T$?" answer="1" hint="First, find a basis for the domain $\mathbb{P}_2$ and determine its dimension. Then, find the matrix representation of $T$ with respect to these bases and calculate its rank." solution="Step 1: Determine the dimension of the domain.
The domain is $\mathbb{P}_2$, the space of polynomials of degree at most 2. A standard basis for $\mathbb{P}_2$ is $\{1, x, x^2\}$.
So, $\dim(\mathbb{P}_2) = 3$.

Step 2: Find the matrix representation of $T$.
Apply $T$ to the basis vectors:

The matrix $A$ for $T$ (with respect to the standard basis of $\mathbb{P}_2$ and $\mathbb{R}^2$) is:

Step 3: Determine the rank of $T$.
Row-reduce $A$:

The row echelon form has two non-zero rows, so $\operatorname{rank}(A) = 2$.

Step 4: Apply the Rank-Nullity Theorem.
$\dim(\mathbb{P}_2) = \operatorname{rank}(T) + \operatorname{nullity}(T)$

Answer: The nullity of $T$ is $\boxed{1}$."
:::

:::question type="MCQ" question="For a linear transformation $T: V \to W$, if $\dim(V) = 5$ and $\operatorname{Ker}(T)$ is spanned by two linearly independent vectors, then $\operatorname{Im}(T)$ has dimension:" options=["$2$","$3$","$4$","$5$"] answer="$3$" hint="The number of linearly independent vectors spanning the kernel is the nullity. Use the Rank-Nullity Theorem." solution="Step 1: Identify the given information.
$\dim(V) = 5$.
The null space $\operatorname{Ker}(T)$ is spanned by two linearly independent vectors, which means $\operatorname{nullity}(T) = 2$.

Step 2: Apply the Rank-Nullity Theorem.
$\dim(V) = \operatorname{rank}(T) + \operatorname{nullity}(T)$

Step 3: Solve for $\operatorname{rank}(T)$.

The dimension of $\operatorname{Im}(T)$ is the rank of $T$.
Answer: $\operatorname{Im}(T)$ has dimension $\boxed{3}$."
:::

:::question type="MSQ" question="Let $A$ be a $4 \times 3$ matrix such that the system $Ax=b$ has a unique solution for some $b \in \mathbb{R}^4$. Which of the following statements MUST be true?" options=["$\operatorname{rank}(A) = 3$.","$\operatorname{nullity}(A) = 0$.","The columns of $A$ are linearly independent.","The rows of $A$ are linearly independent."] answer="$\operatorname{rank}(A) = 3$.,$\operatorname{nullity}(A) = 0$.,The columns of $A$ are linearly independent." hint="A unique solution to $Ax=b$ implies that the null space of $A$ is trivial. Relate this to rank and linear independence of columns." solution="Step 1: Analyze the condition 'unique solution for some $b$'.
For a system $Ax=b$ to have a unique solution, two conditions must be met:

The system must be consistent (i.e., $b$ must be in the column space of $A$).

The null space of $A$ must be trivial, meaning $\operatorname{Ker}(A) = \{0\}$. This ensures that if a solution exists, it is unique. If $\operatorname{Ker}(A)$ were non-trivial, $x_0 + v_h$ (where $v_h \in \operatorname{Ker}(A)$) would yield multiple solutions.

Step 2: Relate to nullity and rank.
From condition 2, $\operatorname{nullity}(A) = \dim(\operatorname{Ker}(A)) = 0$. (Statement B is correct)
Using the Rank-Nullity Theorem: $\dim(\text{domain}) = \operatorname{rank}(A) + \operatorname{nullity}(A)$.
The domain for a $4 \times 3$ matrix is $\mathbb{R}^3$, so $\dim(\text{domain}) = 3$.

(Statement A is correct)

Step 3: Relate to linear independence of columns and rows.
Since $\operatorname{rank}(A) = 3$, and $A$ is a $4 \times 3$ matrix, the number of pivot columns is 3. This means all 3 columns of $A$ are pivot columns, which implies they are linearly independent. (Statement C is correct)
The rank of a matrix is also the number of linearly independent rows. A $4 \times 3$ matrix with rank 3 has 3 linearly independent rows. However, it has 4 rows in total. Therefore, the rows of $A$ cannot be linearly independent, as 4 vectors in $\mathbb{R}^3$ must be linearly dependent. (Statement D is incorrect)

Answer: $\operatorname{rank}(A) = 3$.,$\operatorname{nullity}(A) = 0$.,The columns of $A$ are linearly independent."
:::

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Summary

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What's Next?

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Part 3: Rank-Nullity Theorem

The Rank-Nullity Theorem is a fundamental result in linear algebra, establishing a crucial relationship between the dimensions of the null space (kernel) and the range space (image) of a linear transformation. We utilize this theorem extensively in determining properties of linear mappings and analyzing systems of linear equations, which are recurrent themes in the CUET PG examination.

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Core Concepts

1. Linear Transformation

A mapping $T: V \to W$ between two vector spaces $V$ and $W$ over the same field $F$ is a linear transformation if, for all vectors $\mathbf{u}, \mathbf{v} \in V$ and scalars $c \in F$, it satisfies two conditions:

$T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ (additivity)

$T(c\mathbf{u}) = cT(\mathbf{u})$ (homogeneity)

We frequently represent linear transformations using matrices, where matrix multiplication corresponds to the transformation.

Quick Example:

Consider the transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x,y) = (x+y, x-y)$. We verify its linearity.

Step 1: Check additivity.
Let $\mathbf{u} = (x_1, y_1)$ and $\mathbf{v} = (x_2, y_2)$.
$T(\mathbf{u} + \mathbf{v}) = T(x_1+x_2, y_1+y_2)$

Step 2: Check homogeneity.
Let $c \in \mathbb{R}$.
$T(c\mathbf{u}) = T(cx_1, cy_1)$

Answer: $T$ is a linear transformation.

:::question type="MCQ" question="Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a transformation defined by $T(x,y) = (xy, x+y)$. Is $T$ a linear transformation?" options=["Yes, because it maps from $\mathbb{R}^2$ to $\mathbb{R}^2$.","Yes, because it preserves vector addition.","No, because it does not preserve scalar multiplication.","No, because it does not preserve vector addition."] answer="No, because it does not preserve vector addition." hint="Test the additivity and homogeneity properties with specific vectors and scalars." solution="Let $\mathbf{u} = (1,0)$ and $\mathbf{v} = (0,1)$. Then $\mathbf{u} + \mathbf{v} = (1,1)$.

However,

Since $T(\mathbf{u} + \mathbf{v}) \neq T(\mathbf{u}) + T(\mathbf{v})$, the transformation is not linear. Therefore, it does not preserve vector addition.
Answer: $\boxed{\text{No, because it does not preserve vector addition.}}$"
:::

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2. Null Space (Kernel)

For a matrix $A$, its null space $\operatorname{Null}(A)$ is the set of all solutions to the homogeneous equation $A\mathbf{x} = \mathbf{0}$. The nullity of $A$ is the dimension of this solution space.

Quick Example:

Find the null space and nullity of the matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 6 \end{bmatrix}$.

Step 1: Set up the homogeneous system $A\mathbf{x} = \mathbf{0}$.

Step 2: Reduce the augmented matrix to row echelon form.

Step 3: Write the system of equations from the row echelon form.
$x + 2y = 0 \implies x = -2y$.
Here $y$ is a free variable.

Step 4: Express the general solution.
Let $y = t$ for some scalar $t \in \mathbb{R}$.
Then

Step 5: Identify the null space and nullity.
The null space is

The nullity is the number of free variables, which is 1.
Answer: $\operatorname{Null}(A) = \operatorname{span}\left\{\begin{bmatrix} -2 \\ 1 \end{bmatrix}\right\}$, $\operatorname{nullity}(A) = 1$.

:::question type="NAT" question="Consider the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $T(x,y,z) = (x-y, y-z)$. What is the nullity of $T$?" answer="1" hint="Find the matrix representation of $T$ and then its null space dimension. The null space consists of vectors $(x,y,z)$ such that $x-y=0$ and $y-z=0$." solution="The transformation $T(x,y,z) = (x-y, y-z)$ can be represented by a matrix $A$ such that $T(\mathbf{x}) = A\mathbf{x}$.

To find the null space, we solve $A\mathbf{x} = \mathbf{0}$:

From the second equation, $y = z$.
Substitute into the first equation: $x - z = 0 \implies x = z$.
So, any vector in the null space is of the form $(z, z, z)$.
We can write this as $z(1,1,1)$.
Thus, the null space is $\operatorname{span}\{(1,1,1)\}$.
The basis for the null space is $\{(1,1,1)\}$, which contains one vector.
Therefore, the nullity of $T$ is 1.
Answer: \boxed{1}"
:::

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3. Range Space (Image)

For a matrix $A$, its range space $\operatorname{Range}(A)$ is the column space of $A$, which is the span of its column vectors. The rank of $A$ is the dimension of its column space, which is also equal to the dimension of its row space. We typically find the rank by reducing the matrix to row echelon form and counting the number of non-zero rows (pivot positions).

Quick Example:

Find the range space and rank of the matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 6 \end{bmatrix}$.

Step 1: The range space is the column space of $A$.

Step 2: Determine a basis for the column space.
We observe that the second column is $2$ times the first column.

The columns are linearly dependent. A basis for the column space is

Step 3: Identify the rank.
The rank is the number of linearly independent column vectors, which is 1.
Alternatively, we reduce $A$ to row echelon form:

The number of non-zero rows (pivot positions) is 1.
Answer: $\operatorname{Range}(A) = \operatorname{span}\left\{\begin{bmatrix} 1 \\ 3 \end{bmatrix}\right\}$, $\operatorname{rank}(A) = 1$.

:::question type="MCQ" question="Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(x,y,z) = (x+y, y+z, x-z)$. What is the rank of $T$?" options=["$1$","$2$","$3$","$0$"] answer="2" hint="Form the matrix for $T$ and find the number of pivot positions after row reduction." solution="The matrix representation of $T$ is:

We reduce $A$ to row echelon form:

The row echelon form has two non-zero rows.
Therefore, the rank of $T$ is 2.
Answer: \boxed{2}"
:::

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4. Rank-Nullity Theorem

This theorem applies equally to matrices, where $\operatorname{rank}(A) + \operatorname{nullity}(A) = n$, for an $m \times n$ matrix $A$. Here $n$ is the number of columns (dimension of the domain space).

Quick Example:

Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be a linear transformation with $\operatorname{rank}(T) = 2$. Find $\operatorname{nullity}(T)$.

Step 1: Identify the dimension of the domain space.
The domain is $\mathbb{R}^4$, so $\dim(V) = 4$.

Step 2: Apply the Rank-Nullity Theorem.

Answer: $\operatorname{nullity}(T) = 2$.

:::question type="MCQ" question="A linear transformation $T: P_3(\mathbb{R}) \to \mathbb{R}^3$ has a range space spanned by $\{(1,0,1), (0,1,1)\}$. What is the nullity of $T$?" options=["$1$","$2$","$3$","$4$"] answer="2" hint="First, determine the dimension of the domain space $P_3(\mathbb{R})$ and the rank of $T$ from the given span." solution="The domain space is $P_3(\mathbb{R})$, the space of polynomials of degree at most 3. A basis for $P_3(\mathbb{R})$ is $\{1, x, x^2, x^3\}$, so $\dim(P_3(\mathbb{R})) = 4$.
The range space is spanned by $\{(1,0,1), (0,1,1)\}$. These two vectors are linearly independent, so the dimension of the range space, $\operatorname{rank}(T)$, is 2.
Applying the Rank-Nullity Theorem:

The nullity of $T$ is 2.
Answer: \boxed{2}"
:::

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Advanced Applications

1. Rank and Nullity of a Matrix

For an $m \times n$ matrix $A$, we can view it as the matrix representation of a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$.
The rank of $A$, denoted $\operatorname{rank}(A)$, is the dimension of its column space (or row space). This is equivalent to the number of pivot positions in its row echelon form.
The nullity of $A$, denoted $\operatorname{nullity}(A)$, is the dimension of the solution space to $A\mathbf{x} = \mathbf{0}$. This is equivalent to the number of free variables in the solution to $A\mathbf{x} = \mathbf{0}$.

Worked Example:

Find the rank and nullity of the matrix

Step 1: Reduce the matrix $A$ to row echelon form.

Step 2: Determine the rank.
The row echelon form has one non-zero row.
Therefore, $\operatorname{rank}(A) = 1$.

Step 3: Determine the nullity using the Rank-Nullity Theorem.
The matrix $A$ is $3 \times 3$, so $n=3$.

Alternatively, from the row echelon form, $x_1 + 2x_2 + 3x_3 = 0$. Variables $x_2$ and $x_3$ are free variables. Thus, there are 2 free variables, so $\operatorname{nullity}(A) = 2$.
Answer: $\operatorname{rank}(A) = 1$, $\operatorname{nullity}(A) = 2$.

:::question type="NAT" question="What is the nullity of the matrix $M = \begin{bmatrix} 1 & -1 & 2 & 0 \\ 2 & -2 & 4 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}$?" answer="2" hint="Find the rank of the matrix first by row reduction, then use the Rank-Nullity Theorem." solution="We reduce the matrix $M$ to row echelon form:

The number of non-zero rows (pivot positions) is 2. So, $\operatorname{rank}(M) = 2$.
The matrix $M$ is $3 \times 4$, so the number of columns $n=4$.
Using the Rank-Nullity Theorem:



Answer: \boxed{2}"
:::

2. Systems of Linear Equations

For a homogeneous system of linear equations $A\mathbf{x} = \mathbf{0}$, the solution set is precisely the null space of $A$. Thus, the dimension of the solution space is $\operatorname{nullity}(A)$.

For a non-homogeneous system $A\mathbf{x} = \mathbf{b}$, solutions exist if and only if $\mathbf{b}$ is in the column space (range) of $A$. The number of free variables in the general solution corresponds to $\operatorname{nullity}(A)$.

Worked Example (PYQ 1 type):

The dimension of the general solution space $W$ of the homogeneous system

is required.

Step 1: Form the coefficient matrix $A$.

Step 2: Reduce $A$ to row echelon form.

Step 3: Determine the rank of $A$.
The number of non-zero rows (pivot positions) is 2. So, $\operatorname{rank}(A) = 2$.

Step 4: Determine the nullity of $A$ using the Rank-Nullity Theorem.
The number of columns $n=5$.

The dimension of the general solution space $W$ is the nullity of the coefficient matrix.
Answer: The dimension of the general solution space is 3.

:::question type="MCQ" question="Consider the homogeneous system $x - 2y + z = 0$ and $2x - 4y + 2z = 0$. What is the dimension of its solution space?" options=["$0$","$1$","$2$","$3$"] answer="2" hint="Identify the coefficient matrix, find its rank, and then apply the Rank-Nullity Theorem." solution="The coefficient matrix $A$ for the system is:

We reduce $A$ to row echelon form:

The rank of $A$ is 1 (one non-zero row).
The number of columns (variables) $n=3$.
Using the Rank-Nullity Theorem:



The dimension of the solution space is the nullity of the coefficient matrix.
Answer: \boxed{2}"
:::

3. Properties of $T$ and $T^2$

Consider a linear transformation $T: V \to V$ on a finite-dimensional vector space $V$. We can define $T^2 = T \circ T$.

If $\operatorname{rank}(T) = \operatorname{rank}(T^2)$, this implies that the range space chain has stabilized at $k=1$, i.e., $\operatorname{Range}(T) = \operatorname{Range}(T^2)$.
By the Rank-Nullity Theorem, if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ and $\dim(V)$ is finite, then $\operatorname{nullity}(T) = \operatorname{nullity}(T^2)$. This implies that the null space chain has also stabilized at $k=1$, i.e., $\operatorname{Null}(T) = \operatorname{Null}(T^2)$.

Therefore, if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$, then:

$\operatorname{Null}(T) = \operatorname{Null}(T^2)$

$\operatorname{Range}(T) = \operatorname{Range}(T^2)$

$\operatorname{Null}(T) \cap \operatorname{Range}(T) = \{\mathbf{0}\}$

Worked Example (PYQ 2 type):

Let $T: V \to V$ be a linear transformation on a finite-dimensional vector space $V$. If $\operatorname{rank}(T) = \operatorname{rank}(T^2)$, identify the correct statements among:
A. $\operatorname{Null}(T) = \operatorname{Range}(T)$
B. $\operatorname{Null}(T) = \operatorname{Null}(T^2)$
C. $\operatorname{Null}(T) \cap \operatorname{Range}(T) = \{\mathbf{0}\}$
D. $\operatorname{Range}(T) = \operatorname{Range}(T^2)$

Step 1: Analyze option B.
We know that $\operatorname{Null}(T) \subseteq \operatorname{Null}(T^2)$.
From $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ and $\dim(V) = \operatorname{rank}(T) + \operatorname{nullity}(T)$, it follows that $\operatorname{nullity}(T) = \operatorname{nullity}(T^2)$.
Since $\operatorname{Null}(T)$ is a subspace of $\operatorname{Null}(T^2)$ and they have the same dimension, they must be equal. So, B is correct.

Step 2: Analyze option D.
We know that $\operatorname{Range}(T^2) \subseteq \operatorname{Range}(T)$.
Given $\operatorname{rank}(T) = \operatorname{rank}(T^2)$, and rank is the dimension of the range space, $\dim(\operatorname{Range}(T)) = \dim(\operatorname{Range}(T^2))$.
Since $\operatorname{Range}(T^2)$ is a subspace of $\operatorname{Range}(T)$ and they have the same dimension, they must be equal. So, D is correct.

Step 3: Analyze option C.
Let $\mathbf{v} \in \operatorname{Null}(T) \cap \operatorname{Range}(T)$.
This means $\mathbf{v} \in \operatorname{Null}(T)$, so $T(\mathbf{v}) = \mathbf{0}$.
And $\mathbf{v} \in \operatorname{Range}(T)$, so $\mathbf{v} = T(\mathbf{u})$ for some $\mathbf{u} \in V$.
Substituting the second into the first: $T(T(\mathbf{u})) = \mathbf{0}$, which means $T^2(\mathbf{u}) = \mathbf{0}$.
Thus, $\mathbf{u} \in \operatorname{Null}(T^2)$.
Since we established that $\operatorname{Null}(T) = \operatorname{Null}(T^2)$ (from B), it implies $\mathbf{u} \in \operatorname{Null}(T)$.
If $\mathbf{u} \in \operatorname{Null}(T)$, then $T(\mathbf{u}) = \mathbf{0}$.
Since $\mathbf{v} = T(\mathbf{u})$, this implies $\mathbf{v} = \mathbf{0}$.
Therefore, $\operatorname{Null}(T) \cap \operatorname{Range}(T) = \{\mathbf{0}\}$. So, C is correct.

Step 4: Analyze option A.
$\operatorname{Null}(T) = \operatorname{Range}(T)$ is generally not true. For example, consider $T: \mathbb{R}^2 \to \mathbb{R}^2$ given by $T(x,y) = (y,0)$.
$\operatorname{Null}(T) = \operatorname{span}\{(1,0)\}$ and $\operatorname{Range}(T) = \operatorname{span}\{(1,0)\}$. In this specific case, $N(T)=R(T)$.
However, consider $T(x,y) = (0,x)$.
$\operatorname{Null}(T) = \operatorname{span}\{(0,1)\}$. $\operatorname{Range}(T) = \operatorname{span}\{(0,1)\}$. This is also true here.
Consider $T(x,y,z) = (x,0,0)$. $N(T) = \operatorname{span}\{(0,1,0), (0,0,1)\}$. $R(T) = \operatorname{span}\{(1,0,0)\}$. Here $N(T) \neq R(T)$.
The condition $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ holds for $T(x,y,z) = (x,0,0)$ because $\operatorname{rank}(T)=1$, $T^2(x,y,z) = T(x,0,0) = (x,0,0)$, so $\operatorname{rank}(T^2)=1$.
But $N(T) \neq R(T)$. So, A is not generally correct.

Answer: Options B, C, and D are correct.

:::question type="MSQ" question="Let $T: V \to V$ be a linear transformation on a finite-dimensional vector space $V$. If $\operatorname{nullity}(T) = \operatorname{nullity}(T^2)$, which of the following statements are correct?" options=["$\operatorname{rank}(T) = \operatorname{rank}(T^2)$","$\operatorname{Null}(T) = \operatorname{Null}(T^2)$","$\operatorname{Range}(T) = \operatorname{Range}(T^2)$","$\operatorname{Null}(T) \cap \operatorname{Range}(T) = \{\mathbf{0}\}$"] answer="$\operatorname{rank}(T) = \operatorname{rank}(T^2)$,$\operatorname{Null}(T) = \operatorname{Null}(T^2)$,$\operatorname{Range}(T) = \operatorname{Range}(T^2)$,$\operatorname{Null}(T) \cap \operatorname{Range}(T) = \{\mathbf{0}\}$" hint="The condition $\operatorname{nullity}(T) = \operatorname{nullity}(T^2)$ is equivalent to $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ for finite-dimensional spaces. Analyze the consequences as in the previous example." solution="Given $\operatorname{nullity}(T) = \operatorname{nullity}(T^2)$.
Since $V$ is finite-dimensional, we use the Rank-Nullity Theorem: $\dim(V) = \operatorname{rank}(T) + \operatorname{nullity}(T)$.
Also, $\dim(V) = \operatorname{rank}(T^2) + \operatorname{nullity}(T^2)$.
If $\operatorname{nullity}(T) = \operatorname{nullity}(T^2)$, then by comparing the two equations, we must have $\operatorname{rank}(T) = \operatorname{rank}(T^2)$. So, the first option is correct.

We know that $\operatorname{Null}(T) \subseteq \operatorname{Null}(T^2)$. Since $\operatorname{nullity}(T) = \dim(\operatorname{Null}(T))$ and $\operatorname{nullity}(T^2) = \dim(\operatorname{Null}(T^2))$ are equal, and one subspace is contained within the other, they must be the same subspace. So, $\operatorname{Null}(T) = \operatorname{Null}(T^2)$ is correct.

Similarly, we know that $\operatorname{Range}(T^2) \subseteq \operatorname{Range}(T)$. Since $\operatorname{rank}(T) = \dim(\operatorname{Range}(T))$ and $\operatorname{rank}(T^2) = \dim(\operatorname{Range}(T^2))$ are equal, and one subspace is contained within the other, they must be the same subspace. So, $\operatorname{Range}(T) = \operatorname{Range}(T^2)$ is correct.

For the last option, let $\mathbf{v} \in \operatorname{Null}(T) \cap \operatorname{Range}(T)$.
Then $T(\mathbf{v}) = \mathbf{0}$ and $\mathbf{v} = T(\mathbf{u})$ for some $\mathbf{u} \in V$.
This implies $T(T(\mathbf{u})) = \mathbf{0}$, so $T^2(\mathbf{u}) = \mathbf{0}$.
Thus, $\mathbf{u} \in \operatorname{Null}(T^2)$.
Since $\operatorname{Null}(T) = \operatorname{Null}(T^2)$, it follows that $\mathbf{u} \in \operatorname{Null}(T)$.
Therefore, $T(\mathbf{u}) = \mathbf{0}$.
Since $\mathbf{v} = T(\mathbf{u})$, we have $\mathbf{v} = \mathbf{0}$.
Hence, $\operatorname{Null}(T) \cap \operatorname{Range}(T) = \{\mathbf{0}\}$ is correct.
Answer: \boxed{$\operatorname{rank}(T) = \operatorname{rank}(T^2)$,$\operatorname{Null}(T) = \operatorname{Null}(T^2)$,$\operatorname{Range}(T) = \operatorname{Range}(T^2)$,$\operatorname{Null}(T) \cap \operatorname{Range}(T) = \{\mathbf{0}\}$}"
:::

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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 $5 \times 7$ matrix such that $\operatorname{rank}(A) = 3$. What is the dimension of the null space of the linear transformation $T(\mathbf{x}) = A\mathbf{x}$?" options=["2","3","4","5"] answer="4" hint="Apply the Rank-Nullity Theorem directly. The domain dimension is the number of columns of A." solution="The matrix $A$ is $5 \times 7$, so the domain of the transformation $T(\mathbf{x}) = A\mathbf{x}$ is $\mathbb{R}^7$. Thus, $\dim(V) = 7$.
We are given $\operatorname{rank}(A) = 3$.
By the Rank-Nullity Theorem:

The dimension of the null space is 4. Answer: \boxed{4}"
:::

:::question type="NAT" question="A linear transformation $T: \mathbb{R}^5 \to \mathbb{R}^4$ is given by $T(x_1, x_2, x_3, x_4, x_5) = (x_1+x_2, x_2+x_3, x_3+x_4, x_4+x_5)$. What is the nullity of $T$?" answer="1" hint="Form the matrix for $T$, find its rank by row reduction, then use Rank-Nullity Theorem." solution="The matrix representation $A$ for $T$ is:

This matrix is already in row echelon form.
The number of non-zero rows (pivot positions) is 4. So, $\operatorname{rank}(A) = 4$.
The domain space is $\mathbb{R}^5$, so $n=5$.
Using the Rank-Nullity Theorem:

Answer: \boxed{1}"
:::

:::question type="MCQ" question="Let $T: V \to V$ be a linear transformation where $\dim(V) = 6$. If $\operatorname{nullity}(T) = 2$, which of the following statements is true?" options=["$\dim(\operatorname{Range}(T)) = 2$","$\operatorname{Null}(T) = V$","$\operatorname{Range}(T) = V$","$\operatorname{rank}(T) = 4$"] answer="$\operatorname{rank}(T) = 4$" hint="Apply the Rank-Nullity Theorem to find the rank. Then evaluate the options." solution="Given $\dim(V) = 6$ and $\operatorname{nullity}(T) = 2$.
By the Rank-Nullity Theorem:

Thus, $\dim(\operatorname{Range}(T)) = 4$.
Option 1 states $\dim(\operatorname{Range}(T)) = 2$, which is false.
Option 2 states $\operatorname{Null}(T) = V$. This would mean $\operatorname{nullity}(T) = \dim(V) = 6$, which is false.
Option 3 states $\operatorname{Range}(T) = V$. This would mean $\operatorname{rank}(T) = \dim(V) = 6$, which is false.
Option 4 states $\operatorname{rank}(T) = 4$, which is true. Answer: \boxed{\operatorname{rank}(T) = 4}"
:::

:::question type="MSQ" question="Let $A$ be an $n \times n$ matrix. Which of the following statements are always true?" options=["If $\operatorname{rank}(A) = n$, then $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.","If $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$, then $\operatorname{nullity}(A) = 0$.","If $\operatorname{nullity}(A) > 0$, then $A$ is invertible.","$\operatorname{rank}(A) + \operatorname{nullity}(A) = n$"] answer="If $\operatorname{rank}(A) = n$, then $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.,If $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$, then $\operatorname{nullity}(A) = 0.$,$\operatorname{rank}(A) + \operatorname{nullity}(A) = n$" hint="Recall the properties of invertible matrices and the implications of the Rank-Nullity Theorem for square matrices." solution="Let's analyze each statement:

If $\operatorname{rank}(A) = n$, then $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.

By the Rank-Nullity Theorem, $n = \operatorname{rank}(A) + \operatorname{nullity}(A)$. If $\operatorname{rank}(A) = n$, then $n = n + \operatorname{nullity}(A)$, which implies $\operatorname{nullity}(A) = 0$. A nullity of 0 means the null space contains only the zero vector, so $A\mathbf{x} = \mathbf{0}$ has only the trivial solution ($\mathbf{x} = \mathbf{0}$). This statement is correct.

If $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$, then $\operatorname{nullity}(A) = 0$.

If $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$, it means $A$ is invertible. An invertible $n \times n$ matrix has $\operatorname{rank}(A) = n$. From the Rank-Nullity Theorem, $\operatorname{nullity}(A) = n - \operatorname{rank}(A) = n - n = 0$. This statement is correct.

If $\operatorname{nullity}(A) > 0$, then $A$ is invertible.

If $\operatorname{nullity}(A) > 0$, it means $A\mathbf{x} = \mathbf{0}$ has non-trivial solutions. This implies $A$ is singular (not invertible). So, this statement is incorrect.

$\operatorname{rank}(A) + \operatorname{nullity}(A) = n$.

This is the direct statement of the Rank-Nullity Theorem for an $n \times n$ matrix (where $n$ is the dimension of the domain). This statement is correct.
Answer: \boxed{1, 2, 4}"
:::

:::question type="NAT" question="Consider the matrix $B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}$. What is the sum of its rank and nullity?" answer="3" hint="The sum of rank and nullity for an $n \times n$ matrix is always $n$. Identify $n$ for this matrix." solution="The matrix $B$ is a $3 \times 3$ matrix.
For any $n \times n$ matrix, the Rank-Nullity Theorem states that $\operatorname{rank}(B) + \operatorname{nullity}(B) = n$.
In this case, $n=3$.
Therefore, the sum of its rank and nullity is 3.

(To verify, one could calculate rank:

$\operatorname{rank}(B) = 2$.
Then $\operatorname{nullity}(B) = 3 - 2 = 1$.
$\operatorname{rank}(B) + \operatorname{nullity}(B) = 2 + 1 = 3$.)
Answer: \boxed{3}"
:::

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Summary

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What's Next?

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Part 4: Matrix Representation

Matrix representation provides a concrete means to analyze and compute with linear transformations between finite-dimensional vector spaces. This framework is fundamental for understanding concepts such as change of basis, similarity transformations, and the kernel and image of a linear map, which are central to competitive examinations in linear algebra.

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Core Concepts

1. Definition of a Linear Transformation

A mapping $F: V \to W$ between two vector spaces $V$ and $W$ over the same field $K$ is a linear transformation if, for all vectors $u, v \in V$ and all scalars $c \in K$:

$F(u + v) = F(u) + F(v)$ (Additivity)

$F(cu) = cF(u)$ (Homogeneity)

Quick Example:
Consider $F: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $F(x, y) = (x+y, x-y)$. We verify linearity.

Step 1: Additivity
Let $u = (x_1, y_1)$ and $v = (x_2, y_2)$.

The results are identical, so $F(u+v) = F(u) + F(v)$.

Step 2: Homogeneity
Let $c$ be a scalar and $u = (x, y)$.

The results are identical, so $F(cu) = cF(u)$. Thus, $F$ is a linear transformation.

:::question type="MCQ" question="Let $F: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $F(x, y) = (x^2, y)$. Is $F$ a linear transformation?" options=["Yes, it satisfies both additivity and homogeneity.","No, it fails additivity.","No, it fails homogeneity.","No, it fails both additivity and homogeneity."] answer="No, it fails homogeneity." hint="Check the homogeneity condition with a non-zero scalar." solution="We test the homogeneity condition. Let $u = (1, 1)$ and $c=2$.

Since $F(cu) = (4, 2)$ and $cF(u) = (2, 2)$, we observe $F(cu) \ne cF(u)$. Thus, $F$ is not a linear transformation because it fails homogeneity. It also fails additivity. Answer: \boxed{\text{No, it fails homogeneity.}}"
:::

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2. Matrix Representation with Respect to Standard Bases

Consider a linear transformation $F: V \to W$, where $V = \mathbb{R}^n$ and $W = \mathbb{R}^m$. Let $E_n = \{e_1, \dots, e_n\}$ be the standard basis for $\mathbb{R}^n$ and $E_m = \{e'_1, \dots, e'_m\}$ be the standard basis for $\mathbb{R}^m$. The matrix representation of $F$ with respect to these standard bases, denoted $M_{E_m, E_n}(F)$ or simply $[F]$, is an $m \times n$ matrix whose columns are the images of the standard basis vectors of $V$, expressed in terms of the standard basis vectors of $W$.

Quick Example:
Let $F: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $F(x, y) = (3x + 4y, 2x - 5y)$. We determine the matrix representation of $F$ relative to the standard basis $E = \{(1, 0), (0, 1)\}$.

Step 1: Apply $F$ to each standard basis vector of the domain.

Step 2: Form the matrix using these images as columns.

Answer: The matrix representation is

.

:::question type="MCQ" question="Let $F: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation defined by $F(x, y, z) = (x - 2y + z, 3x + y - 4z)$. What is the matrix representation of $F$ with respect to the standard bases of $\mathbb{R}^3$ and $\mathbb{R}^2$?" options=["

",
"",
"",
""] answer="" hint="The columns of the matrix are the images of the standard basis vectors of the domain." solution="Step 1: Apply $F$ to the standard basis vectors of $\mathbb{R}^3$: $e_1 = (1, 0, 0)$, $e_2 = (0, 1, 0)$, $e_3 = (0, 0, 1)$.

Step 2: Form the matrix using these image vectors as columns.

"
:::

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3. Matrix Representation with Respect to Arbitrary Bases

Let $F: V \to W$ be a linear transformation, where $\operatorname{dim}(V) = n$ and $\operatorname{dim}(W) = m$. Let $B = \{v_1, v_2, \dots, v_n\}$ be an ordered basis for $V$ and $C = \{w_1, w_2, \dots, w_m\}$ be an ordered basis for $W$. The matrix representation of $F$ with respect to bases $B$ and $C$, denoted $M_{C,B}(F)$ or $[F]_{C \leftarrow B}$, is an $m \times n$ matrix. Its $j$-th column is the coordinate vector of $F(v_j)$ with respect to the basis $C$, i.e., $[F(v_j)]_C$.

Quick Example:
Let $F: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $F(x, y) = (3x + 4y, 2x - 5y)$. Consider the domain basis $B = \{(1, 0), (0, 1)\}$ (standard basis $E$) and codomain basis $C = \{(1, 2), (2, 3)\}$. We find $M_{C,B}(F)$.

Step 1: Apply $F$ to each vector in the domain basis $B$.

Step 2: Express each resulting vector in terms of the codomain basis $C$.
For $F(1, 0) = (3, 2)$:
We need to find $a, b$ such that

Multiplying the first equation by 2: $6 = 2a + 4b$.
Subtracting the second equation from this:

Substitute $b=4$ into $3 = a + 2b \implies 3 = a + 2(4) \implies a = 3 - 8 = -5$.
So, .

For $F(0, 1) = (4, -5)$:
We need to find $c, d$ such that

Multiplying the first equation by 2: $8 = 2c + 4d$.
Subtracting the second equation from this:

Substitute $d=13$ into $4 = c + 2d \implies 4 = c + 2(13) \implies c = 4 - 26 = -22$.
So, .

Step 3: Form the matrix $M_{C,B}(F)$ using these coordinate vectors as columns.

Answer: The matrix representation is

.

:::question type="MCQ" question="Let $F: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $F(x, y) = (x+2y, 3x-y)$. Let $B = \{(1, 1), (1, -1)\}$ be a basis for the domain and $C = \{(1, 0), (0, 1)\}$ be the standard basis for the codomain. Find $M_{C,B}(F)$." options=["

",
"",
"",
""] answer="" hint="Apply $F$ to each vector in $B$, then express the result in terms of $C$ (which is the standard basis)." solution="Step 1: Apply $F$ to each vector in the domain basis $B = \{(1, 1), (1, -1)\}$.

Step 2: Express each resulting vector in terms of the codomain basis $C = \{(1, 0), (0, 1)\}$. Since $C$ is the standard basis, the coordinate vector is simply the vector itself.

Step 3: Form the matrix $M_{C,B}(F)$ using these coordinate vectors as columns.

"
:::

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4. Change of Basis Matrix

A change of basis matrix transforms coordinate vectors from one basis to another within the same vector space. Let $B = \{v_1, \dots, v_n\}$ and $B' = \{v'_1, \dots, v'_n\}$ be two ordered bases for a vector space $V$. The change of basis matrix from $B'$ to $B$, denoted $P_{B \leftarrow B'}$ or $[I]_{B,B'}$, is an $n \times n$ matrix whose $j$-th column is the coordinate vector of $v'_j$ with respect to the basis $B$.

Quick Example:
Consider two bases for $\mathbb{R}^2$: $B = \{(1, 0), (0, 1)\}$ (standard basis) and $B' = \{(1, 2), (2, 3)\}$. We find the change of basis matrix from $B'$ to $B$, i.e., $P_{B \leftarrow B'}$.

Step 1: Express each vector in $B'$ as a linear combination of vectors in $B$.
For $v'_1 = (1, 2)$:

For $v'_2 = (2, 3)$:

Step 2: Form the matrix $P_{B \leftarrow B'}$ using these coordinate vectors as columns.

Answer: The change of basis matrix is

.

:::question type="MCQ" question="Let $B = \{(1, -1), (2, 0)\}$ and $B' = \{(1, 0), (0, 1)\}$ be bases for $\mathbb{R}^2$. Find the change of basis matrix from $B$ to $B'$, i.e., $P_{B' \leftarrow B}$." options=["

",
"",
"",
""] answer="" hint="Express vectors of $B$ in terms of $B'$. Since $B'$ is the standard basis, this is straightforward." solution="Step 1: Express each vector in $B = \{(1, -1), (2, 0)\}$ in terms of the standard basis $B' = \{(1, 0), (0, 1)\}$.
For $v_1 = (1, -1)$:

For $v_2 = (2, 0)$:

Step 2: Form the matrix $P_{B' \leftarrow B}$ using these coordinate vectors as columns.

"
:::

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5. Matrix of a Linear Operator in Different Bases (Similarity)

If $F: V \to V$ is a linear operator and $B$ and $B'$ are two bases for $V$, then the matrix representation of $F$ with respect to $B'$, $M_{B',B'}(F)$, is related to the matrix representation with respect to $B$, $M_{B,B}(F)$, by a similarity transformation.

Quick Example:
Let $F: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $F(x, y) = (2x+y, x-y)$.
Let $E = \{(1,0), (0,1)\}$ be the standard basis and $B = \{(1,1), (1,2)\}$ be another basis.
The matrix of $F$ with respect to $E$ is

.
We find the matrix of $F$ with respect to $B$, i.e., $A' = M_{B,B}(F)$.

Step 1: Find the change of basis matrix $P_{E \leftarrow B}$ (from $B$ to $E$).
The columns of $P_{E \leftarrow B}$ are the vectors of $B$ expressed in terms of $E$.

Step 2: Compute $P_{B \leftarrow E} = (P_{E \leftarrow B})^{-1}$.

Step 3: Compute $A' = P_{B \leftarrow E} A P_{E \leftarrow B}$.

Step 4: Perform matrix multiplication.
First, $M_1 = A P_{E \leftarrow B}$:

Next, $A' = P_{B \leftarrow E} M_1$:

Answer: The matrix of $F$ with respect to basis $B$ is

.

:::question type="MCQ" question="Let $F: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear operator with matrix $A = M_{E,E}(F) = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ with respect to the standard basis $E = \{(1,0), (0,1)\}$. If $B = \{(1,1), (0,1)\}$ is another basis for $\mathbb{R}^2$, find $M_{B,B}(F)$." options=["

",
"",
"",
""] answer="" hint="Calculate the change of basis matrix from $B$ to $E$ and its inverse. Then apply the similarity formula $A' = P^{-1}AP$." solution="Step 1: Find the change of basis matrix $P_{E \leftarrow B}$ (from $B$ to $E$).
The columns of $P_{E \leftarrow B}$ are the vectors of $B$ expressed in terms of $E$.

Step 2: Compute $P_{B \leftarrow E} = (P_{E \leftarrow B})^{-1}$.

Step 3: Compute $M_{B,B}(F) = P_{B \leftarrow E} M_{E,E}(F) P_{E \leftarrow B}$.

Step 4: Perform matrix multiplication.
First, $M_1 = M_{E,E}(F) P_{E \leftarrow B}$:

Next, $M_{B,B}(F) = P_{B \leftarrow E} M_1$:

"
:::

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Advanced Applications

Matrix Representation for Transformations between Higher Dimensional Spaces

We extend the concept of matrix representation to transformations between vector spaces of different dimensions, with non-standard bases for both domain and codomain.

Worked Example (based on PYQ 4):
Let $F: \mathbf{R}^3 \to \mathbf{R}^2$ be the linear map defined by $F(x, y, z) = (3x + 2y - 4z, x - 5y + 3z)$. The basis of $\mathbf{R}^3$ is $S = \{(1, 1, 1), (1, 1, 0), (1, 0, 0)\}$ and basis of $\mathbf{R}^2$ is $S' = \{(1, 3), (2, 5)\}$. We find the matrix of $F$ in the bases $S$ (domain) and $S'$ (codomain), i.e., $M_{S',S}(F)$.

Step 1: Apply $F$ to each vector in the domain basis $S$.

Step 2: Express each resulting vector in terms of the codomain basis $S' = \{(1, 3), (2, 5)\}$.
For $F(1, 1, 1) = (1, -1)$:
We need $a, b$ such that

Multiply first equation by 3: $3 = 3a + 6b$.
Subtract second from this:

Substitute $b=4$ into $1 = a + 2b \implies 1 = a + 2(4) \implies a = 1 - 8 = -7$.
So, .

For $F(1, 1, 0) = (5, -4)$:
We need $c, d$ such that

Multiply first equation by 3: $15 = 3c + 6d$.
Subtract second from this:

Substitute $d=19$ into $5 = c + 2d \implies 5 = c + 2(19) \implies c = 5 - 38 = -33$.
So, .

For $F(1, 0, 0) = (3, 1)$:
We need $e, f$ such that

Multiply first equation by 3: $9 = 3e + 6f$.
Subtract second from this:

Substitute $f=8$ into $3 = e + 2f \implies 3 = e + 2(8) \implies e = 3 - 16 = -13$.
So, .

Step 3: Form the matrix $M_{S',S}(F)$ using these coordinate vectors as columns.

Answer: The matrix of $F$ in the given bases is

.

:::question type="NAT" question="Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $T(x, y, z) = (x-y, y-z)$. Let $B = \{(1,0,0), (0,1,0), (0,0,1)\}$ be the standard basis for $\mathbb{R}^3$ and $C = \{(1,1), (1,-1)\}$ be a basis for $\mathbb{R}^2$. Find the entry in the second row, second column of $M_{C,B}(T)$." answer="-1" hint="Calculate $T(e_2)$ and express it in terms of $C$. The second component of this coordinate vector is the answer." solution="Step 1: Identify the relevant basis vector and its image under $T$.
The second column of $M_{C,B}(T)$ corresponds to $T(v_2)$, where $v_2$ is the second vector in basis $B$.
Here, $B = \{e_1, e_2, e_3\} = \{(1,0,0), (0,1,0), (0,0,1)\}$, so $v_2 = e_2 = (0,1,0)$.

Step 2: Express $T(v_2)$ in terms of the codomain basis $C = \{(1,1), (1,-1)\}$.
We need to find $a, b$ such that

Adding the two equations:

Substitute $a=0$ into $-1 = a+b \implies -1 = 0+b \implies b = -1$.
So, .

Step 3: Identify the required entry.
The second row, second column entry of $M_{C,B}(T)$ is the second component of $[T(v_2)]_C$, which is $b$.
The value is $-1$."
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $F: \mathbb{P}_1 \to \mathbb{R}^2$ be defined by $F(a+bx) = (a-b, 2a)$. Let $B = \{1, x\}$ be the standard basis for $\mathbb{P}_1$ and $C = \{(1, 0), (1, 1)\}$ be a basis for $\mathbb{R}^2$. Find $M_{C,B}(F)$." options=["

",
"",
"",
""] answer="" hint="Apply $F$ to $1$ and $x$, then express the results in terms of $C = \{(1, 0), (1, 1)\}$." solution="Step 1: Apply $F$ to each vector in the domain basis $B = \{1, x\}$.
$F(1)$ (here $a=1, b=0$)

$F(x)$ (here $a=0, b=1$)

Step 2: Express each resulting vector in terms of the codomain basis $C = \{(1, 0), (1, 1)\}$.
For $F(1) = (1, 2)$:
We need $c_1, c_2$ such that

Substitute $c_2=2$ into $1 = c_1 + c_2 \implies 1 = c_1 + 2 \implies c_1 = -1$.
So, .

For $F(x) = (-1, 0)$:
We need $d_1, d_2$ such that

Substitute $d_2=0$ into $-1 = d_1 + d_2 \implies -1 = d_1 + 0 \implies d_1 = -1$.
So, .

Step 3: Form the matrix $M_{C,B}(F)$ using these coordinate vectors as columns.

"
:::

:::question type="NAT" question="Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be a linear transformation given by $T(x,y) = (x+y, x-y, 2y)$. Let $B = \{(1,2), (2,1)\}$ be a basis for $\mathbb{R}^2$ and $E_3$ be the standard basis for $\mathbb{R}^3$. Find the sum of all entries in the first column of $M_{E_3,B}(T)$." answer="6" hint="Calculate $T(v_1)$ for $v_1=(1,2)$ and sum its components." solution="Step 1: Identify the first vector in the domain basis $B$.
The first vector in $B = \{(1,2), (2,1)\}$ is $v_1 = (1,2)$.

Step 2: Apply the transformation $T$ to $v_1$.

Step 3: Express $T(v_1)$ in terms of the codomain basis $E_3 = \{(1,0,0), (0,1,0), (0,0,1)\}$.
Since $E_3$ is the standard basis, the coordinate vector $[T(v_1)]_{E_3}$ is simply $(3, -1, 4)$.
So the first column of $M_{E_3,B}(T)$ is

.

Step 4: Sum the entries in the first column.

"
:::

:::question type="MCQ" question="Let $L: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear operator such that $L(1, 0) = (2, 1)$ and $L(0, 1) = (-1, 3)$. Find the matrix of $L$ with respect to the basis $B = \{(1, 1), (-1, 1)\}$. " options=["

",
"",
"",
""] answer="" hint="First find the matrix in the standard basis, then use the similarity transformation $A' = P^{-1}AP$." solution="Step 1: Find the matrix of $L$ with respect to the standard basis $E = \{(1,0), (0,1)\}$.
Given $L(1, 0) = (2, 1)$ and $L(0, 1) = (-1, 3)$, the matrix $A = M_{E,E}(L)$ is:

Step 2: Find the change of basis matrix $P_{E \leftarrow B}$ (from $B$ to $E$).
The columns of $P_{E \leftarrow B}$ are the vectors of $B = \{(1, 1), (-1, 1)\}$ expressed in terms of $E$.

Step 3: Compute $P_{B \leftarrow E} = (P_{E \leftarrow B})^{-1}$.

Step 4: Compute $M_{B,B}(L) = P_{B \leftarrow E} A P_{E \leftarrow B}$.

Step 5: Perform matrix multiplication.
First, $M_1 = A P_{E \leftarrow B}$:

Next, $M_{B,B}(L) = P_{B \leftarrow E} M_1$:

"
:::

:::question type="MSQ" question="Let $S = \{v_1, v_2\}$ and $S' = \{u_1, u_2\}$ be two bases for $\mathbb{R}^2$, where $v_1 = (1, 2)$, $v_2 = (3, 4)$, $u_1 = (1, 1)$, $u_2 = (0, 1)$. Which of the following statements are true?" options=["The change of basis matrix $P_{S' \leftarrow S}$ is $\begin{bmatrix} -2 & -2 \\ 3 & 4 \end{bmatrix}$",
"The change of basis matrix $P_{S \leftarrow S'}$ is $\begin{bmatrix} -2 & -2 \\ 3 & 4 \end{bmatrix}$",
"The change of basis matrix $P_{S' \leftarrow S}$ is $\begin{bmatrix} 1 & 3 \\ 1 & 1 \end{bmatrix}$",
"The change of basis matrix $P_{S \leftarrow S'}$ is $\begin{bmatrix} -1 & 3 \\ 1 & -2 \end{bmatrix}$"] answer="The change of basis matrix $P_{S' \leftarrow S}$ is $\begin{bmatrix} 1 & 3 \\ 1 & 1 \end{bmatrix}$" hint="Calculate $P_{S' \leftarrow S}$ and $P_{S \leftarrow S'}$ separately by expressing basis vectors of the 'from' basis in terms of the 'to' basis." solution="Step 1: Calculate $P_{S' \leftarrow S}$.
This matrix transforms coordinates from $S$ to $S'$. Its columns are the vectors of $S$ expressed in terms of $S'$.
$v_1 = (1, 2)$. We need $a, b$ such that

So .

$v_2 = (3, 4)$. We need $c, d$ such that

So .
Therefore, .
This matches option 3.

Step 2: Calculate $P_{S \leftarrow S'}$.
This matrix transforms coordinates from $S'$ to $S$. Its columns are the vectors of $S'$ expressed in terms of $S$.
$u_1 = (1, 1)$. We need $e, f$ such that

Multiply first by 2: $2 = 2e + 6f$.
Subtract second from this:

Substitute $f=1/2$ into $1 = e + 3f \implies 1 = e + 3(1/2) \implies e = 1 - 3/2 = -1/2$.
So .

$u_2 = (0, 1)$. We need $g, h$ such that

From first, $g = -3h$. Substitute into second:

Then .
So .
Therefore, .
This does not match option 4.

Therefore, only option 3 is true."
:::

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Summary

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What's Next?

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be a linear transformation defined by $T(x,y,z,w) = (x+2y-z, y+w, x-w)$. What is the dimension of the null space of $T$?" options=["1", "2", "3", "4"] answer="1" hint="Form the matrix representation of $T$ and find its rank. Then apply the Rank-Nullity Theorem." solution="The matrix representation of $T$ is

.
Row reducing $A$:

The rank of $A$ is 3 (number of non-zero rows).
By the Rank-Nullity Theorem, .
Thus, the dimension of the null space of $T$ is 1."
:::

:::question type="NAT" question="Let $P_2$ be the vector space of polynomials of degree at most 2. Let $T: P_2 \to P_2$ be a linear transformation defined by $T(p(x)) = p'(x) + p(x)$. Find the sum of the entries in the second column of the matrix representation of $T$ with respect to the standard basis $B = \{1, x, x^2\}$." answer="2" hint="Apply the transformation to each basis vector to find its coordinates in the standard basis. The coordinates form the columns of the matrix." solution="Applying $T$ to the basis vectors:

The matrix representation $[T]_B$ is:

The entries in the second column are 1, 1, and 0. Their sum is $1 + 1 + 0 = 2$."
:::

:::question type="MCQ" question="Let $V$ and $W$ be finite-dimensional vector spaces, and let $T: V \to W$ be a linear transformation. Which of the following statements is always true?" options=["If $T$ is injective, then $\operatorname{dim}(V) = \operatorname{dim}(W)$.", "If $T$ is surjective, then $\operatorname{nullity}(T) = 0$.", "$T$ is an isomorphism if and only if $\operatorname{nullity}(T) = 0$ and $\operatorname{rank}(T) = \operatorname{dim}(W)$.", "If $\operatorname{nullity}(T) > 0$, then $T$ is surjective."] answer="$T$ is an isomorphism if and only if $\operatorname{nullity}(T) = 0$ and $\operatorname{rank}(T) = \operatorname{dim}(W)$." hint="Recall the definitions of injective, surjective, and isomorphism in terms of kernel and range, and relate them to dimensions using the Rank-Nullity Theorem." solution="

False. If $T$ is injective, then $\operatorname{dim}(V) \le \operatorname{dim}(W)$. For example, $T: \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(x,y) = (x,y,0)$ is injective, but $\operatorname{dim}(\mathbb{R}^2) \ne \operatorname{dim}(\mathbb{R}^3)$.

False. If $T$ is surjective, then $\operatorname{rank}(T) = \operatorname{dim}(W)$. By Rank-Nullity Theorem,

$$\operatorname{nullity}(T) = \operatorname{dim}(V) - \operatorname{rank}(T) = \operatorname{dim}(V) - \operatorname{dim}(W)$$

. This is only 0 if $\operatorname{dim}(V) = \operatorname{dim}(W)$. For example, $T: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $T(x,y,z) = (x,y)$ is surjective, but $\operatorname{nullity}(T) = 1$.

True. An isomorphism is both injective and surjective. $T$ is injective if and only if $\operatorname{nullity}(T) = 0$. $T$ is surjective if and only if $\operatorname{rank}(T) = \operatorname{dim}(W)$. Therefore, $T$ is an isomorphism if and only if both conditions hold.

False. If $\operatorname{nullity}(T) > 0$, $T$ is not injective. This does not imply surjectivity. For example, $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $T(x,y,z) = (x,y,0)$ has $\operatorname{nullity}(T)=1$ but is not surjective.

"
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What's Next?