Rank, Invertibility and Linear Systems Notes for GATE DA

Rank, Invertibility and Linear Systems notes for GATE DA: 13 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Rank, Invertibility and Linear Systems

Chapter Roadmap

Rank, Invertibility and Linear Systems

Rank, Nullity and Matrix Polynomials
Solution spaces, Rank-Nullity Theorem, and matrix powers.

Invertibility and Rank-One Updates
Outer products

x x^{T}

, invertibility shifts, and quick eigenvalue tricks.

Goal: Move from calculating determinants to understanding the structure of linear transformations.

Rank, Nullity and the Solution Space

For any $m \times n$ matrix $A$ , the fundamental accounting rule of linear algebra is:

Rank (A) + Nullity (A) = n

Rank: Dimension of the column space (independent columns).

Nullity: Dimension of the null space (free variables in

A x = 0

$n$ : Total number of columns (variables).

Exam Insight: If

Rank (A) < n

, the system

A x = 0

has non-trivial (infinite) solutions.

Matrix Polynomials: When $p(A) = 0$

Matrix Polynomials: When $p (A) = 0$

A matrix $A$ satisfies a polynomial equation if:

c_{k} A^{k} + c_{k - 1} A^{k - 1} + \dots + c_{1} A + c_{0} I = 0

Key Consequence for Eigenvalues

If $λ$ is an eigenvalue of $A$ , then $λ$ must be a root of the scalar polynomial:

c_{k} λ^{k} + c_{k - 1} λ^{k - 1} + \dots + c_{1} λ + c_{0} = 0

Example: If

A^{3} - A = 0

, then

λ^{3} - λ = 0

.
Roots:

λ (λ - 1) (λ + 1) = 0 ⟹ λ \in {0, 1, - 1}

The matrix may not have all these eigenvalues, but it cannot have any other eigenvalues.

Solving $A^3 = A$: Step-by-Step

Solving $A^{3} = A$ : Step-by-Step

Given: $A^{3} = A$ for $A \in R^{n \times n}$ .

Step 1: Form the polynomial

A^{3} - A = 0 ⟹ A (A - I) (A + I) = 0

Step 2: Identify Possible Eigenvalues
Roots of

λ^{3} - λ = 0

are

λ \in {0, 1, - 1}

Step 3: Check Diagonalizability
Since

x^{3} - x

has distinct roots, the minimal polynomial has no repeated factors. Thus, $A$ is diagonalizable.

Step 4: Analyze Trace and Determinant

Tr (A) = \sum λ_{i}

and

det (A) = \prod λ_{i}

Conclusion:

A

is diagonalizable, but not necessarily invertible (0 is a possible eigenvalue).

Rank, Invertibility and Linear Systems: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Linear Algebra MCQ

For an $m \times n$ matrix $A$ , if the rank of $A$ is strictly less than $n$ , then the homogeneous system of linear equations $A x = 0$ has:

A.
No solution
B.
Exactly one solution
C.
Exactly $n$ solutions
D.
Infinitely many solutions

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the fundamental implication of the Rank-Nullity Theorem on linear systems.

Step 1: For an $m \times n$ matrix $A$ , the Rank-Nullity Theorem states that $Rank (A) + Nullity (A) = n$ .

Step 2: We are given that $Rank (A) < n$ .

Step 3: This implies that $Nullity (A) = n - Rank (A) > 0$ .

Step 4: A positive nullity means the null space contains non-zero vectors, so the homogeneous system $A x = 0$ has non-trivial (infinitely many) solutions.

Answer: Infinitely many solutions

Question 2 · Linear Algebra MCQ

Let $A$ be a square matrix satisfying $A^{2} - 4 I = 0$ . Which of the following CANNOT be an eigenvalue of $A$ ?

A.
2
B.
-2
C.
0
D.
Both 2 and -2

Correct Answer:

Step-by-Step Solution

Key idea: This is a matrix polynomial question. Recognisable because it gives a polynomial equation that $A$ satisfies and asks about the possible eigenvalues.

Step 1: Recall the property of matrix polynomials. If a matrix $A$ satisfies $p (A) = 0$ , then every eigenvalue $λ$ of $A$ must be a root of the scalar polynomial $p (x) = 0$ .

Step 2: Identify the polynomial from the given equation.

The equation is $A^{2} - 4 I = 0$ . The corresponding scalar polynomial is:

$p (x) = x^{2} - 4$

Step 3: Find the roots of the polynomial.

Set $x^{2} - 4 = 0$ , which gives $x^{2} = 4$ .

The roots are $x = 2$ and $x = - 2$ .

Step 4: Determine the possible eigenvalues.

The only possible eigenvalues for $A$ are 2 and -2. Any other number cannot be an eigenvalue.

Step 5: Evaluate the options.

2 is a possible eigenvalue.
-2 is a possible eigenvalue.
0 is NOT a root of $x^{2} - 4 = 0$ , so it CANNOT be an eigenvalue.

Answer: Option C is correct.

Question 3 · Linear Algebra MCQ

If a square matrix $A$ satisfies the equation $A^{2} - 5 A + 6 I = 0$ , which of the following is a possible eigenvalue of $A$ ?

A.
$0$
B.
$5$
C.
$6$
D.
$2$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct application of the matrix polynomial eigenvalue property. Recognisable because it gives a polynomial equation that $A$ satisfies and asks for a possible eigenvalue.

Step 1: Recall the property that if a matrix $A$ satisfies $p (A) = 0$ , then every eigenvalue $λ$ of $A$ must be a root of the scalar polynomial $p (x) = 0$ .

Step 2: Identify the polynomial from the given equation.

The equation is $A^{2} - 5 A + 6 I = 0$ . The corresponding scalar polynomial is:

$p (x) = x^{2} - 5 x + 6$

Step 3: Find the roots of the polynomial by setting it to zero.

$x^{2} - 5 x + 6 = 0$

Factor the quadratic:

$(x - 2) (x - 3) = 0$

The roots are $x = 2$ and $x = 3$ .

Step 4: Determine the possible eigenvalues.

The only possible eigenvalues for $A$ are 2 and 3.

Step 5: Evaluate the options.

0 is not a root.
5 is not a root.
6 is not a root.
2 is a root, so it is a possible eigenvalue.

Answer: Option D is correct.

Question 4 · Linear Algebra MCQ

Let $A$ be a square matrix satisfying $A^{2} + 2 A = 0$ . If $λ$ is an eigenvalue of $A$ , which of the following could be a value of $λ$ ?

A.
1
B.
2
C.
-2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: This is a matrix polynomial question. When a matrix satisfies a polynomial equation $p (A) = 0$ , its eigenvalues must satisfy the same scalar polynomial equation $p (λ) = 0$ .

Step 1: The given matrix equation is $A^{2} + 2 A = 0$ .

Step 2: Replace $A$ with the scalar eigenvalue $λ$ to form the characteristic scalar equation: $λ^{2} + 2 λ = 0$ .

Step 3: Factor the equation: $λ (λ + 2) = 0$ .

Step 4: Solve for $λ$ : the possible eigenvalues are $λ = 0$ and $λ = - 2$ .

Step 5: Compare with the given options. The value $- 2$ is among the possible eigenvalues.

Answer: -2

Question 5 · Linear Algebra MCQ

Let $A$ be a square matrix satisfying the equation $A^{3} = A$ . If $λ$ is an eigenvalue of $A$ , which of the following scalar equations must $λ$ satisfy?

A.
$λ^{3} = 1$
B.
$λ^{2} = 1$
C.
$λ^{3} = λ$
D.
$λ^{3} = 0$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct pattern recognition question for matrix polynomials. Recognisable because it gives a simple polynomial equation $A^{3} = A$ and asks for the corresponding scalar equation for the eigenvalues.

Step 1: Recall the fundamental property of matrix polynomials. If a matrix $A$ satisfies a polynomial equation $p (A) = 0$ , then every eigenvalue $λ$ of $A$ must satisfy the exact same scalar polynomial equation $p (λ) = 0$ .

Step 2: Identify the polynomial equation from the given matrix equation. The equation is $A^{3} = A$ , which can be rewritten as $A^{3} - A = 0$ .

Step 3: Apply the property to the eigenvalues. Replace the matrix $A$ with the scalar eigenvalue $λ$ . The equation becomes $λ^{3} - λ = 0$ , or simply $λ^{3} = λ$ .

Step 4: Check the options. The equation $λ^{3} = λ$ matches Option C.

Answer: Option C is correct.

Question 6 · Linear Algebra NAT

Let $A$ be a $4 \times 7$ matrix. If the null space of $A$ has dimension 3, what is the rank of $A$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct application of the Rank-Nullity Theorem. Recognisable because it gives the matrix dimensions and the nullity, and asks for the rank.

Step 1: Recall the Rank-Nullity Theorem for an $m \times n$ matrix:

$Rank (A) + Nullity (A) = n$

where $n$ is the number of columns.

Step 2: Identify the given values. The matrix is $4 \times 7$ , so the number of columns is $n = 7$ . The nullity (dimension of the null space) is given as 3.

Step 3: Substitute the values into the theorem:

$Rank (A) + 3 = 7$

Step 4: Solve for the rank:

$Rank (A) = 7 - 3 = 4$

Answer: The rank of $A$ is 4.

Question 7 · Linear Algebra NAT

Let $A$ be a $5 \times 5$ idempotent matrix (i.e., $A^{2} = A$ ). If the trace of $A$ is 3, what is the rank of $A$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a property question about idempotent matrices. Recognisable because it gives the condition $A^{2} = A$ and relates the trace to the rank.

Step 1: Recall the eigenvalue property of idempotent matrices. If $A^{2} = A$ , then for any eigenvalue $λ$ , we have $λ^{2} = λ$ .

Step 2: Solve the scalar equation $λ^{2} - λ = 0$ , which factors to $λ (λ - 1) = 0$ . The only possible eigenvalues for an idempotent matrix are 0 and 1.

Step 3: Recall the relationship between trace, rank, and eigenvalues for idempotent matrices.

The trace of a matrix is the sum of its eigenvalues.
Since the eigenvalues are only 0s and 1s, the sum of the eigenvalues is exactly equal to the number of 1s.
For an idempotent matrix, it is diagonalizable, and its rank is exactly equal to the number of non-zero eigenvalues (the number of 1s).

Step 4: Conclude the fundamental property: For any idempotent matrix, $Trace (A) = Rank (A)$ .

Step 5: Apply the given value. We are given $Trace (A) = 3$ . Therefore, $Rank (A) = 3$ .

Answer: The rank of $A$ is 3.

Question 8 · Linear Algebra MCQ

Let $A$ be a $6 \times 6$ matrix. If the rank of $A$ is $4$ , how many free variables are there in the general solution to the homogeneous system $A x = 0$ ?

A.
$2$
B.
$4$
C.
$6$
D.
$0$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct application of the Rank-Nullity Theorem. Recognisable because it gives the matrix dimensions and the rank, and asks for the number of free variables (which is the nullity).

Step 1: Recall that the number of free variables in the solution to $A x = 0$ is exactly the nullity of $A$ .

Step 2: Recall the Rank-Nullity Theorem for an $m \times n$ matrix:

$Rank (A) + Nullity (A) = n$

where $n$ is the number of columns.

Step 3: Identify the values from the question. The matrix is $6 \times 6$ , so $n = 6$ . The rank is given as $4$ .

Step 4: Substitute the values into the theorem:

$4 + Nullity (A) = 6$

Step 5: Solve for the nullity:

$Nullity (A) = 6 - 4 = 2$

Answer: There are 2 free variables.

Question 9 · Linear Algebra MCQ

If a square matrix $A$ satisfies the equation $A^{2} - 3 A + 2 I = 0$ , then $A$ is guaranteed to be:

A.
Nilpotent
B.
Orthogonal
C.
Diagonalizable
D.
Skew-symmetric

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct recall question testing the diagonalizability condition for matrix polynomials.

Step 1: The matrix satisfies $p (A) = A^{2} - 3 A + 2 I = 0$ .

Step 2: The corresponding scalar polynomial is $p (x) = x^{2} - 3 x + 2$ .

Step 3: Find the roots of $p (x)$ : $(x - 1) (x - 2) = 0 ⟹ x = 1, 2$ .

Step 4: Since the roots are distinct, the matrix $A$ is guaranteed to be diagonalizable.

Answer: Diagonalizable

Question 10 · Linear Algebra MCQ

If a matrix $A$ satisfies the equation $A^{3} = A$ , it is guaranteed to be diagonalizable. This is because the roots of the corresponding polynomial equation $x^{3} - x = 0$ are:

A.
All zero
B.
Repeated
C.
Distinct
D.
Complex

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the structural consequence of the minimal polynomial having distinct roots.

Step 1: The matrix satisfies $A^{3} = A$ , which can be rewritten as $A^{3} - A = 0$ .

Step 2: The corresponding scalar polynomial is $p (x) = x^{3} - x$ .

Step 3: Factor the polynomial: $x (x^{2} - 1) = x (x - 1) (x + 1) = 0$ .

Step 4: The roots are $0, 1, - 1$ . Since these roots are all distinct, the matrix $A$ is guaranteed to be diagonalizable.

Answer: Distinct