Rank, Invertibility and Linear Systems Notes for GATE DA: Concepts, Formulas, Worked Examples & Practice

Rank, Invertibility and Linear Systems notes for GATE DA: 17 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 (2 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.

More notes in this unit

chapter

Rank, Invertibility and Linear Systems Notes for GATE DA: Concepts, Formulas, Worked Examples & Practice

Rank, Invertibility and Linear Systems notes for GATE DA: 17 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions

A question from this chapter

Question 1

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:

Question 2

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

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Rank, Invertibility and Linear Systems Notes for GATE DA: Concepts, Formulas, Worked Examples & Practice

Chapter Roadmap: Rank, Invertibility and Linear Systems

Chapter Roadmap

Rank, Nullity and the Solution Space

Rank, Nullity and the Solution Space

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

Matrix Polynomials: When p(A)=0

Key Consequence for Eigenvalues

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

Solving A3=A: Step-by-Step

Rank, Invertibility and Linear Systems: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

More notes in this unit

A question from this chapter

Login to view the complete notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

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

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