Eigenvalues and Eigenvectors PYQs for GATE CS

Solve 6+ Eigenvalues and Eigenvectors previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Linear Algebra: Eigenvalues and Eigenvectors Roadmap

1. Eigenvalues, Determinants and Multiplicity

The foundation: characteristic equations, trace, determinant, and root multiplicity. Current Topic

2. Eigenvector Verification

Test vectors and compute null spaces to verify eigenvectors.

3. Spectra of Graph Adjacency Matrices

Apply eigenvalues to graph theory and symmetric matrices.

4. Eigenvalues of Matrix Powers

Predict long-term behavior and compute high matrix powers efficiently.

The Core Idea: Eigenvalues, Determinants, and Multiplicity

When a matrix multiplies a vector, it usually changes both the length and the direction of that vector.

However, for any square matrix, there are special directions that only stretch or shrink, without rotating.

Eigenvectors are the vectors that point in these special directions.
Eigenvalues are the pure scaling factors (how much they stretch or shrink).

The determinant is the overall volume scaling factor of the entire space. Multiplicity tells us how many independent directions share the exact same scaling factor.

Eigenvalues and Eigenvectors: Solved Questions with Step-by-Step Explanations (6 Problems)

Question 1 · Engineering Mathematics · 2026_Set1 MCQ

For $n > 1$ , the maximum multiplicity of any eigenvalue of an $n \times n$ matrix with elements from $R$ is

A.
$n$
B.
$n - 1$
C.
1
D.
$n + 1$

Correct Answer:

Step-by-Step Solution

Insight: The characteristic polynomial of an $n \times n$ matrix has degree exactly $n$ , so no eigenvalue can repeat more than $n$ times — and the identity matrix achieves this bound.

Exam route: The characteristic equation $det (A - λ I) = 0$ is a polynomial of degree $n$ . The maximum multiplicity of any root of a degree- $n$ polynomial is $n$ . The identity matrix $I_{n}$ has characteristic polynomial $(1 - λ)^{n} = 0$ , so eigenvalue $1$ has multiplicity $n$ . Answer: $n$ .

Learning route: This is a theoretical maximum-multiplicity question, recognisable because no specific matrix is given and the answer is in terms of $n$ .

Step 1: For any $n \times n$ matrix $A$ , the characteristic equation $det (A - λ I) = 0$ expands to a polynomial in $λ$ of degree exactly $n$ .

Step 2: By the Fundamental Theorem of Algebra, this polynomial has exactly $n$ roots counting multiplicity. The algebraic multiplicity of a single eigenvalue is the number of times it appears as a root.

Step 3: Since the total count of all roots (with multiplicity) is $n$ , no single eigenvalue can have algebraic multiplicity exceeding $n$ .

Step 4: To confirm $n$ is achievable, consider $A = I_{n}$ . Its characteristic polynomial is $(1 - λ)^{n} = 0$ , giving $λ = 1$ with algebraic multiplicity exactly $n$ .

Wrong path — Option B ( $n - 1$ ): A student confuses this with the rank-nullity theorem or thinks "at least one eigenvalue must differ." This produces $n - 1$ . It breaks at Step 4: the identity matrix is a direct counterexample where all $n$ eigenvalues are identical.

Wrong path — Option C ( $1$ ): A student assumes all eigenvalues must be distinct, or confuses algebraic multiplicity with the minimum geometric multiplicity. This produces $1$ . It breaks at Step 3: nothing prevents all $n$ roots from coinciding.

Wrong path — Option D ( $n + 1$ ): A student does not realise the characteristic polynomial has degree exactly $n$ and thinks multiplicity can exceed the matrix size. This produces $n + 1$ . It breaks at Step 1: a degree- $n$ polynomial cannot have a root of multiplicity $n + 1$ .

Generalization: The sum of algebraic multiplicities of all eigenvalues of an $n \times n$ matrix always equals $n$ , so the maximum any single eigenvalue can claim is the entire sum.

Verification: For $n = 3$ , $I_{3}$ has characteristic polynomial $(1 - λ)^{3} = 0$ , giving eigenvalue $1$ with multiplicity $3 = n$ . Confirmed.

Question 2 · Engineering Mathematics · 2025_Set1 MCQ

Let

A

be a

2 \times 2

matrix as given.

A = [11 1 - 1]

What are the eigenvalues of the matrix

A^{13}

A.
$1, - 1$
B.
$22, - 22$
C.
$42, - 42$
D.
$642, - 642$

Correct Answer:

Step-by-Step Solution

Insight: Eigenvalues of $A^{k}$ are $λ^{k}$ where $λ$ are eigenvalues of $A$ — find eigenvalues of $A$ first, then raise to the 13th power.

Exam route: Compute $det (A - λ I) = (1 - λ) (- 1 - λ) - 1 = λ^{2} - 2 = 0$ , giving $λ = \pm 2$ . Then eigenvalues of $A^{13}$ are $(2)^{13} = 642$ and $(- 2)^{13} = - 642$ . Answer: $642, - 642$ .

Learning route: This is a matrix-power eigenvalue question, recognisable because it asks for eigenvalues of $A^{13}$ given a specific $2 \times 2$ matrix.

Step 1: Find eigenvalues of $A = [11 1 - 1]$ .

The characteristic equation is:

$det (A - λ I) = 1 - λ 1 1 - 1 - λ = (1 - λ) (- 1 - λ) - (1) (1) = 0$

Expanding: $- (1 - λ^{2}) - 1 = λ^{2} - 1 - 1 = λ^{2} - 2 = 0$ .

So $λ_{1} = 2$ and $λ_{2} = - 2$ .

Step 2: Apply the power rule. If $λ$ is an eigenvalue of $A$ , then $λ^{k}$ is an eigenvalue of $A^{k}$ .

For $k = 13$ :

$(2)^{13} = 2^{13/2} = 2^{6} \cdot 2^{1/2} = 642$

$(- 2)^{13} = - (2)^{13} = - 642$

(The negative sign is preserved because $13$ is odd.)

Step 3: The eigenvalues of $A^{13}$ are $642$ and $- 642$ .

Wrong path — Option A ( $1, - 1$ ): A student assumes $A^{2} = I$ (so $A^{13} = A$ ), giving eigenvalues $\pm 1$ . This breaks at Step 1: $A^{2} = [2002] = 2 I \neq = I$ . The eigenvalues of $A$ are $\pm 2$ , not $\pm 1$ .

Wrong path — Option B ( $22, - 22$ ): A student computes $(2)^{3} = 22$ instead of $(2)^{13}$ . This produces $22$ . The error is using $k = 3$ instead of $k = 13$ .

Wrong path — Option C ( $42, - 42$ ): A student computes $(2)^{5} = 42$ instead of $(2)^{13}$ . This produces $42$ . The error is using $k = 5$ instead of $k = 13$ .

Generalization: For any matrix $A$ with eigenvalues $λ_{i}$ , the eigenvalues of $A^{k}$ are $λ_{i}^{k}$ . Always check the parity of $k$ when eigenvalues are negative.

Verification: $det (A) = (1) (- 1) - (1) (1) = - 2$ . So $det (A^{13}) = (- 2)^{13} = - 8192$ . Product of our eigenvalues: $(642) (- 642) = - 4096 \times 2 = - 8192$ . Matches.

Question 3 · Engineering Mathematics · 2024_Set1 MCQ

The product of all eigenvalues of the matrix $147258369$ is

A.
$- 1$
B.
$0$
C.
$1$
D.
$2$

Correct Answer:

Step-by-Step Solution

Insight: Product of all eigenvalues equals the determinant. The columns of this matrix are linearly dependent, so the determinant is $0$ .

Exam route: Observe that Column 3 $= 2 \times$ Column 2 $-$ Column 1 (check: $2 (2) - 1 = 3$ , $2 (5) - 4 = 6$ , $2 (8) - 7 = 9$ ). Linearly dependent columns means $det = 0$ . Product of eigenvalues $= 0$ . Answer: $0$ .

Learning route: This is a product-of-eigenvalues question, recognisable because it asks for "the product of all eigenvalues" of a given matrix.

Step 1: Recall the fundamental property: for any $n \times n$ matrix $A$ , the product of all eigenvalues equals $det (A)$ .

Step 2: Compute $det (A)$ for $A = 147258369$ .

Method A (column dependency): Notice that $Col_{1} - 2 Col_{2} + Col_{3} = 1 - 4 + 3 4 - 10 + 6 7 - 16 + 9 = 000$ . The columns are linearly dependent, so $det (A) = 0$ .

Method B (direct expansion):

$det (A) = 1 (5 \cdot 9 - 6 \cdot 8) - 2 (4 \cdot 9 - 6 \cdot 7) + 3 (4 \cdot 8 - 5 \cdot 7)$

$= 1 (45 - 48) - 2 (36 - 42) + 3 (32 - 35)$

$= 1 (- 3) - 2 (- 6) + 3 (- 3) = - 3 + 12 - 9 = 0$

Step 3: Product of eigenvalues $= det (A) = 0$ .

Wrong path — Option A ( $- 1$ ): A student makes an arithmetic error in the determinant expansion, perhaps computing $1 (- 3) - 2 (- 6) + 3 (- 3) = - 3 - 12 - 9 = - 24$ or some other miscalculation that eventually yields $- 1$ . The break point is in the sign handling of the cofactor expansion.

Wrong path — Option C ( $1$ ): A student guesses or assumes the product of eigenvalues of a matrix with consecutive integer entries is $1$ . This is a comprehension error — no such rule exists.

Wrong path — Option D ( $2$ ): A student makes an arithmetic error in the determinant, perhaps computing $1 (45 - 48) - 2 (36 - 42) + 3 (32 - 35) = - 3 + 12 + 9 = 18$ or similar, eventually arriving at $2$ . The break point is arithmetic in the cofactor terms.

Generalization: Whenever a question asks for the product of eigenvalues, compute the determinant instead. Look for linear dependencies among rows or columns first — they give $det = 0$ instantly.

Verification: Since $det (A) = 0$ , at least one eigenvalue is $0$ . Indeed, the vector $1 - 2 1$ is in the null space: $A 1 - 2 1 = 1 - 4 + 3 4 - 10 + 6 7 - 16 + 9 = 000$ , confirming $λ = 0$ is an eigenvalue, so the product is $0$ .

Question 4 · Engineering Mathematics · 2023 NAT

Let

A

be the adjacency matrix of the graph with vertices

{1, 2, 3, 4, 5}

.

Let

λ_{1}, λ_{2}, λ_{3}, λ_{4},

and

λ_{5}

be the five eigenvalues of

A

. Note that these eigenvalues need not be distinct.
The value of

λ_{1} + λ_{2} + λ_{3} + λ_{4} + λ_{5} =

__________.

Correct Answer:

0.00

Step-by-Step Solution

Insight: This is an adjacency matrix invariant question, recognizable because it asks for the sum of eigenvalues of a graph's adjacency matrix.

Exam route: The sum of eigenvalues equals the trace. A simple graph has no self-loops, so its adjacency matrix has $0$ s on the diagonal. Thus, the trace is $0$ , and the sum of eigenvalues is $0$ .

Learning route:

Step 1: Recall the fundamental property of square matrices: the sum of all eigenvalues $\sum_{i = 1}^{n} λ_{i}$ equals the trace of the matrix, $Tr (A)$ .

Step 2: The trace is the sum of main diagonal elements: $Tr (A) = \sum_{i = 1}^{n} A_{ii}$ .

Step 3: Analyze the given matrix $A$ . It is the adjacency matrix of the provided simple graph.

Step 4: The diagonal entry $A_{ii}$ represents the number of self-loops at vertex $i$ . The graph has no self-loops.

Step 5: Therefore, all diagonal entries of $A$ are exactly $0$ .

Step 6: Calculate the trace: $Tr (A) = 0 + 0 + 0 + 0 + 0 = 0$ .

Step 7: Conclude that the sum of the eigenvalues is $0$ .

Tempting wrong path: A student might confuse the sum of eigenvalues (trace) with the sum of the degrees of the vertices (which equals the sum of all entries in the adjacency matrix, or twice the number of edges). Counting the edges in the diagram (there are 5 edges), they might calculate $2 \times 5 = 10$ and answer $10$ . This breaks at Step 1, as the sum of eigenvalues is strictly the sum of the diagonal entries, not all entries.

Verification: The adjacency matrix has $0$ s on the diagonal, so the trace is unambiguously $0$ . The sum of eigenvalues must match the trace.

Question 5 · Engineering Mathematics · 2022 MSQ

Which of the following is/are the eigenvector(s) for the matrix given below?

$- 9 - 8 2032 - 6 - 6 1521 - 2 - 3 87 - 4 - 1 512$

A.
$- 1 101$
B.
$10 - 1 0$
C.
$- 1 022$
D.
$01 - 3 0$

Correct Answer:

["A","C","D"]

Step-by-Step Solution

Insight: For MSQ eigenvector questions with candidate vectors given, never solve the characteristic equation — just multiply each candidate by the matrix and check proportionality.

Exam route: Compute $M v$ for each option. If $M v = c v$ for some scalar $c$ , it is an eigenvector. Options A, C, D pass; B fails.

Learning route: This is an eigenvector verification MSQ, recognisable because a large matrix is given with candidate vectors and the question asks "which of the following is/are eigenvector(s)."

Let $M = - 9 - 8 2032 - 6 - 6 1521 - 2 - 3 87 - 4 - 1 512$ .

Step 1: Test Option A, $v_{A} = - 1 101$ .

$M v_{A} = (- 9) (- 1) + (- 6) (1) + (- 2) (0) + (- 4) (1) (- 8) (- 1) + (- 6) (1) + (- 3) (0) + (- 1) (1) (20) (- 1) + (15) (1) + (8) (0) + (5) (1) (32) (- 1) + (21) (1) + (7) (0) + (12) (1) = 9 - 6 + 0 - 4 8 - 6 + 0 - 1 - 20 + 15 + 0 + 5 - 32 + 21 + 0 + 12 = - 1 101$

This equals $1 \cdot v_{A}$ . So $v_{A}$ is an eigenvector with $λ = 1$ . Select A.

Step 2: Test Option B, $v_{B} = 10 - 1 0$ .

$M v_{B} = (- 9) (1) + (- 6) (0) + (- 2) (- 1) + (- 4) (0) (- 8) (1) + (- 6) (0) + (- 3) (- 1) + (- 1) (0) (20) (1) + (15) (0) + (8) (- 1) + (5) (0) (32) (1) + (21) (0) + (7) (- 1) + (12) (0) = - 9 + 0 + 2 + 0 - 8 + 0 + 3 + 0 20 + 0 - 8 + 0 32 + 0 - 7 + 0 = - 7 - 5 1225$

Is this proportional to $10 - 1 0$ ? The second component is $- 5 \neq = 0$ , so no scalar $c$ can satisfy $c \cdot 0 = - 5$ . Not an eigenvector. Reject B.

Step 3: Test Option C, $v_{C} = - 1 022$ .

$M v_{C} = 9 + 0 - 4 - 8 8 + 0 - 6 - 2 - 20 + 0 + 16 + 10 - 32 + 0 + 14 + 24 = - 3 066 = 3 - 1 022$

This equals $3 \cdot v_{C}$ . Eigenvector with $λ = 3$ . Select C.

Step 4: Test Option D, $v_{D} = 01 - 3 0$ .

$M v_{D} = 0 - 6 + 6 + 0 0 - 6 + 9 + 0 0 + 15 - 24 + 0 0 + 21 - 21 + 0 = 03 - 9 0 = 3 01 - 3 0$

This equals $3 \cdot v_{D}$ . Eigenvector with $λ = 3$ . Select D.

Wrong path — Selecting B: A student makes an arithmetic error in computing $M v_{B}$ (e.g., getting $0$ in the second row) and incorrectly concludes proportionality. The break point is Row 2: $(- 8) (1) + (- 3) (- 1) = - 8 + 3 = - 5 \neq = 0$ .

Generalization: When candidate vectors are provided, the multiply-and-check method is always faster than solving the characteristic equation. Each check is just a matrix-vector product.

Verification: Options A, C, D each satisfy $M v = c v$ with $c = 1, 3, 3$ respectively. Option B fails the proportionality test.

Question 6 · Engineering Mathematics · 2021_Set1 NAT

Consider the following matrix.

0111101111011110

The largest eigenvalue of the above matrix is __________.

Correct Answer:

3.00

Step-by-Step Solution

Insight: This is a shifted all-ones matrix question, recognizable because all off-diagonal entries are identical (1) and all diagonal entries are identical (0).

Exam route: The matrix is $M = J - I$ . The eigenvalues of the $4 \times 4$ all-ones matrix $J$ are $4, 0, 0, 0$ . Shifting by $- 1$ gives eigenvalues $3, - 1, - 1, - 1$ . The largest is $3$ .

Learning route:

Step 1: Express the matrix algebraically as $M = J - I$ , where $J$ is the $4 \times 4$ matrix of all ones, and $I$ is the $4 \times 4$ identity matrix.

Step 2: Recall the spectral properties of $J$ . Its rank is $1$ , so $λ = 0$ has algebraic multiplicity $3$ . The trace of $J$ is $4$ , so the remaining eigenvalue is $4$ . The spectrum is ${4, 0, 0, 0}$ .

Step 3: Apply the eigenvalue shift theorem. If $λ$ is an eigenvalue of $J$ , then $λ - 1$ is an eigenvalue of $J - I$ .

Step 4: Shift each eigenvalue: $4 - 1 = 3$ and $0 - 1 = - 1$ (multiplicity 3).

Step 5: The largest eigenvalue is $3$ .

Tempting wrong path: A student might look at the diagonal of $M$ (which is all $0$ s) and incorrectly assume the eigenvalues are just the diagonal entries ( $0$ ). Or they might forget the $- I$ shift and answer $4$ . This breaks at Step 2, where the base matrix $J$ must be evaluated and shifted.

Verification: The sum of eigenvalues of $M$ is $3 + (- 1) + (- 1) + (- 1) = 0$ . The trace of $M$ is $0 + 0 + 0 + 0 = 0$ . The sums match, verifying the spectrum.

Try a question

Login to view the complete previous-year questions and solutions

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

Eigenvalues and Eigenvectors PYQs for GATE CS

Linear Algebra: Eigenvalues and Eigenvectors Roadmap

Linear Algebra: Eigenvalues and Eigenvectors Roadmap

The Core Idea: Eigenvalues, Determinants, and Multiplicity

The Core Idea: Eigenvalues, Determinants, and Multiplicity

Eigenvalues and Eigenvectors: Solved Questions with Step-by-Step Explanations (6 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit