GATE DA

Eigenvalues, eigenvectors and determinants

The determinant of a square matrix is one number that says whether the matrix is invertible and how it scales volume. An eigenvector is a non-zero vector that the matrix maps to a scalar multiple of itself, and that scalar is the eigenvalue. The two ideas are tied together: the trace is the sum and the determinant is the product of the eigenvalues. These notes cover determinant rules, the characteristic equation, eigenvalues of related and special matrices, diagonalisability and the Cayley-Hamilton theorem.

6 previous-year questions · 6 solved · 5 practice questions

Conventions

Matrices are real and square of size $n \times n$ unless stated. Vectors are column vectors in bold, $x$ ; $A^{T}$ is the transpose and $I$ is the $n \times n$ identity. Eigenvalues are written $λ$ , are counted with algebraic multiplicity unless stated, and may be complex even for a real matrix. $∥ x ∥$ is the Euclidean norm and $tr (A)$ is the trace, the sum of the diagonal entries. The characteristic polynomial is taken as $p (λ) = det (A - λ I)$ . The other common convention, $det (λ I - A)$ , differs from it by the factor $(- 1)^{n}$ and has the same roots.

Determinants and their properties

The determinant $det (A)$ is defined only for a square matrix. It is zero exactly when $A$ is singular. Each of the following is equivalent to that: the rows (or columns) of $A$ are linearly dependent, $A x = 0$ has a non-zero solution, and $A$ has rank less than $n$ .

DETERMINANT OF A 2 BY 2 MATRIX

det [a c b d] = a d - b c

Always valid for a 2 by 2 matrix.

COFACTOR EXPANSION ALONG ROW I

det (A) = j = 1 \sum n a_{ij} (- 1)^{i + j} M_{ij}

A is n by n and i is any fixed row. M_ij is the determinant of the (n-1) by (n-1) matrix left after deleting row i and column j. The same works along any column. Expand along the row or column with the most zeros.

Swapping two rows (or two columns) multiplies the determinant by $- 1$ .
Multiplying one row by $k$ multiplies the determinant by $k$ .
Adding a multiple of one row to another row leaves the determinant unchanged.
Two equal rows, one row a multiple of another, or a zero row give determinant $0$ .
The determinant of a triangular (upper or lower) or diagonal matrix is the product of its diagonal entries.
If a matrix is block upper or lower triangular with square diagonal blocks, its determinant is the product of the determinants of those blocks.

PRODUCT, TRANSPOSE AND INVERSE

det (A B) = det (A) det (B), det (A^{T}) = det (A), det (A^{- 1}) = \frac{1}{det ( A )}

A and B are n by n. The inverse rule needs det(A) not equal to 0. Applying the product rule repeatedly gives det(A^k) = (det A)^k for every integer k >= 1.

SCALAR MULTIPLE

det (c A) = c^{n} det (A)

A is n by n. The factor c comes out of each of the n rows, so the power is n, the size of the matrix.

ADJOINT (ADJUGATE)

A adj (A) = adj (A) A = det (A) I, det (adj (A)) = (det A)^{n - 1}

A is n by n with n >= 2, and adj(A) is the transpose of the cofactor matrix. If det(A) is not 0, then A^{-1} = adj(A)/det(A). Also adj(adj A) = (det A)^{n-2} A for n >= 3, and adj(adj A) = A for n = 2.

VANDERMONDE DETERMINANT

det 111 x_{1} x_{2} x_{3} x_{1}^{2} x_{2}^{2} x_{3}^{2} = (x_{2} - x_{1}) (x_{3} - x_{1}) (x_{3} - x_{2})

In general, the n by n matrix whose i-th row is 1, x_i, x_i^2, ..., x_i^{n-1} has determinant equal to the product of (x_j - x_i) over all i < j. It is non-zero exactly when the x_i are distinct.

EXAM TIP

In general $det (A + B) \neq = det (A) + det (B)$ . The determinant is multiplicative, not additive. Also $det (- A) = (- 1)^{n} det (A)$ , so for a $3 \times 3$ matrix the sign flips but for a $4 \times 4$ matrix it does not.

Eigenvalues, eigenvectors and the characteristic polynomial

A number $λ$ is an eigenvalue of $A$ if there is a non-zero vector $x$ with $A x = λ x$ ; this $x$ is an eigenvector for $λ$ . Any non-zero multiple of an eigenvector is again an eigenvector, so eigenvectors are never unique. When the eigenspace of $λ$ is one-dimensional, its eigenvector is fixed up to a non-zero scalar multiple. The zero vector is never an eigenvector, but $λ = 0$ is allowed as an eigenvalue. A real matrix can have non-real eigenvalues, and these occur in complex-conjugate pairs.

CHARACTERISTIC EQUATION

det (A - λ I) = 0

A is n by n. The n roots of this equation, counted with multiplicity over the complex numbers, are the eigenvalues. For each root, solve (A - lambda I)x = 0 to find the eigenvectors.

CHARACTERISTIC EQUATION OF A 2 BY 2 MATRIX

λ^{2} - tr (A) λ + det (A) = 0

Valid for any 2 by 2 matrix; it is det(A - lambda I) = 0 written out.

CHARACTERISTIC EQUATION OF A 3 BY 3 MATRIX

λ^{3} - tr (A) λ^{2} + c_{2} λ - det (A) = 0

c_2 is the sum of the three principal 2 by 2 minors: for each i, delete row i and column i, take the determinant, and add the three results. This is det(A - lambda I) = 0 multiplied through by -1.

TRACE AND DETERMINANT FROM THE EIGENVALUES

i = 1 \sum n λ_{i} = tr (A), i = 1 \prod n λ_{i} = det (A)

A is n by n, and the eigenvalues are counted with algebraic multiplicity. Hence A is singular if and only if 0 is an eigenvalue. Applying the trace rule to A^k gives tr(A^k) = lambda_1^k + ... + lambda_n^k for every integer k >= 1.

Form $A - λ I$ and expand $det (A - λ I)$ to get the characteristic polynomial.
Solve the polynomial for the eigenvalues, and check them against the trace and the determinant.
For each eigenvalue $λ$ , row-reduce $A - λ I$ and solve $(A - λ I) x = 0$ .
The non-zero solutions are the eigenvectors for $λ$ . Together with $0$ they form the eigenspace, whose dimension is the geometric multiplicity.

For a triangular or diagonal matrix the eigenvalues are the diagonal entries. Row operations generally change the eigenvalues, so never row-reduce $A$ first and then read eigenvalues off the reduced matrix. $A$ and $A^{T}$ have the same characteristic polynomial and so the same eigenvalues, although their eigenvectors can differ.

Eigenvalues of related matrices

Let $A$ be $n \times n$ with $A x = λ x$ and $x \neq = 0$ . The table gives the matching eigenvalue and eigenvector of each related matrix. Where the eigenvector column says $x$ , the same vector works.

Matrix	Eigenvalue	Eigenvector	Condition
$k A$	$k λ$	$x$	any scalar $k$
$A + k I$	$λ + k$	$x$	any scalar $k$
$A^{m}$	$λ^{m}$	$x$	integer $m \geq 1$ (any integer $m$ if $A$ is invertible)
$A^{- 1}$	$1/ λ$	$x$	$A$ invertible (so $λ \neq = 0$ )
$p (A)$ , a polynomial in $A$	$p (λ)$	$x$	$p$ any polynomial
$adj (A)$	$det (A) / λ$	$x$	$A$ invertible
$A^{T}$	$λ$ (same set of eigenvalues)	may differ from $x$	none
$P^{- 1} A P$	$λ$ (same set of eigenvalues)	$P^{- 1} x$	$P$ an invertible $n \times n$ matrix (similar matrices)

Type of real matrix	What is known about its eigenvalues
Triangular or diagonal	the diagonal entries
Block triangular with square diagonal blocks	the eigenvalues of the diagonal blocks, taken together
Symmetric ( $A^{T} = A$ )	all real; eigenvectors for distinct eigenvalues are orthogonal
Symmetric positive definite ( $A^{T} = A$ and $x^{T} A x > 0$ for every $x \neq = 0$ )	all positive (for positive semidefinite, all $\geq 0$ )
Skew-symmetric ( $A^{T} = - A$ )	zero or purely imaginary; for odd $n$ , $0$ is an eigenvalue and $det A = 0$
Orthogonal ( $A^{T} A = I$ )	magnitude $1$ (so any real ones are $\pm 1$ ); $det A = \pm 1$
Idempotent ( $A^{2} = A$ )	each eigenvalue is $0$ or $1$
Nilpotent ( $A^{k} = 0$ for some $k$ )	all equal to $0$
Singular	at least one eigenvalue is $0$
Rank-one $u v^{T}$ ( $u, v$ non-zero $n \times 1$ vectors)	$v^{T} u$ once and $0$ repeated $n - 1$ times (algebraic multiplicity)

EXAM TIP

Let $J$ be the $n \times n$ matrix of all ones. It is the rank-one matrix $1 1^{T}$ with $1^{T} 1 = n$ , so its eigenvalues are $n$ (once) and $0$ ( $n - 1$ times). Hence $a I + b J$ has eigenvalues $a + nb$ (once, eigenvector $1$ ) and $a$ (repeated $n - 1$ times).

AB AND BA

det (A B - λ I) = det (B A - λ I) for A, B both n \times n

So AB and BA have the same eigenvalues with the same multiplicities, although their eigenvectors can differ. If A is m by n and B is n by m, AB and BA have the same non-zero eigenvalues with the same multiplicities, and the larger product has extra zero eigenvalues.

EXAM TIP

In general the eigenvalues of $A + B$ and $A B$ are not sums or products of eigenvalues of $A$ and $B$ . The exception is a common eigenvector: if $A x = λ x$ and $B x = μ x$ , then $x$ is an eigenvector of $A + B$ with eigenvalue $λ + μ$ and of $A B$ with eigenvalue $λ μ$ . What always holds is $tr (A + B) = tr A + tr B$ and $det (A B) = det A det B$ .

Multiplicity and diagonalisation

Let $A$ be $n \times n$ and let $λ$ be one of its eigenvalues. The algebraic multiplicity of $λ$ is the number of times it is a root of the characteristic polynomial. Its geometric multiplicity is $dim ker (A - λ I) = n - rank (A - λ I)$ . Always $1 \leq$ geometric $\leq$ algebraic.

DIAGONALISATION

A = P D P^{- 1}, D = diag (λ_{1}, \dots, λ_{n}), A^{k} = P D^{k} P^{- 1}

A, P and D are n by n and P is invertible. The columns of P are n linearly independent eigenvectors, in the same order as the eigenvalues on the diagonal of D. This factorisation exists if and only if A has n linearly independent eigenvectors.

Eigenvectors belonging to distinct eigenvalues are linearly independent.
If $A$ has $n$ distinct eigenvalues, it is diagonalisable (sufficient, not necessary).
$A$ is diagonalisable (over the complex numbers) if and only if, for every eigenvalue, the geometric multiplicity equals the algebraic multiplicity. For a diagonalisation with real $P$ and $D$ , all eigenvalues must also be real.
A repeated eigenvalue does not by itself stop diagonalisation. The $n \times n$ identity $I$ has the eigenvalue $1$ repeated $n$ times and is already diagonal. By contrast, $[1011]$ has the repeated eigenvalue $1$ but only one independent eigenvector, so it is not diagonalisable.
The eigenvalue $0$ has geometric multiplicity $n - rank (A)$ . Hence $rank (A)$ is at least the number of non-zero eigenvalues counted with algebraic multiplicity, with equality when $A$ is diagonalisable (for example, when $A$ is real symmetric).

SPECTRAL THEOREM FOR REAL SYMMETRIC MATRICES

A = Q Λ Q^{T}, Q^{T} Q = I

A is an n by n real symmetric matrix, Q is an n by n orthogonal matrix whose columns are orthonormal eigenvectors, and Lambda is the diagonal matrix of the matching eigenvalues. A real symmetric matrix has real eigenvalues, its eigenvectors for distinct eigenvalues are orthogonal, and it is always diagonalisable by an orthogonal matrix.

RAYLEIGH QUOTIENT BOUNDS

λ_{m i n} \leq \frac{x ^{T} A x}{x ^{T} x} \leq λ_{m a x} (x \neq = 0)

A is an n by n real symmetric matrix with smallest eigenvalue lambda_min and largest eigenvalue lambda_max. Each bound is attained at an eigenvector for that eigenvalue, so the maximum of x^T A x over unit vectors (||x|| = 1) is lambda_max and the minimum is lambda_min.

WORKED EXAMPLE

Let $A = [4213]$ . Find its eigenvalues and eigenvectors, and compute $det (A^{2} - 3 A + I)$ .

Solution

$tr (A) = 7$ and $det (A) = 12 - 2 = 10$ , so $λ^{2} - 7 λ + 10 = 0$ , giving $λ = 5$ and $λ = 2$ .
For $λ = 5$ : $A - 5 I = [- 1 2 1 - 2]$ , so $- x_{1} + x_{2} = 0$ and an eigenvector is $(1, 1)^{T}$ .
For $λ = 2$ : $A - 2 I = [2211]$ , so $2 x_{1} + x_{2} = 0$ and an eigenvector is $(1, - 2)^{T}$ .
The matrix $A^{2} - 3 A + I$ has eigenvalues $λ^{2} - 3 λ + 1$ : $11$ for $λ = 5$ and $- 1$ for $λ = 2$ .
The determinant is the product of these eigenvalues. Check by direct computation: $A^{2} - 3 A + I = [7843]$ , whose determinant is $21 - 32 = - 11$ .

Answer: Eigenvalues 5 and 2 with eigenvectors $(1, 1)^{T}$ and $(1, - 2)^{T}$ (or any non-zero multiples); $det (A^{2} - 3 A + I) = 11 \times (- 1) = - 11$ .

WORKED EXAMPLE

A real $3 \times 3$ matrix $A$ has eigenvalues $2$ , $- 1$ and $x$ , and $tr (A) = 5$ . Find $det (A)$ , $det (adj A)$ and the eigenvalues of $adj A$ .

Solution

The trace gives $2 - 1 + x = 5$ , so $x = 4$ .
$det (A) = 2 \times (- 1) \times 4 = - 8$ . This is non-zero, so $A$ is invertible and $adj A = det (A) A^{- 1}$ .
$det (adj A) = (det A)^{n - 1} = (- 8)^{2} = 64$ .
The eigenvalues of $adj A$ are $det (A) / λ$ : $- 8/2 = - 4$ , $- 8/ (- 1) = 8$ and $- 8/4 = - 2$ .
Check: $(- 4) (8) (- 2) = 64$ .

Answer: $det A = - 8$ , $det (adj A) = 64$ , and the eigenvalues of $adj A$ are $- 4, 8, - 2$ .

Cayley-Hamilton theorem

CAYLEY-HAMILTON THEOREM

p (λ) = det (A - λ I) ⟹ p (A) = 0

A is n by n, and the constant term c of p is replaced by cI. Every square matrix satisfies its own characteristic equation. For a 2 by 2 matrix: A^2 - tr(A) A + det(A) I = 0.

The theorem has two common uses. First, any power of $A$ can be reduced to a polynomial in $A$ of degree at most $n - 1$ , by repeatedly replacing the highest power using the relation. Second, if $det (A) \neq = 0$ , multiplying the relation by $A^{- 1}$ expresses $A^{- 1}$ as a polynomial in $A$ . For a $2 \times 2$ matrix, $A^{- 1} = \frac{1}{det A} (tr (A) I - A)$ .

EXAM TIP

Two quick checks on a computed characteristic polynomial. With $p (λ) = det (A - λ I)$ , the constant term is $p (0) = det A$ , the leading coefficient is $(- 1)^{n}$ , and the coefficient of $λ^{n - 1}$ is $(- 1)^{n - 1} tr (A)$ . With the monic convention $det (λ I - A)$ , the coefficient of $λ^{n - 1}$ is $- tr (A)$ and the constant term is $(- 1)^{n} det A$ . In either convention, a non-zero constant term means the matrix is invertible.

PREVIOUS-YEAR QUESTIONS

6 questions from the DA papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE DA 2026 · GATE DA 2025 · GATE DA 2024

GATE 2026 DA Q.46

GATE 2026 · DAQ.46 · 2 marks · MCQ

Let $γ_{1}, γ_{2}, γ_{3}$ be the eigenvalues of the matrix $100 0 cos t - sin t 0 sin t cos t$ , where $t \in [- π, π]$ is in radians.

Which one of the following options lists all the possible values of $t$ satisfying $γ_{1} + γ_{2} + γ_{3} = 1 + 2$ ?

(A)
${\frac{π}{3}, - \frac{π}{4}}$
(B)
${\frac{π}{4}, - \frac{π}{3}}$
(C)
${\frac{π}{4}, - \frac{π}{4}}$
(D)
${\frac{π}{3}, - \frac{π}{3}}$

Answer and solution

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GATE 2025 DA Q.13

GATE 2025 · DAQ.13 · 1 mark · MCQ

The sum of the elements in each row of $A \in R^{n \times n}$ is 1. If $B = A^{3} - 2 A^{2} + A$ , which one of the following statements is correct (for $x \in R^{n}$ )?

(A)
The equation $B x = 0$ has no solution
(B)
The equation $B x = 0$ has exactly two solutions
(C)
The equation $B x = 0$ has infinitely many solutions
(D)
The equation $B x = 0$ has a unique solution

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GATE 2025 DA Q.28

GATE 2025 · DAQ.28 · 1 mark · MSQ

Let $A = I_{n} + x x^{⊤}$ , where $I_{n}$ is the $n \times n$ identity matrix and $x \in R^{n}$ , $x^{⊤} x = 1$ . Which of the following options is/are correct?

(A)
Rank of $A$ is $n$
(B)
$A$ is invertible
(C)
$0$ is an eigenvalue of $A$
(D)
$A^{- 1}$ has a negative eigenvalue

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GATE 2025 DA Q.37

GATE 2025 · DAQ.37 · 2 marks · MCQ

Let $A \in R^{n \times n}$ be such that $A^{3} = A$ . Which one of the following statements is ALWAYS correct?

(A)
$A$ is invertible
(B)
Determinant of $A$ is $0$
(C)
The sum of the diagonal elements of $A$ is $1$
(D)
$A$ and $A^{2}$ have the same rank

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GATE 2024 DA Q.13

GATE 2024 · DAQ.13 · 1 mark · MCQ

Consider the matrix $M = [23 - 1 1]$ .

Which ONE of the following statements is TRUE?

(A)
The eigenvalues of $M$ are non-negative and real.
(B)
The eigenvalues of $M$ are complex conjugate pairs.
(C)
One eigenvalue of $M$ is positive and real, and another eigenvalue of $M$ is zero.
(D)
One eigenvalue of $M$ is non-negative and real, and another eigenvalue of $M$ is negative and real.

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GATE 2024 DA Q.35

GATE 2024 · DAQ.35 · 1 mark · NAT

Consider the $3 \times 3$ matrix $M = 134213336$ . The determinant of $(M^{2} + 12 M)$ is ______.

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PRACTICE

5 questions to test yourself

PRACTICE 11 mark · NAT

A real $3 \times 3$ matrix $A$ has eigenvalues $1$ , $2$ and $3$ . Find the value of $det (A^{2} + A)$ .

Answer and solution

Answer: $144$

The matrix $A^{2} + A$ has eigenvalues $λ^{2} + λ = λ (λ + 1)$ .
For $λ = 1, 2, 3$ these are $2, 6, 12$ .
The determinant is the product of the eigenvalues: $2 \times 6 \times 12 = 144$ .

PRACTICE 21 mark · MCQ

A real $3 \times 3$ matrix $A$ has characteristic equation $(λ - 2)^{2} (λ - 5) = 0$ and $rank (A - 2 I) = 2$ . Which statement is correct?

(A)
$A$ is diagonalisable because its eigenvalues are real
(B)
$A$ is not diagonalisable
(C)
$A$ is singular
(D)
$det (A) = 10$

Answer and solution

Answer: (B)

The eigenvalue $2$ has algebraic multiplicity $2$ .
Its geometric multiplicity is $3 - rank (A - 2 I) = 3 - 2 = 1$ , which is less than $2$ .
So $A$ does not have three linearly independent eigenvectors and is not diagonalisable. B is true and A is false.
The eigenvalues are $2, 2, 5$ and none is zero, so $A$ is non-singular (C false) and $det A = 2 \times 2 \times 5 = 20$ , not $10$ (D false).

PRACTICE 32 marks · MSQ

Let $A = 211121112$ . Which of the following statements are correct?

(A)
$A$ is diagonalisable
(B)
$det (A) = 4$
(C)
An eigenvector for the eigenvalue $4$ is orthogonal to every eigenvector for the eigenvalue $1$
(D)
$A^{- 1}$ has an eigenvalue $4$

Answer and solution

Answer: (A), (B), (C)

Every row sums to $4$ , so $(1, 1, 1)^{T}$ is an eigenvector with eigenvalue $4$ .
$A - I$ is the $3 \times 3$ all-ones matrix, which has rank $1$ . So the eigenvalue $1$ has geometric multiplicity $3 - 1 = 2$ , and therefore algebraic multiplicity at least $2$ .
With the eigenvalue $4$ as well, the eigenvalues are $4, 1, 1$ , which agrees with the trace $2 + 2 + 2 = 6$ .
$A$ is real symmetric, so it is diagonalisable (A true), and eigenvectors for distinct eigenvalues are orthogonal (C true). Directly: the eigenvectors for $1$ satisfy $x_{1} + x_{2} + x_{3} = 0$ , which says they are orthogonal to $(1, 1, 1)^{T}$ .
$det A = 4 \times 1 \times 1 = 4$ (B true).
The eigenvalues of $A^{- 1}$ are $1/4, 1, 1$ , so $4$ is not one of them (D false).

PRACTICE 42 marks · NAT

Let $A = [1322]$ . By the Cayley-Hamilton theorem, $A^{3} = p A + q I$ for real numbers $p$ and $q$ . Find $p + q$ .

Answer and solution

Answer: $25$

$tr (A) = 3$ and $det (A) = 2 - 6 = - 4$ , so $A^{2} - 3 A - 4 I = 0$ , that is $A^{2} = 3 A + 4 I$ .
$A^{3} = A \cdot A^{2} = 3 A^{2} + 4 A = 3 (3 A + 4 I) + 4 A = 13 A + 12 I$ .
$A$ is not a multiple of $I$ , so $A$ and $I$ are linearly independent and $p$ , $q$ are unique: $p = 13$ , $q = 12$ , and $p + q = 25$ .

PRACTICE 52 marks · MCQ

Let $B = 111123149$ . What is $det (2 B^{- 1})$ ?

(A)
1
(B)
2
(C)
4
(D)
8

Answer and solution

Answer: (C)

Expanding along the first row: $det B = 1 (18 - 12) - 1 (9 - 4) + 1 (3 - 2) = 6 - 5 + 1 = 2$ . As a check, $B$ is a Vandermonde matrix with $x = 1, 2, 3$ , so $det B = (2 - 1) (3 - 1) (3 - 2) = 2$ .
$B^{- 1}$ is $3 \times 3$ , so $det (2 B^{- 1}) = 2^{3} det (B^{- 1})$ .
$det (B^{- 1}) = 1/2$ , so the result is $8 \times \frac{1}{2} = 4$ .

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GATE DA

eigenvalues and eigenvectors, determinant

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