Eigenvalues and Eigenvectors Questions and Answers

Q: The product of all eigenvalues of the matrix \(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\) is

The product of the eigenvalues is 0.The product of the eigenvalues of a square matrix equals its determinant. Here the rows are in arithmetic progression: row 2 − row 1 = (3, 3, 3) = row 3 − row 2, so row 1 − 2(row 2) + row 3 = 0 and the rows are linearly dependent:\[ \begin{aligned} \det &= 1(45 - 48) - 2(36 - 42) + 3(32 - 35) \\ &= -3 + 12 - 9 = 0 \end{aligned} \]So one eigenvalue is 0 and the product is 0. The values −1, 1 and 2 would require a non-singular matrix.

Q: Let \(A\) be a \(2 \times 2\) matrix as given.\(A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\)What are the eigenvalues of the matrix \(A^{13}\)?

The eigenvalues of \(A^{13}\) are \(64\sqrt{2}, -64\sqrt{2}\).Squaring A gives a multiple of the identity:\[ \begin{aligned} A^{2} &= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2I \\ A^{13} &= (A^{2})^{6}A = 64A \end{aligned} \]The eigenvalues of A solve \(\lambda^{2} - 2 = 0\) (trace 0, determinant -2), so \(\lambda = \pm\sqrt{2}\). Those of \(64A\) are \(\pm 64\sqrt{2}\); equivalently \((\sqrt{2})^{13} = 2^{6}\sqrt{2} = 64\sqrt{2}\).\(\pm 1\) ignores the \(\sqrt{2}\) scale of A's eigenvalues, and \(\pm 2\sqrt{2}\) and \(\pm 4\sqrt{2}\) are the eigenvalues of \(A^{3}\) and \(A^{5}\), not \(A^{13}\).

Q: For \(n \gt 1\), the maximum multiplicity of any eigenvalue of an n × n matrix with elements from \(\mathbb{R}\) is

The maximum multiplicity is n.The characteristic polynomial of an n × n matrix has degree n, so no eigenvalue can be a root more than n times, which rules out n + 1.The bound n is reached: the identity matrix, or any \(cI\), has the single eigenvalue c with multiplicity n. So the maximum is n, not n - 1 or 1.

Q: Consider a matrix P whose only eigenvectors are the multiples of \(\begin{bmatrix} 1 \\ 4 \end{bmatrix}\).Consider the following statements.(I) P does not have an inverse(II) P has a repeated eigenvalue(III) P cannot be diagonalizedWhich one of the following options is correct?

Only II and III are necessarily true.The eigenvectors live in two dimensions, so \(P\) is a 2 by 2 matrix. If it had two different eigenvalues, each would bring its own eigenvector, and the two would be independent. Since every eigenvector is a multiple of \((1, 4)\), there is only one eigenvalue, repeated (II).A 2 by 2 matrix is diagonalizable only when it has two independent eigenvectors. Here there is only one direction, so \(P\) cannot be diagonalized (III).Statement I need not hold. For example \(P = \begin{bmatrix} 5 & -1 \\ 16 & -3 \end{bmatrix}\) has the single eigenvalue 1, its eigenvectors are the multiples of \((1, 4)\), and its determinant is \(-15 + 16 = 1\), so it is invertible. That rules out the options that include I and the option with II alone.

Q: Suppose that the characteristic equation of \(M \in \mathbb{C}^{3 \times 3}\) is\[ \lambda^{3} + \alpha\lambda^{2} + \beta\lambda - 1 = 0, \]where \(\alpha, \beta \in \mathbb{C}\) with \(\alpha + \beta \neq 0\).Which of the following statements is TRUE?

The true statement is \(M^{-1}(M^{-1} - \beta I) = M + \alpha I\).By the Cayley-Hamilton theorem M satisfies its own characteristic equation. The constant term is -1, so det M is non-zero and M is invertible:\[ M^{3} + \alpha M^{2} + \beta M - I = 0 \]Multiply on the left by \(M^{-2}\) and rearrange:\[ \begin{aligned} M + \alpha I + \beta M^{-1} - M^{-2} &= 0 \\ M^{-2} - \beta M^{-1} &= M + \alpha I \\ M^{-1}(M^{-1} - \beta I) &= M + \alpha I \end{aligned} \]Clearing the inverses in the other options gives cubic relations with the signs of the coefficients changed, for example \(M^{3} - \alpha M^{2} - \beta M - I = 0\) from the third option. Such a relation is not implied by the characteristic equation, so those options do not hold in general.

Q: Let \(A\) be a \(2 \times 2\) real symmetric matrix such that the trace of \(A\) is \(6\) and the determinant of \(A\) is \(5\). Which one of the following statements is true?

The true statement is \(A^{2} - 6A\) is negative definite.A real symmetric matrix is described by its eigenvalues. Here \(\lambda_{1} + \lambda_{2} = 6\) and \(\lambda_{1}\lambda_{2} = 5\), so the eigenvalues are the roots of \(\lambda^{2} - 6\lambda + 5 = 0\), namely \(1\) and \(5\). A polynomial \(p(A)\) has eigenvalues \(p(1)\) and \(p(5)\):\(A - I_{2}\): \(0\) and \(4\). Positive semi-definite only, because of the zero eigenvalue.\(A - 3I_{2}\): \(-2\) and \(2\). Indefinite.\(A^{2} + A - 3I_{2}\): \(-1\) and \(27\). Indefinite.\(A^{2} - 6A\): \(1 - 6 = -5\) and \(25 - 30 = -5\). Both negative, so negative definite.In fact \(A^{2} - 6A + 5I_{2} = 0\) by the Cayley-Hamilton theorem, so \(A^{2} - 6A = -5I_{2}\).

Q: Consider a \(2 \times 2\) matrix \(A\). The determinant of \(A\) is \(-1\) and the trace\((A) = 1\). Which of the following options gives the eigenvalues of \(A\)?

The eigenvalues are \(\frac{1+\sqrt{5}}{2}\) and \(\frac{1-\sqrt{5}}{2}\).For a \(2 \times 2\) matrix the characteristic equation is \(\lambda^2 - (\text{tr}\,A)\lambda + \det A = 0\):\[ \begin{aligned} \lambda^{2} - \lambda - 1 &= 0 \\ \lambda &= \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \end{aligned} \]Check: the two roots add to 1 (the trace) and multiply to \(\frac{1 - 5}{4} = -1\) (the determinant). The pair 1, 0 has product 0, not \(-1\). Dividing by 4 instead of 2 gives roots adding to \(\frac{1}{2}\). \(\frac{\sqrt{5}}{2}\) and \(1 - \frac{\sqrt{5}}{2}\) add to 1 but multiply to \(\frac{\sqrt{5}}{2} - \frac{5}{4}\), not \(-1\).

Q: Let T be a linear operator defined on a vector space V; \(T : V \mapsto V\). If dim ker \(T \gt 0\), then :

The conclusion is 0 is an eigenvalue of T.\(\dim \ker T \gt 0\) means there is a non-zero vector v with \(Tv = 0 = 0 \cdot v\). That is exactly the definition of 0 being an eigenvalue, with v as an eigenvector.The other statements fail. A map with a non-zero kernel is not one-to-one, so T is not invertible. It need not be nilpotent: the projection \(T(x, y) = (x, 0)\) has kernel the y-axis but \(T^{k} = T \ne 0\) for every k. And Im T = V is false for a finite-dimensional V, because rank-nullity gives \(\dim \text{Im}\, T = \dim V - \dim \ker T \lt \dim V\).

The product of all eigenvalues of the matrix \(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\) is

A.\(-1\)
B.0
C.1
D.2

Exam Year(s)

GATE CS - 2024,

Answer: B

The product of the eigenvalues is 0.

The product of the eigenvalues of a square matrix equals its determinant. Here the rows are in arithmetic progression: row 2 − row 1 = (3, 3, 3) = row 3 − row 2, so row 1 − 2(row 2) + row 3 = 0 and the rows are linearly dependent:

\[ \begin{aligned} \det &= 1(45 - 48) - 2(36 - 42) + 3(32 - 35) \\ &= -3 + 12 - 9 = 0 \end{aligned} \]

So one eigenvalue is 0 and the product is 0. The values −1, 1 and 2 would require a non-singular matrix.

Discuss in Forum

Let \(A\) be a \(2 \times 2\) matrix as given.

\(A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\)

What are the eigenvalues of the matrix \(A^{13}\)?

A.\(1, -1\)
B.\(2\sqrt{2}, -2\sqrt{2}\)
C.\(4\sqrt{2}, -4\sqrt{2}\)
D.\(64\sqrt{2}, -64\sqrt{2}\)

Exam Year(s)

GATE CS - 2025,

Answer: D

The eigenvalues of \(A^{13}\) are \(64\sqrt{2}, -64\sqrt{2}\).

Squaring A gives a multiple of the identity:

\[ \begin{aligned} A^{2} &= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2I \\ A^{13} &= (A^{2})^{6}A = 64A \end{aligned} \]

The eigenvalues of A solve \(\lambda^{2} - 2 = 0\) (trace 0, determinant -2), so \(\lambda = \pm\sqrt{2}\). Those of \(64A\) are \(\pm 64\sqrt{2}\); equivalently \((\sqrt{2})^{13} = 2^{6}\sqrt{2} = 64\sqrt{2}\).

\(\pm 1\) ignores the \(\sqrt{2}\) scale of A's eigenvalues, and \(\pm 2\sqrt{2}\) and \(\pm 4\sqrt{2}\) are the eigenvalues of \(A^{3}\) and \(A^{5}\), not \(A^{13}\).

Discuss in Forum

For \(n \gt 1\), the maximum multiplicity of any eigenvalue of an n × n matrix with elements from \(\mathbb{R}\) is

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

Exam Year(s)

GATE CS - 2026,

Answer: A

The maximum multiplicity is n.

The characteristic polynomial of an n × n matrix has degree n, so no eigenvalue can be a root more than n times, which rules out n + 1.

The bound n is reached: the identity matrix, or any \(cI\), has the single eigenvalue c with multiplicity n. So the maximum is n, not n - 1 or 1.

Discuss in Forum

Consider a matrix P whose only eigenvectors are the multiples of \(\begin{bmatrix} 1 \\ 4 \end{bmatrix}\).

Consider the following statements.

(I) P does not have an inverse
(II) P has a repeated eigenvalue
(III) P cannot be diagonalized

Which one of the following options is correct?

A.Only I and III are necessarily true
B.Only II is necessarily true
C.Only I and II are necessarily true
D.Only II and III are necessarily true

Exam Year(s)

GATE CS - 2018,

Answer: D

Only II and III are necessarily true.

The eigenvectors live in two dimensions, so \(P\) is a 2 by 2 matrix. If it had two different eigenvalues, each would bring its own eigenvector, and the two would be independent. Since every eigenvector is a multiple of \((1, 4)\), there is only one eigenvalue, repeated (II).

A 2 by 2 matrix is diagonalizable only when it has two independent eigenvectors. Here there is only one direction, so \(P\) cannot be diagonalized (III).

Statement I need not hold. For example \(P = \begin{bmatrix} 5 & -1 \\ 16 & -3 \end{bmatrix}\) has the single eigenvalue 1, its eigenvectors are the multiples of \((1, 4)\), and its determinant is \(-15 + 16 = 1\), so it is invertible. That rules out the options that include I and the option with II alone.

Discuss in Forum

Let \(M\) be an invertible Hermitian matrix and let \(x, y \in \mathbb{R}\) be such that \(x^{2} \lt 4y\). Then

A.both \(M^{2} + xM + yI\) and \(M^{2} - xM + yI\) are singular
B.\(M^{2} + xM + yI\) is singular but \(M^{2} - xM + yI\) is non-singular
C.\(M^{2} + xM + yI\) is non-singular but \(M^{2} - xM + yI\) is singular
D.both \(M^{2} + xM + yI\) and \(M^{2} - xM + yI\) are non-singular

Exam Year(s)

GATE Mathematics - 2015,

Answer: D

Both \(M^{2} + xM + yI\) and \(M^{2} - xM + yI\) are non-singular.

A Hermitian matrix is diagonalisable with real eigenvalues \(\lambda\). The eigenvalues of \(M^{2} \pm xM + yI\) are \(\lambda^{2} \pm x\lambda + y\).

The quadratic \(t^{2} \pm xt + y\) has discriminant \(x^{2} - 4y \lt 0\), so it has no real roots and is never zero at a real \(\lambda\). Hence no eigenvalue of either matrix is 0, and both are non-singular.

Invertibility of \(M\) is not even needed; the argument uses only the real eigenvalues and the sign of the discriminant.

Discuss in Forum

Suppose that the characteristic equation of \(M \in \mathbb{C}^{3 \times 3}\) is

\[ \lambda^{3} + \alpha\lambda^{2} + \beta\lambda - 1 = 0, \]

where \(\alpha, \beta \in \mathbb{C}\) with \(\alpha + \beta \neq 0\).

Which of the following statements is TRUE?

A.\(M(I - \beta M) = M^{-1}(M + \alpha I)\)
B.\(M(I + \beta M) = M^{-1}(M - \alpha I)\)
C.\(M^{-1}(M^{-1} + \beta I) = M - \alpha I\)
D.\(M^{-1}(M^{-1} - \beta I) = M + \alpha I\)

Exam Year(s)

GATE Mathematics - 2022,

Answer: D

The true statement is \(M^{-1}(M^{-1} - \beta I) = M + \alpha I\).

By the Cayley-Hamilton theorem M satisfies its own characteristic equation. The constant term is -1, so det M is non-zero and M is invertible:

\[ M^{3} + \alpha M^{2} + \beta M - I = 0 \]

Multiply on the left by \(M^{-2}\) and rearrange:

\[ \begin{aligned} M + \alpha I + \beta M^{-1} - M^{-2} &= 0 \\ M^{-2} - \beta M^{-1} &= M + \alpha I \\ M^{-1}(M^{-1} - \beta I) &= M + \alpha I \end{aligned} \]

Clearing the inverses in the other options gives cubic relations with the signs of the coefficients changed, for example \(M^{3} - \alpha M^{2} - \beta M - I = 0\) from the third option. Such a relation is not implied by the characteristic equation, so those options do not hold in general.

Discuss in Forum

Let \(P\) be a nonzero complex \(n \times n\) matrix such that \(v^{*}Pv \geq 0\) for all column vectors \(v\) in \(\mathbb{C}^{n}\).

Which ONE of the following statements is necessarily TRUE?

A.All eigenvalues of \(P\) are negative real numbers.
B.If \(v_1, v_2\) are column vectors in \(\mathbb{C}^{n}\) such that \(Pv_1 = v_2\) and \(Pv_2 = v_1\), then \(v_1 = v_2\).
C.\(v^{*}Pv \neq 0\) for all nonzero column vectors \(v\) in \(\mathbb{C}^{n}\).
D.All eigenvalues of \(P^{*}\) are negative real numbers.

Exam Year(s)

IIT JAM - 2026,

Answer: B

The necessarily true statement is: if \(Pv_1 = v_2\) and \(Pv_2 = v_1\), then \(v_1 = v_2\).

Over \(\mathbb{C}\), a matrix with \(v^{*}Pv\) real for every \(v\) is Hermitian, so here \(P\) is Hermitian positive semidefinite: it has an orthonormal eigenbasis \(u_i\) with eigenvalues \(\lambda_i \geq 0\).

From the two equations, \(P^{2}v_1 = Pv_2 = v_1\). Writing \(v_1 = \sum c_i u_i\) gives \(\lambda_i^{2} = 1\) whenever \(c_i \neq 0\), and since \(\lambda_i \geq 0\), \(\lambda_i = 1\). So \(Pv_1 = v_1\), that is \(v_2 = v_1\). Without semidefiniteness this fails: the swap matrix \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) exchanges \(e_1\) and \(e_2\), but it has eigenvalue \(-1\).

The eigenvalues are non-negative, and at least one is positive since \(P \neq 0\), so neither \(P\) nor \(P^{*} = P\) has all eigenvalues negative. Semidefinite is not definite: \(P = \text{diag}(1, 0)\) gives \(v^{*}Pv = 0\) for \(v = e_2\), so \(v^{*}Pv \neq 0\) is not guaranteed.

Discuss in Forum

Let \(A\) be a \(2 \times 2\) real symmetric matrix such that the trace of \(A\) is \(6\) and the determinant of \(A\) is \(5\). Which one of the following statements is true?

A.\(A - I_{2}\) is positive definite
B.\(A - 3I_{2}\) is negative definite
C.\(A^{2} + A - 3I_{2}\) is positive definite
D.\(A^{2} - 6A\) is negative definite

Exam Year(s)

IIT JAM - 2026,

Answer: D

The true statement is \(A^{2} - 6A\) is negative definite.

A real symmetric matrix is described by its eigenvalues. Here \(\lambda_{1} + \lambda_{2} = 6\) and \(\lambda_{1}\lambda_{2} = 5\), so the eigenvalues are the roots of \(\lambda^{2} - 6\lambda + 5 = 0\), namely \(1\) and \(5\). A polynomial \(p(A)\) has eigenvalues \(p(1)\) and \(p(5)\):

\(A - I_{2}\): \(0\) and \(4\). Positive semi-definite only, because of the zero eigenvalue.
\(A - 3I_{2}\): \(-2\) and \(2\). Indefinite.
\(A^{2} + A - 3I_{2}\): \(-1\) and \(27\). Indefinite.
\(A^{2} - 6A\): \(1 - 6 = -5\) and \(25 - 30 = -5\). Both negative, so negative definite.

In fact \(A^{2} - 6A + 5I_{2} = 0\) by the Cayley-Hamilton theorem, so \(A^{2} - 6A = -5I_{2}\).

Discuss in Forum

Consider a \(2 \times 2\) matrix \(A\). The determinant of \(A\) is \(-1\) and the trace\((A) = 1\). Which of the following options gives the eigenvalues of \(A\)?

A.1, 0
B.\(\frac{1+\sqrt{5}}{2},\ \frac{1-\sqrt{5}}{2}\)
C.\(\frac{1+\sqrt{5}}{4},\ \frac{1-\sqrt{5}}{4}\)
D.\(\frac{\sqrt{5}}{2},\ 1 - \frac{\sqrt{5}}{2}\)

Exam Year(s)

IIT JAM - 2026,

Answer: B

The eigenvalues are \(\frac{1+\sqrt{5}}{2}\) and \(\frac{1-\sqrt{5}}{2}\).

For a \(2 \times 2\) matrix the characteristic equation is \(\lambda^2 - (\text{tr}\,A)\lambda + \det A = 0\):

\[ \begin{aligned} \lambda^{2} - \lambda - 1 &= 0 \\ \lambda &= \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \end{aligned} \]

Check: the two roots add to 1 (the trace) and multiply to \(\frac{1 - 5}{4} = -1\) (the determinant). The pair 1, 0 has product 0, not \(-1\). Dividing by 4 instead of 2 gives roots adding to \(\frac{1}{2}\). \(\frac{\sqrt{5}}{2}\) and \(1 - \frac{\sqrt{5}}{2}\) add to 1 but multiply to \(\frac{\sqrt{5}}{2} - \frac{5}{4}\), not \(-1\).

Discuss in Forum

10.

Let T be a linear operator defined on a vector space V; \(T : V \mapsto V\). If dim ker \(T \gt 0\), then :

A.0 is an eigenvalue of T
B.T is invertible
C.T is nilpotent
D.\(I_{m}\) T = V

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The conclusion is 0 is an eigenvalue of T.

\(\dim \ker T \gt 0\) means there is a non-zero vector v with \(Tv = 0 = 0 \cdot v\). That is exactly the definition of 0 being an eigenvalue, with v as an eigenvector.

The other statements fail. A map with a non-zero kernel is not one-to-one, so T is not invertible. It need not be nilpotent: the projection \(T(x, y) = (x, 0)\) has kernel the y-axis but \(T^{k} = T \ne 0\) for every k. And Im T = V is false for a finite-dimensional V, because rank-nullity gives \(\dim \text{Im}\, T = \dim V - \dim \ker T \lt \dim V\).

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Eigenvalues and Eigenvectors Questions and Answers || Eigenvalues and Eigenvectors MCQ Questions and Answers || Eigenvalues and Eigenvectors GK Questions and Answers || Eigenvalues and Eigenvectors GK MCQ Questions || Eigenvalues and Eigenvectors Multiple Choice Questions and Answers || Eigenvalues and Eigenvectors GK || GK on Eigenvalues and Eigenvectors || Linear Algebra Questions and Answers || Linear Algebra MCQ Questions and Answers || Linear Algebra GK Questions and Answers || GK on Linear Algebra

Eigenvalues and Eigenvectors Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications