Eigenvalues and Eigenvectors Questions and Answers

Q: If A is a \(2 \times 2\) real matrix with trace zero and determinant 1, then the eigenvalues of A are :

The eigenvalues are purely imaginary.For a \(2 \times 2\) matrix the characteristic polynomial is \(\lambda^{2} - (\text{trace})\lambda + \det\). With trace 0 and determinant 1:\[ \lambda^{2} + 1 = 0 \quad \Rightarrow \quad \lambda = \pm i \]An example is the rotation \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\).Real eigenvalues would need the discriminant \(\text{tr}^{2} - 4\det = -4\) to be non-negative, which it is not, so they are neither real and distinct nor real and repeated. The real part of the eigenvalues is half the trace, which is 0, so 'complex with non-zero real part' is also ruled out.

Q: Eigen values of a nilpotent matrix A, i.e., a matrix satisfying \(A^{K} = 0\) for some integer K are :

The eigenvalues of a nilpotent matrix are all zero.Let \(A\vec{v} = \lambda\vec{v}\) with \(\vec{v} \neq 0\). Applying A repeatedly gives \(A^{K}\vec{v} = \lambda^{K}\vec{v}\). Since \(A^{K} = 0\), this means \(\lambda^{K}\vec{v} = 0\), and because \(\vec{v}\) is not the zero vector, \(\lambda^{K} = 0\), so \(\lambda = 0\).Thus every eigenvalue is 0; equivalently the characteristic polynomial is \(\lambda^{n}\). They cannot be distinct (all are equal), and they are neither purely imaginary nor negative. A simple example is the matrix with a single 1 above the diagonal, whose square is zero.

Q: Trace of a 3 × 3 matrix is 6. Two of its eigen values are 1 and 2. The third eigen value is :

The third eigenvalue is 3.The trace of a square matrix equals the sum of its eigenvalues (counted with multiplicity). So\[ \begin{aligned} \lambda_{1} + \lambda_{2} + \lambda_{3} &= 6 \\ 1 + 2 + \lambda_{3} &= 6 \\ \lambda_{3} &= 3 \end{aligned} \]The option 6 is the trace itself, mistaken for an eigenvalue; 1 and 2 repeat the eigenvalues already given and would make the trace 4 or 5. Only 3 makes the three eigenvalues add up to the trace.

Q: The eigen values of an anti-Hermitian matrix are :

The eigenvalues of an anti-Hermitian matrix are purely imaginary (or zero).An anti-Hermitian matrix satisfies \( A^{\dagger} = -A \). If \( Av = \lambda v \), then\[ \begin{aligned} v^{\dagger}Av &= \lambda\, v^{\dagger}v \\ (v^{\dagger}Av)^{*} &= v^{\dagger}A^{\dagger}v = -\lambda\, v^{\dagger}v \end{aligned} \]so \( \lambda^{*} = -\lambda \), which means the real part of \( \lambda \) is zero. Equivalently, \( iA \) is Hermitian and has real eigenvalues.Real eigenvalues, positive or negative, belong to Hermitian matrices, and a non-zero real part is exactly what the condition rules out.

Q: If a real matrix A has characteristic polynomial \((x - 1)(x^{2} + 1)\), then which of the following statements about A is true ?

A is invertible.The determinant equals the product of the eigenvalues, the roots of \((x - 1)(x^{2} + 1)\): \(1 \cdot i \cdot (-i) = 1 \ne 0\). Equivalently, 0 is not a root of the characteristic polynomial, so A is invertible.The others fail. A is not diagonalizable or triangulable over \(\mathbf{R}\), because both need the characteristic polynomial to split into real linear factors, and \(x^{2} + 1\) does not. A is not nilpotent, since a nilpotent matrix has characteristic polynomial \(x^{3}\).

Q: Let V be a finite dimensional vector space over \(\mathbf{R}\) and \(T \in L(V)\). If \(\lambda\) is an eigenvalue of T, then which of the following statements is false ?

The false statement is \(\ker(T - \lambda I) = \{0\}\).\(\lambda\) is an eigenvalue exactly when there is a non-zero \(v\) with \(Tv = \lambda v\), that is, a non-zero vector in \(\ker(T - \lambda I)\). So the kernel is not \(\{0\}\), which makes (A) true and (B) false.On a finite dimensional space, a non-trivial kernel means \(T - \lambda I\) is not injective, so by rank-nullity its rank is less than \(\dim V\) (C), and a singular operator has determinant 0 (D). Those three are equivalent forms of the same fact.

Q: If A is a real 3 × 3 matrix, which of the following statements can not be true about A ?

The impossible statement is that the minimum polynomial of A is \(x^{2} + 1\).The characteristic polynomial of a real 3 × 3 matrix has real coefficients and odd degree, so it has a real root \(\lambda\). That real eigenvalue must also be a root of the minimal polynomial. But \(x^{2} + 1\) has no real roots, so it cannot be the minimal polynomial.The others can happen. A companion matrix has minimal polynomial equal to its characteristic polynomial (A). Every real eigenvalue is also a complex number, so A always has a complex eigenvalue (B). And \(\mathrm{diag}(1, -1, 1)\) has minimal polynomial \(x^{2} - 1\) (C).

Q: Let A be a square matrix and let B = A - 0.2 I, where I is the identity matrix. Which of the following is always true ?

The statement that always holds is A and B have the same eigen vectors.If \(Av = \lambda v\), then \(Bv = Av - 0.2v = (\lambda - 0.2)v\), and conversely. So every eigenvector of A is an eigenvector of B and vice versa; only the eigenvalues move, each one shifted down by 0.2.That shift is exactly why 'same eigen values' fails. Rank can change: if 0.2 is an eigenvalue of A, then B is singular while A may be invertible (take A = 0.2 I, which has full rank, while B = 0 has rank 0). The same example shows that \(B^{-1}\) need not exist. So only the eigenvector statement is true for every square matrix A.

Q: Suppose the eigen values of a 3 × 3 matrix A are 0, 1 and 2. Then, the determinant of \((I + A^{2})^{-1}\).

The determinant is 1/10.If A has eigenvalues 0, 1, 2, then \(A^{2}\) has eigenvalues 0, 1, 4 and \(I + A^{2}\) has eigenvalues \(1 + 0 = 1\), \(1 + 1 = 2\), \(1 + 4 = 5\). This holds for any matrix, diagonalizable or not, because eigenvalues of a polynomial in A are the polynomial evaluated at the eigenvalues of A.\[ \begin{aligned} \det(I + A^{2}) &= 1 \times 2 \times 5 = 10 \\ \det(I + A^{2})^{-1} &= \frac{1}{10} \end{aligned} \]So the information is enough, and the answer is not 0: \(I + A^{2}\) is invertible because none of its eigenvalues is zero. The value 1/8 comes from wrongly taking the eigenvalues of \(I + A^{2}\) as 1, 2, 4 (forgetting to square the 2).

Q: Suppose the two matrices A and B have same characteristic polynomial. Then :

The conclusion that always follows is A is invertible if and only if B is invertible.The characteristic polynomial \(\det(xI - A)\) has constant term \((-1)^{n}\det A\). Equal characteristic polynomials therefore give \(\det A = \det B\), so one determinant is zero exactly when the other is. Equivalently, 0 is a root of the shared polynomial for both or for neither.The other options fail. The matrices \(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\) and the zero matrix both have characteristic polynomial \(x^{2}\), but minimal polynomials \(x^{2}\) and \(x\); they are not similar, and neither is the transpose of the other. Same characteristic polynomial says nothing about the Jordan structure, so similarity, equal minimal polynomials and \(A = B^{t}\) are all too strong.

11.

If A is a \(2 \times 2\) real matrix with trace zero and determinant 1, then the eigenvalues of A are :

A.real and distinct
B.real and repeated
C.complex with non-zero real part
D.purely imaginary

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The eigenvalues are purely imaginary.

For a \(2 \times 2\) matrix the characteristic polynomial is \(\lambda^{2} - (\text{trace})\lambda + \det\). With trace 0 and determinant 1:

\[ \lambda^{2} + 1 = 0 \quad \Rightarrow \quad \lambda = \pm i \]

An example is the rotation \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\).

Real eigenvalues would need the discriminant \(\text{tr}^{2} - 4\det = -4\) to be non-negative, which it is not, so they are neither real and distinct nor real and repeated. The real part of the eigenvalues is half the trace, which is 0, so 'complex with non-zero real part' is also ruled out.

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12.

Eigen values of a nilpotent matrix A, i.e., a matrix satisfying \(A^{K} = 0\) for some integer K are :

A.Real and distinct
B.Zero
C.Purely imaginary
D.Real and negative

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The eigenvalues of a nilpotent matrix are all zero.

Let \(A\vec{v} = \lambda\vec{v}\) with \(\vec{v} \neq 0\). Applying A repeatedly gives \(A^{K}\vec{v} = \lambda^{K}\vec{v}\). Since \(A^{K} = 0\), this means \(\lambda^{K}\vec{v} = 0\), and because \(\vec{v}\) is not the zero vector, \(\lambda^{K} = 0\), so \(\lambda = 0\).

Thus every eigenvalue is 0; equivalently the characteristic polynomial is \(\lambda^{n}\). They cannot be distinct (all are equal), and they are neither purely imaginary nor negative. A simple example is the matrix with a single 1 above the diagonal, whose square is zero.

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13.

Trace of a 3 × 3 matrix is 6. Two of its eigen values are 1 and 2. The third eigen value is :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The third eigenvalue is 3.

The trace of a square matrix equals the sum of its eigenvalues (counted with multiplicity). So

\[ \begin{aligned} \lambda_{1} + \lambda_{2} + \lambda_{3} &= 6 \\ 1 + 2 + \lambda_{3} &= 6 \\ \lambda_{3} &= 3 \end{aligned} \]

The option 6 is the trace itself, mistaken for an eigenvalue; 1 and 2 repeat the eigenvalues already given and would make the trace 4 or 5. Only 3 makes the three eigenvalues add up to the trace.

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14.

The eigen values of an anti-Hermitian matrix are :

A.Real positive
B.Real negative
C.Purely imaginary
D.Have non-zero real part

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The eigenvalues of an anti-Hermitian matrix are purely imaginary (or zero).

An anti-Hermitian matrix satisfies \( A^{\dagger} = -A \). If \( Av = \lambda v \), then

\[ \begin{aligned} v^{\dagger}Av &= \lambda\, v^{\dagger}v \\ (v^{\dagger}Av)^{*} &= v^{\dagger}A^{\dagger}v = -\lambda\, v^{\dagger}v \end{aligned} \]

so \( \lambda^{*} = -\lambda \), which means the real part of \( \lambda \) is zero. Equivalently, \( iA \) is Hermitian and has real eigenvalues.

Real eigenvalues, positive or negative, belong to Hermitian matrices, and a non-zero real part is exactly what the condition rules out.

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15.

If a real matrix A has characteristic polynomial \((x - 1)(x^{2} + 1)\), then which of the following statements about A is true ?

A.A is diagonalizable over \(\mathbf{R}\)
B.A is triangulable over \(\mathbf{R}\)
C.A is nilpotent
D.A is invertible

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

A is invertible.

The determinant equals the product of the eigenvalues, the roots of \((x - 1)(x^{2} + 1)\): \(1 \cdot i \cdot (-i) = 1 \ne 0\). Equivalently, 0 is not a root of the characteristic polynomial, so A is invertible.

The others fail. A is not diagonalizable or triangulable over \(\mathbf{R}\), because both need the characteristic polynomial to split into real linear factors, and \(x^{2} + 1\) does not. A is not nilpotent, since a nilpotent matrix has characteristic polynomial \(x^{3}\).

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16.

Let V be a finite dimensional vector space over \(\mathbf{R}\) and \(T \in L(V)\). If \(\lambda\) is an eigenvalue of T, then which of the following statements is false ?

A.\(\ker(T - \lambda I) \ne \{0\}\)
B.\(\ker(T - \lambda I) = \{0\}\)
C.\(\mathrm{rank}(T - \lambda I) \lt \dim V\)
D.\(\det(T - \lambda I) = 0\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The false statement is \(\ker(T - \lambda I) = \{0\}\).

\(\lambda\) is an eigenvalue exactly when there is a non-zero \(v\) with \(Tv = \lambda v\), that is, a non-zero vector in \(\ker(T - \lambda I)\). So the kernel is not \(\{0\}\), which makes (A) true and (B) false.

On a finite dimensional space, a non-trivial kernel means \(T - \lambda I\) is not injective, so by rank-nullity its rank is less than \(\dim V\) (C), and a singular operator has determinant 0 (D). Those three are equivalent forms of the same fact.

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17.

If A is a real 3 × 3 matrix, which of the following statements can not be true about A ?

A.Minimum polynomial = characteristic polynomial
B.A has at least one complex eigen value
C.Minimum polynomial of A = \(x^{2} - 1\)
D.Minimum polynomial of A = \(x^{2} + 1\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The impossible statement is that the minimum polynomial of A is \(x^{2} + 1\).

The characteristic polynomial of a real 3 × 3 matrix has real coefficients and odd degree, so it has a real root \(\lambda\). That real eigenvalue must also be a root of the minimal polynomial. But \(x^{2} + 1\) has no real roots, so it cannot be the minimal polynomial.

The others can happen. A companion matrix has minimal polynomial equal to its characteristic polynomial (A). Every real eigenvalue is also a complex number, so A always has a complex eigenvalue (B). And \(\mathrm{diag}(1, -1, 1)\) has minimal polynomial \(x^{2} - 1\) (C).

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18.

Let A be a square matrix and let B = A - 0.2 I, where I is the identity matrix. Which of the following is always true ?

A.A and B have the same rank
B.\(B^{-1}\) exists
C.A and B have the same eigen vectors
D.A and B have the same eigen values

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The statement that always holds is A and B have the same eigen vectors.

If \(Av = \lambda v\), then \(Bv = Av - 0.2v = (\lambda - 0.2)v\), and conversely. So every eigenvector of A is an eigenvector of B and vice versa; only the eigenvalues move, each one shifted down by 0.2.

That shift is exactly why 'same eigen values' fails. Rank can change: if 0.2 is an eigenvalue of A, then B is singular while A may be invertible (take A = 0.2 I, which has full rank, while B = 0 has rank 0). The same example shows that \(B^{-1}\) need not exist. So only the eigenvector statement is true for every square matrix A.

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19.

Suppose the eigen values of a 3 × 3 matrix A are 0, 1 and 2. Then, the determinant of \((I + A^{2})^{-1}\).

A.Cannot be found from the given information
B.is 1/10
C.0
D.1/8

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The determinant is 1/10.

If A has eigenvalues 0, 1, 2, then \(A^{2}\) has eigenvalues 0, 1, 4 and \(I + A^{2}\) has eigenvalues \(1 + 0 = 1\), \(1 + 1 = 2\), \(1 + 4 = 5\). This holds for any matrix, diagonalizable or not, because eigenvalues of a polynomial in A are the polynomial evaluated at the eigenvalues of A.

\[ \begin{aligned} \det(I + A^{2}) &= 1 \times 2 \times 5 = 10 \\ \det(I + A^{2})^{-1} &= \frac{1}{10} \end{aligned} \]

So the information is enough, and the answer is not 0: \(I + A^{2}\) is invertible because none of its eigenvalues is zero. The value 1/8 comes from wrongly taking the eigenvalues of \(I + A^{2}\) as 1, 2, 4 (forgetting to square the 2).

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20.

Suppose the two matrices A and B have same characteristic polynomial. Then :

A.A and B have same minimal polynomial
B.\(A = B^{t}\)
C.A and B are similar
D.A is invertible if and only if B is invertible

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The conclusion that always follows is A is invertible if and only if B is invertible.

The characteristic polynomial \(\det(xI - A)\) has constant term \((-1)^{n}\det A\). Equal characteristic polynomials therefore give \(\det A = \det B\), so one determinant is zero exactly when the other is. Equivalently, 0 is a root of the shared polynomial for both or for neither.

The other options fail. The matrices \(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\) and the zero matrix both have characteristic polynomial \(x^{2}\), but minimal polynomials \(x^{2}\) and \(x\); they are not similar, and neither is the transpose of the other. Same characteristic polynomial says nothing about the Jordan structure, so similarity, equal minimal polynomials and \(A = B^{t}\) are all too strong.

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