Diagonalization Questions and Answers

Q: Let \(M = \begin{bmatrix} 4 & -3 \\ 1 & 0 \end{bmatrix}\).Consider the following statements:P: \(M^{8} + M^{12}\) is diagonalizable.Q: \(M^{7} + M^{9}\) is diagonalizable.Which of the following statements is correct?

Both P and Q are TRUE.The characteristic polynomial of M is\[ \lambda^{2} - 4\lambda + 3 = (\lambda-1)(\lambda-3), \]so M has the distinct eigenvalues 1 and 3 and is diagonalizable: \(M = PDP^{-1}\) with D = diag(1, 3). Any polynomial in M is then \(p(M) = P\,p(D)\,P^{-1}\), which is diagonal in the same basis. Both \(M^{8}+M^{12}\) and \(M^{7}+M^{9}\) are polynomials in M, so both are diagonalizable.The options declaring either sum non-diagonalizable overlook that powers and sums of a diagonalizable matrix share its eigenvectors.

Q: Let \(A \in M_{n}(\mathbb{C})\) be a normal matrix. Consider the following statements.I. If all the eigenvalues of \(A\) are real, then \(A\) is Hermitian.II. If all the eigenvalues of \(A\) have absolute value 1, then \(A\) is unitary.Which one of the following is correct?

Both I and II are TRUE.A normal matrix is unitarily diagonalisable: \(A = UDU^{*}\) with \(U\) unitary and \(D\) diagonal holding the eigenvalues.I: if the eigenvalues are real, \(D^{*} = D\), so \(A^{*} = UD^{*}U^{*} = A\) and \(A\) is Hermitian. II: if every eigenvalue has modulus 1, then \(D^{*}D = I\), so \(A^{*}A = UD^{*}DU^{*} = I\) and \(A\) is unitary.Normality is essential: without it, neither statement holds (a non-normal upper triangular matrix with real diagonal is not Hermitian). With it, both are true, so the other three options are wrong.

Q: Let \(A\) be a \(2 \times 2\) non-diagonalizable real matrix with a real eigenvalue \(\lambda\) and \(v\) be an eigenvector of \(A\) corresponding to \(\lambda\). Which one of the following is the general solution of the system \(y' = Ay\) of first-order linear differential equations?

The general solution is \(c_{1}e^{\lambda t}v + c_{2}e^{\lambda t}(tv + u)\), with \((A - \lambda I_{2})u = v\).A non-diagonalisable \(2 \times 2\) matrix has a single eigenvector direction \(v\), and a generalised eigenvector \(u\) with \((A - \lambda I)u = v\). Check the second solution \(y = e^{\lambda t}(tv + u)\):\[ \begin{aligned} y' &= \lambda e^{\lambda t}(tv + u) + e^{\lambda t}v \\ Ay &= e^{\lambda t}(t\lambda v + \lambda u + v) \end{aligned} \]These agree. \(te^{\lambda t}v\) alone is not a solution (its derivative has an extra \(e^{\lambda t}v\) that \(A\) cannot produce), which rules out the first option; \(t^{2}e^{\lambda t}v\) fails the same way; and \(te^{\lambda t}(v + u)\) misplaces the factor \(t\), so it does not satisfy the system either.

Q: Matrix\[ A = \begin{pmatrix} 1 & 3 & -2 \\ 0 & 4 & -2 \\ 0 & 3 & -1 \end{pmatrix} \]is similar to :

A is similar to the diagonal matrix diag(1, 1, 2).Expanding along the first column, the characteristic polynomial is \((1 - \lambda)\) times that of the block \(\begin{pmatrix} 4 & -2 \\ 3 & -1 \end{pmatrix}\), whose trace is 3 and determinant is \(-4 + 6 = 2\). So the block has eigenvalues 1 and 2, and A has eigenvalues 1, 1, 2.To decide between the diagonal form and a Jordan block for 1, compute\[ A - I = \begin{pmatrix} 0 & 3 & -2 \\ 0 & 3 & -2 \\ 0 & 3 & -2 \end{pmatrix} \]This has rank 1, so the eigenspace for 1 has dimension \(3 - 1 = 2\), equal to the algebraic multiplicity. A is diagonalisable, so it is similar to diag(1, 1, 2).The matrix with a 1 above the diagonal in the eigenvalue-1 block would need a one-dimensional eigenspace. The options with eigenvalues 1, 2, 2 have the wrong trace: trace A = 1 + 4 - 1 = 4, not 5.

Q: If characteristic equation of matrix A is same as minimum polynomial which reads as \((x - 1)^{3}(x - 4)\), then the Jordan canonical form of A is :

The Jordan form is \(\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&4\end{pmatrix}\).The exponent of \((x - \lambda)\) in the minimal polynomial is the size of the largest Jordan block for \(\lambda\), and the exponent in the characteristic polynomial is the total size of all blocks for \(\lambda\). Here both are 3 for \(\lambda = 1\), so there is a single \(3 \times 3\) block for the eigenvalue 1, and a single \(1 \times 1\) block for 4.The first option has blocks of sizes 2 and 1 for eigenvalue 1, so its minimal polynomial is \((x - 1)^{2}(x - 4)\). The second has eigenvalue 1 only twice and 4 twice, so its characteristic polynomial is wrong. The last puts a 1 above the diagonal between the eigenvalues 1 and 4, which is not a Jordan form at all.

Q: Let T be a linear operator on \(\mathbf{R}^{4}\) having the minimal polynomial \((x + 1)^{3}\). Then which of the following statements is true ?

The true statement is T is triangulable over \(\mathbf{R}\).An operator is triangulable exactly when its minimal polynomial splits into linear factors over the field. Here \((x + 1)^{3}\) splits over \(\mathbf{R}\), so T has an upper triangular matrix.The others are false. T is diagonalizable only if the minimal polynomial has distinct linear factors, but \((x + 1)\) is repeated. T is not nilpotent, since its only eigenvalue is -1, not 0. The characteristic polynomial has degree 4 on \(\mathbf{R}^{4}\), and with the same roots as the minimal polynomial it must be \((x + 1)^{4}\), not \((x + 1)^{3}\).

Q: Matrix \(A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end{pmatrix}\) is :

The matrix is diagonalizable over C.The characteristic polynomial is\[ (1 - \lambda)[(1 - \lambda)^{2} + 1] = 0 \]giving eigenvalues \(1, 1 + i, 1 - i\). These are three distinct eigenvalues, so A is diagonalizable over C.Over R it is neither diagonalizable nor triangulable, because triangulating over R requires all eigenvalues to be real, and \(1 \pm i\) are not. That rules out A and B, and C is wrong because distinct eigenvalues guarantee diagonalizability over C.

Q: The Jordan canonical form of a real symmetric matrix with characteristic polynomial \((x - 1)^{4}\) is :

The Jordan form is the 4 × 4 identity matrix.A real symmetric matrix is orthogonally diagonalizable (the spectral theorem), so its Jordan canonical form is diagonal, with its eigenvalues on the diagonal. The characteristic polynomial \((x - 1)^{4}\) says the only eigenvalue is 1, with algebraic multiplicity 4. A diagonal matrix with four 1s is I. In fact the symmetric matrix itself must be I, since \(Q I Q^{T} = I\).The other three matrices have 1s above the diagonal, that is, Jordan blocks of size 2, 4 and so on, and such matrices are not diagonalizable. The last one is not even in Jordan form, since its off-diagonal 1 is not on the superdiagonal.

Q: The quadratic form corresponding to the matrix\[ \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} \]is :

The form is positive definite.Use Sylvester's criterion on the leading principal minors:\[ \begin{aligned} \Delta_{1} &= 2 \gt 0 \\ \Delta_{2} &= 2 \cdot 2 - 1 \cdot 1 = 3 \gt 0 \\ \Delta_{3} &= 2(2 - 1) - 1(1 - 0) + 0 = 1 \gt 0 \end{aligned} \]All three are positive, so the quadratic form \(2x^{2} + 2y^{2} + z^{2} + 2xy + 2yz\) is positive for every non-zero vector.A negative definite form needs minors alternating in sign starting negative, and a semidefinite form needs a zero determinant; neither fits a matrix whose determinant is 1 and whose first minor is 2.

Q: The quadratic form\[ 4X_{1}^{2}+X_{2}^{2}+X_{3}^{2}-2X_{1}X_{2}+2X_{1}X_{3} \]is :

The quadratic form is positive definite.Its symmetric matrix and leading principal minors are\[ \begin{aligned} &\begin{bmatrix} 4 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \\ &\Delta_{1} = 4,\ \Delta_{2} = 4 - 1 = 3 \\ &\Delta_{3} = 4(1) + 1(-1) + 1(-1) = 2 \end{aligned} \]All three are positive, so by Sylvester's criterion the form is positive definite. Each off-diagonal entry is half the coefficient of the corresponding cross term.A semidefinite form would need \(\Delta_{3} = 0\), and the negative options need \(\Delta_{1} \lt 0\).

Let \(M = \begin{bmatrix} 4 & -3 \\ 1 & 0 \end{bmatrix}\).

Consider the following statements:

P: \(M^{8} + M^{12}\) is diagonalizable.
Q: \(M^{7} + M^{9}\) is diagonalizable.

Which of the following statements is correct?

A.P is TRUE and Q is FALSE
B.P is FALSE and Q is TRUE
C.Both P and Q are FALSE
D.Both P and Q are TRUE

Exam Year(s)

GATE Mathematics - 2023,

Answer: D

Both P and Q are TRUE.

The characteristic polynomial of M is

\[ \lambda^{2} - 4\lambda + 3 = (\lambda-1)(\lambda-3), \]

so M has the distinct eigenvalues 1 and 3 and is diagonalizable: \(M = PDP^{-1}\) with D = diag(1, 3). Any polynomial in M is then \(p(M) = P\,p(D)\,P^{-1}\), which is diagonal in the same basis. Both \(M^{8}+M^{12}\) and \(M^{7}+M^{9}\) are polynomials in M, so both are diagonalizable.

The options declaring either sum non-diagonalizable overlook that powers and sums of a diagonalizable matrix share its eigenvectors.

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Let \(A \in M_{n}(\mathbb{C})\) be a normal matrix. Consider the following statements.

I. If all the eigenvalues of \(A\) are real, then \(A\) is Hermitian.
II. If all the eigenvalues of \(A\) have absolute value 1, then \(A\) is unitary.

Which one of the following is correct?

A.Both I and II are TRUE
B.I is TRUE and II is FALSE
C.I is FALSE and II is TRUE
D.Both I and II are FALSE

Exam Year(s)

GATE Mathematics - 2024,

Answer: A

Both I and II are TRUE.

A normal matrix is unitarily diagonalisable: \(A = UDU^{*}\) with \(U\) unitary and \(D\) diagonal holding the eigenvalues.

I: if the eigenvalues are real, \(D^{*} = D\), so \(A^{*} = UD^{*}U^{*} = A\) and \(A\) is Hermitian. II: if every eigenvalue has modulus 1, then \(D^{*}D = I\), so \(A^{*}A = UD^{*}DU^{*} = I\) and \(A\) is unitary.

Normality is essential: without it, neither statement holds (a non-normal upper triangular matrix with real diagonal is not Hermitian). With it, both are true, so the other three options are wrong.

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Let \(A\) be a \(2 \times 2\) non-diagonalizable real matrix with a real eigenvalue \(\lambda\) and \(v\) be an eigenvector of \(A\) corresponding to \(\lambda\). Which one of the following is the general solution of the system \(y' = Ay\) of first-order linear differential equations?

A.\(c_{1}e^{\lambda t}v + c_{2}te^{\lambda t}v\), where \(c_{1}, c_{2} \in \mathbb{R}\)
B.\(c_{1}e^{\lambda t}v + c_{2}t^{2}e^{\lambda t}v\), where \(c_{1}, c_{2} \in \mathbb{R}\)
C.\(c_{1}e^{\lambda t}v + c_{2}e^{\lambda t}(tv + u)\), where \(c_{1}, c_{2} \in \mathbb{R}\) and \(u\) is a \(2 \times 1\) real column vector such that \((A - \lambda I_{2})u = v\)
D.\(c_{1}e^{\lambda t}v + c_{2}te^{\lambda t}(v + u)\), where \(c_{1}, c_{2} \in \mathbb{R}\) and \(u\) is a \(2 \times 1\) real column vector such that \((A - \lambda I_{2})u = v\)

Exam Year(s)

GATE Mathematics - 2024,

Answer: C

The general solution is \(c_{1}e^{\lambda t}v + c_{2}e^{\lambda t}(tv + u)\), with \((A - \lambda I_{2})u = v\).

A non-diagonalisable \(2 \times 2\) matrix has a single eigenvector direction \(v\), and a generalised eigenvector \(u\) with \((A - \lambda I)u = v\). Check the second solution \(y = e^{\lambda t}(tv + u)\):

\[ \begin{aligned} y' &= \lambda e^{\lambda t}(tv + u) + e^{\lambda t}v \\ Ay &= e^{\lambda t}(t\lambda v + \lambda u + v) \end{aligned} \]

These agree. \(te^{\lambda t}v\) alone is not a solution (its derivative has an extra \(e^{\lambda t}v\) that \(A\) cannot produce), which rules out the first option; \(t^{2}e^{\lambda t}v\) fails the same way; and \(te^{\lambda t}(v + u)\) misplaces the factor \(t\), so it does not satisfy the system either.

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Matrix

\[ A = \begin{pmatrix} 1 & 3 & -2 \\ 0 & 4 & -2 \\ 0 & 3 & -1 \end{pmatrix} \]

is similar to :

A.\(\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}\)
B.\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}\)
C.\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}\)
D.\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

A is similar to the diagonal matrix diag(1, 1, 2).

Expanding along the first column, the characteristic polynomial is \((1 - \lambda)\) times that of the block \(\begin{pmatrix} 4 & -2 \\ 3 & -1 \end{pmatrix}\), whose trace is 3 and determinant is \(-4 + 6 = 2\). So the block has eigenvalues 1 and 2, and A has eigenvalues 1, 1, 2.

To decide between the diagonal form and a Jordan block for 1, compute

\[ A - I = \begin{pmatrix} 0 & 3 & -2 \\ 0 & 3 & -2 \\ 0 & 3 & -2 \end{pmatrix} \]

This has rank 1, so the eigenspace for 1 has dimension \(3 - 1 = 2\), equal to the algebraic multiplicity. A is diagonalisable, so it is similar to diag(1, 1, 2).

The matrix with a 1 above the diagonal in the eigenvalue-1 block would need a one-dimensional eigenspace. The options with eigenvalues 1, 2, 2 have the wrong trace: trace A = 1 + 4 - 1 = 4, not 5.

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If characteristic equation of matrix A is same as minimum polynomial which reads as \((x - 1)^{3}(x - 4)\), then the Jordan canonical form of A is :

A.\(\begin{pmatrix}1&1&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&4\end{pmatrix}\)
B.\(\begin{pmatrix}1&1&0&0\\0&1&0&0\\0&0&4&0\\0&0&0&4\end{pmatrix}\)
C.\(\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&4\end{pmatrix}\)
D.\(\begin{pmatrix}1&1&0&0\\0&1&0&0\\0&0&1&1\\0&0&0&4\end{pmatrix}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The Jordan form is \(\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&4\end{pmatrix}\).

The exponent of \((x - \lambda)\) in the minimal polynomial is the size of the largest Jordan block for \(\lambda\), and the exponent in the characteristic polynomial is the total size of all blocks for \(\lambda\). Here both are 3 for \(\lambda = 1\), so there is a single \(3 \times 3\) block for the eigenvalue 1, and a single \(1 \times 1\) block for 4.

The first option has blocks of sizes 2 and 1 for eigenvalue 1, so its minimal polynomial is \((x - 1)^{2}(x - 4)\). The second has eigenvalue 1 only twice and 4 twice, so its characteristic polynomial is wrong. The last puts a 1 above the diagonal between the eigenvalues 1 and 4, which is not a Jordan form at all.

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Let T be a linear operator on \(\mathbf{R}^{4}\) having the minimal polynomial \((x + 1)^{3}\). Then which of the following statements is true ?

A.T is diagonalizable over \(\mathbf{R}\)
B.T is nilpotent
C.T is triangulable over \(\mathbf{R}\)
D.characteristic polynomial of T is \((x + 1)^{3}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The true statement is T is triangulable over \(\mathbf{R}\).

An operator is triangulable exactly when its minimal polynomial splits into linear factors over the field. Here \((x + 1)^{3}\) splits over \(\mathbf{R}\), so T has an upper triangular matrix.

The others are false. T is diagonalizable only if the minimal polynomial has distinct linear factors, but \((x + 1)\) is repeated. T is not nilpotent, since its only eigenvalue is -1, not 0. The characteristic polynomial has degree 4 on \(\mathbf{R}^{4}\), and with the same roots as the minimal polynomial it must be \((x + 1)^{4}\), not \((x + 1)^{3}\).

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Matrix \(A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end{pmatrix}\) is :

A.Diagonalizable over R
B.Triangulable over R but not diagonalizable over R
C.Triangulable over C but not diagonalizable over C
D.Diagonalizable over C

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The matrix is diagonalizable over C.

The characteristic polynomial is

\[ (1 - \lambda)[(1 - \lambda)^{2} + 1] = 0 \]

giving eigenvalues \(1, 1 + i, 1 - i\). These are three distinct eigenvalues, so A is diagonalizable over C.

Over R it is neither diagonalizable nor triangulable, because triangulating over R requires all eigenvalues to be real, and \(1 \pm i\) are not. That rules out A and B, and C is wrong because distinct eigenvalues guarantee diagonalizability over C.

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The Jordan canonical form of a real symmetric matrix with characteristic polynomial \((x - 1)^{4}\) is :

A.\[ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
B.\[ \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
C.\[ \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
D.\[ \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The Jordan form is the 4 × 4 identity matrix.

A real symmetric matrix is orthogonally diagonalizable (the spectral theorem), so its Jordan canonical form is diagonal, with its eigenvalues on the diagonal. The characteristic polynomial \((x - 1)^{4}\) says the only eigenvalue is 1, with algebraic multiplicity 4. A diagonal matrix with four 1s is I. In fact the symmetric matrix itself must be I, since \(Q I Q^{T} = I\).

The other three matrices have 1s above the diagonal, that is, Jordan blocks of size 2, 4 and so on, and such matrices are not diagonalizable. The last one is not even in Jordan form, since its off-diagonal 1 is not on the superdiagonal.

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The quadratic form corresponding to the matrix

\[ \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} \]

is :

A.Positive definite
B.Negative definite
C.Positive semidefinite
D.Negative semidefinite

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The form is positive definite.

Use Sylvester's criterion on the leading principal minors:

\[ \begin{aligned} \Delta_{1} &= 2 \gt 0 \\ \Delta_{2} &= 2 \cdot 2 - 1 \cdot 1 = 3 \gt 0 \\ \Delta_{3} &= 2(2 - 1) - 1(1 - 0) + 0 = 1 \gt 0 \end{aligned} \]

All three are positive, so the quadratic form \(2x^{2} + 2y^{2} + z^{2} + 2xy + 2yz\) is positive for every non-zero vector.

A negative definite form needs minors alternating in sign starting negative, and a semidefinite form needs a zero determinant; neither fits a matrix whose determinant is 1 and whose first minor is 2.

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10.

The quadratic form

\[ 4X_{1}^{2}+X_{2}^{2}+X_{3}^{2}-2X_{1}X_{2}+2X_{1}X_{3} \]

is :

A.negative definite
B.negative semidefinite
C.positive definite
D.positive semidefinite

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The quadratic form is positive definite.

Its symmetric matrix and leading principal minors are

\[ \begin{aligned} &\begin{bmatrix} 4 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \\ &\Delta_{1} = 4,\ \Delta_{2} = 4 - 1 = 3 \\ &\Delta_{3} = 4(1) + 1(-1) + 1(-1) = 2 \end{aligned} \]

All three are positive, so by Sylvester's criterion the form is positive definite. Each off-diagonal entry is half the coefficient of the corresponding cross term.

A semidefinite form would need \(\Delta_{3} = 0\), and the negative options need \(\Delta_{1} \lt 0\).

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