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.

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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