GATE CS 2012 Q.11 (CS Q.11): Linear Algebra

Let $A$ be the $2 \times 2$ matrix with elements $a_{11} = a_{12} = a_{21} = +1$ and $a_{22} = -1$ . Then the eigenvalues of the matrix $A^{19}$ are

1024 and $-1024$

$1024\sqrt{2}$ and $-1024\sqrt{2}$

$4\sqrt{2}$ and $-4\sqrt{2}$

$512\sqrt{2}$ and $-512\sqrt{2}$

Official answer and solution

Solution

Find the eigenvalues of matrix A using its characteristic equation, and then compute the eigenvalues of A^19 using spectral mapping.

The given matrix is:

A = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

Find eigenvalues of $A$ :
$\det(A - \lambda I) = \det\begin{pmatrix} 1 - \lambda & 1 \\ 1 & -1 - \lambda \end{pmatrix} = (1 - \lambda)(-1 - \lambda) - 1 = \lambda^2 - 1 - 1 = \lambda^2 - 2 = 0$ $\lambda^2 = 2 \implies \lambda = \pm\sqrt{2}$
Eigenvalues of $A^{19}$ :
By the spectral theorem, if $\lambda$ is an eigenvalue of $A$ , then $\lambda^{19}$ is an eigenvalue of $A^{19}$ :
$\lambda_1^{19} = (\sqrt{2})^{19} = (\sqrt{2})^{18} \cdot \sqrt{2} = (2^9) \sqrt{2} = 512\sqrt{2}$ $\lambda_2^{19} = (-\sqrt{2})^{19} = -(\sqrt{2})^{19} = -512\sqrt{2}$

Therefore, the eigenvalues of $A^{19}$ are $512\sqrt{2}$ and $-512\sqrt{2}$ , which corresponds to option (D).

Final answer

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