Discussion
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 \(A^{13}\)?
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 \(A^{13}\).
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