GATE CS 2012 Q11 — Eigenvalues and Eigenvectors, LU Decomposition

Given the elements of the $2 \times 2$ matrix $A$:

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

To find the eigenvalues of $A$, we solve the characteristic equation $\det(A - \lambda I) = 0$:

$$\det \begin{pmatrix} 1 - \lambda & 1 \\ 1 & -1 - \lambda \end{pmatrix} = 0$$

$$(1 - \lambda)(-1 - \lambda) - 1 = 0$$

$$\lambda^2 - 1 - 1 = 0 \implies \lambda^2 = 2$$

$$\lambda = \pm\sqrt{2}$$

So, the eigenvalues of $A$ are $\lambda_1 = \sqrt{2}$ and $\lambda_2 = -\sqrt{2}$.

By the properties of eigenvalues, if $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigenvalue of $A^k$. Therefore, the eigenvalues of $A^{19}$ are:

$$\lambda_1^{19} = (\sqrt{2})^{19} = (\sqrt{2})^{18} \cdot \sqrt{2} = 2^9 \cdot \sqrt{2} = 512\sqrt{2}$$

$$\lambda_2^{19} = (-\sqrt{2})^{19} = -(\sqrt{2})^{19} = -512\sqrt{2}$$

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