GATE CS 1993 Q01.1 — Eigenvalues and Eigenvectors, LU Decomposition

The given matrix is:
$$A = \begin{bmatrix} 0 & 0 & \alpha \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ where $\alpha \neq 0$. Since $A$ is an upper triangular matrix, its eigenvalues are its diagonal elements, which are all $\lambda = 0$.

To find the eigenvectors corresponding to $\lambda = 0$, we solve the system $Ax = 0x = 0$ for a non-zero vector $x = [x_1, x_2, x_3]^T$:

Since $\alpha \neq 0$, we must have $x_3 = 0$. The variables $x_1$ and $x_2$ are free variables. Thus, any non-zero vector of the form $[x_1, x_2, 0]^T$ is an eigenvector of $A$.

Evaluating the options:

• (B) $(\alpha, 0, 0)$: Has $x_3 = 0$ and is non-zero (since $\alpha \neq 0$). Hence, it is an eigenvector.
• (D) $(0, \alpha, 0)$: Has $x_3 = 0$ and is non-zero (since $\alpha \neq 0$). Hence, it is an eigenvector.