GATE CS 2015 Set 3 Q.25 (CS Q.15): Linear Algebra

In the given matrix

\begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}

one of the eigenvalues is 1. The eigenvectors corresponding to the eigenvalue 1 are

$\{\alpha(4, 2, 1) \mid \alpha \ne 0, \alpha \in \mathbb{R}\}$

$\{\alpha(-4, 2, 1) \mid \alpha \ne 0, \alpha \in \mathbb{R}\}$

$\{\alpha(\sqrt{2}, 0, 1) \mid \alpha \ne 0, \alpha \in \mathbb{R}\}$

$\{\alpha(-\sqrt{2}, 0, 1) \mid \alpha \ne 0, \alpha \in \mathbb{R}\}$

Official answer and solution

Solution

Solve the homogeneous linear system (A - lambda * I)v = 0 for lambda = 1.

Given matrix:

A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}

For the eigenvalue $\lambda = 1$ , the eigenvector $v = [x, y, z]^T$ satisfies $(A - 1 \cdot I)v = 0$ :

\begin{bmatrix} 1 - 1 & -1 & 2 \\ 0 & 1 - 1 & 0 \\ 1 & 2 & 1 - 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \implies \begin{bmatrix} 0 & -1 & 2 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

From row 1:

-y + 2z = 0 \implies y = 2z

From row 3:

x + 2y = 0 \implies x = -2y = -2(2z) = -4z

Setting $z = \alpha$ (where $\alpha \in \mathbb{R}, \alpha \ne 0$ ):

x = -4\alpha, \quad y = 2\alpha, \quad z = \alpha

v = \alpha \begin{bmatrix} -4 \\ 2 \\ 1 \end{bmatrix}

Therefore, the set of eigenvectors corresponding to $\lambda = 1$ is $\{\alpha(-4, 2, 1) \mid \alpha \ne 0, \alpha \in \mathbb{R}\}$ (Option B).

Final answer

Common trap

Sign error when solving x + 2y = 0, wrongly setting x = +4z instead of x = -4z.

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