GATE CS 2022 Q43 — Eigenvalues and Eigenvectors, LU Decomposition

A non-zero vector $x$ is an eigenvector of a matrix $A$ if it satisfies the relation:

where $\lambda$ is a scalar (the eigenvalue). We can verify each option by computing the matrix-vector product $Ax$:

• Option A: Let $x_1 = \begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix}$. $$Ax_1 = \begin{pmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 9-6-4 \\ 8-6-1 \\ -20+15+5 \\ -32+21+12 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = 1 \cdot x_1$$ Since $Ax_1 = 1 \cdot x_1$, this is an eigenvector with eigenvalue $\lambda = 1$.
• Option B: Let $x_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}$. $$Ax_2 = \begin{pmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -7 \\ -5 \\ 12 \\ 25 \end{pmatrix}$$ This is not a scalar multiple of $x_2$, so it is not an eigenvector.
• Option C: Let $x_3 = \begin{pmatrix} -1 \\ 0 \\ 2 \\ 2 \end{pmatrix}$. $$Ax_3 = \begin{pmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix} \begin{pmatrix} -1 \\ 0 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 9-4-8 \\ 8-6-2 \\ -20+16+10 \\ -32+14+24 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ 6 \\ 6 \end{pmatrix} = 3 \cdot x_3$$ Since $Ax_3 = 3 \cdot x_3$, this is an eigenvector with eigenvalue $\lambda = 3$.
• Option D: Let $x_4 = \begin{pmatrix} 0 \\ 1 \\ -3 \\ 0 \end{pmatrix}$. $$Ax_4 = \begin{pmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ -3 \\ 0 \end{pmatrix} = \begin{pmatrix} -6+6 \\ -6+9 \\ 15-24 \\ 21-21 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -9 \\ 0 \end{pmatrix} = 3 \cdot x_4$$ Since $Ax_4 = 3 \cdot x_4$, this is an eigenvector with eigenvalue $\lambda = 3$.