GATE CS 2014 Set 2 Q47 — Eigenvalues and Eigenvectors, LU Decomposition

We can find the non-zero eigenvalues of the matrix $A$ by observing its row structure. Notice that $A$ has only two linearly independent rows:

• $\mathbf{r_1} = [1, 0, 0, 0, 1]$ (identical to row 5)
• $\mathbf{r_2} = [0, 1, 1, 1, 0]$ (identical to rows 2, 3, and 4)

Since the rank of $A$ is $2$, there are exactly two non-zero eigenvalues, and the remaining three eigenvalues are $0$.

Let us test the vectors corresponding to these independent row patterns as potential eigenvectors:

1. For $\mathbf{v_1} = [1, 0, 0, 0, 1]^T$:
   $$A\mathbf{v_1} = \begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0 \\ 2 \end{pmatrix} = 2\mathbf{v_1}$$
   Thus, $\lambda_1 = 2$ is an eigenvalue.
2. For $\mathbf{v_2} = [0, 1, 1, 1, 0]^T$:
   $$A\mathbf{v_2} = \begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ 3 \\ 3 \\ 0 \end{pmatrix} = 3\mathbf{v_2}$$
   Thus, $\lambda_2 = 3$ is an eigenvalue.

The two non-zero eigenvalues are $2$ and $3$. Their product is:

$$2 \times 3 = 6$$