Question

In matrix algebra, what is the eigenvalue of the matrix \(\begin{pmatrix} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix}\)?

(a) 1

(b) 2

(c) 3

(d) 4

The question was posed to me in an interview for job.

Question is taken from Eigen Value and Time Dependent Problems topic in section Single Variable Problems of Finite Element Method

Answer 1

The correct answer is (c) 3

To elaborate: The eigenvalue, L of a matrix is equal to the root (factor) of the equation |K-LI|=0.

Let the given matrix be denoted by K then K-LI = \(\begin{pmatrix} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix} – L \begin{pmatrix}1&0&0 \\ 0&1&0 \\0&0&1 \end{pmatrix}\)

= \(\begin{pmatrix} 1-L&1&1 \\ 1&1-L&1 \\ 1&1&1-L \end{pmatrix}\)

|K-LI| = \(\begin{vmatrix} 1-L&1&1 \\ 1&1-L&1 \\ 1&1&1-L \end{vmatrix}\)

= (1-L)((1-L)^2-1)-1(-L)+1(L)

= (1-L)(L^2-2L) + 2L

= -L^3 + 3L^2

= -L^2 (L-3).

Given -L^2 (L-3) = 0

On simplification L = 0, 0 and 3.