Finding the eigenvector of a 3x3 matrix

Question

I'm trying to work on finding eigenvectors. For the following matrix,

$$ \begin{bmatrix} b & 0 & -a \\ 0 & 0 & 0 \\ a & 0 & -b \\ \end{bmatrix} $$

I get the system $$ \left\{ \begin{array}{c} bx-az=0 \\ ax-bz=0 \end{array} \right. $$

which becomes $\frac{b}{a} = \frac{a}{b}$. Because it was part of a larger question, I know \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}

is a solution. But why?

Answer 1

Substituting $x=0, y=1, z=0$ into the equations,

\begin{align} b\cdot0-a\cdot0=0\\ a\cdot0-b\cdot0=0 \end{align}

we see that it fulfills the conditions. Hence it is a solution.

If you multiply that vector to the original matrix, we can see that it is in the nullspace, the corresponding eigenvalue is $0$.

Answer 2

Well, if you multiply your matrix by the vector $(0,1,0)$, you get $0$ times the original vector, so $(0,1,0)$ is an eigenvector with eigenvalue 0.

Answer 3

The columns of a matrix are the images of the basis vectors. The second column of your matrix $A$ is zero, so you know at a glance that $(0,1,0)^T$ is a solution to $A\mathbf x=0$.