$\displaystyle

A = \left[\begin{matrix}-3 & 0 & 5 \\ 0 & 2 & 0 \\ 5 & 0 & -3\end{matrix}\right]

$

I have found the eigenvalues for the above matrix to be, $\displaystyle \lambda = 2 $, it occurs twice.

When i plug my eigenvalue of 2 back into the matrix to find the eigenvectors i get;

$\displaystyle y=0, x = z $

So i said my chose my first eigenvector, v1 to be;

$\displaystyle

A = \left[\begin{matrix}1 \\ 0 \\ 1 \end{matrix}\right]

$

Trouble is i need another 2 vectors V2 and V3 such that:

v1.v2 = 0

v1.v3 = 0

v2.v3 = 0

where v1, v2, v3 are all unit vectors.

Im hoping to find the matrix O where v1,v2,v3 are O's columns

I posted this question earlier but i think its phrasing was a bit messy so i tried using LaTex for the first time...