I expect that when we have a $n \times n$ matrix with non-degenerate eigenvalues, that is to say a matrix for which none of its $n$ eigenvalues have the same value, that there is a unique set of n normalized eigenvectors.
However for this matrix:
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0
\end{pmatrix}
With eigenvalues {$\lambda_1$,$\lambda_2$,$\lambda_3$} = {1, i, -i} its unnormalized eigenvectors can be:
$x_1$ = $(1,0,0)$, $x_2$ = $(0,i,1)$, $x_3$ = $(0,-i,1)$
Or
$x_1$ = $(1,0,0)$, $x_2$ = $(0,1,-i)$, $x_3$ = $(0,1,i)$
How come we're able to find two different sets of eignvectors?
Best Answer
Eigenvectors are never unique. In your case, all you can say is that each eigenspace is one-dimensional. Can you find the respective scalars that relate the two elements of each eigenspace that you found?