Question:
Let $ r \in \mathbb{R} $ and let $ T \in \mathcal{L}(\mathbb{C}) $ be the linear map with canonical matrix $$ T = \begin{pmatrix} 1 & -1 \\ -1 & r \\ \end{pmatrix} $$
Given this matrix, find an orthonormal basis of $ \mathbb{C^2} $ consisting of eigenvectors of $ T $
Approach:
I found the eigenvalues of $ T $ which came out to be
$$ \lambda = \frac{1 + r \pm \sqrt{r^2 -2r + 5}}{2} $$
The eigenvectors I found came out to be
$$ v_1 = (1, \frac{1 - r- \sqrt{r^2 -2r + 5}}{2}), v_2 = (1, \frac{1 - r + \sqrt{r^2 -2r + 5}}{2}) $$
At this point I'd like to apply Gram-Schmidt however, the computation of the length of each vector is quite large:
$$ g_1 = \frac{(1, \frac{1 - r- \sqrt{r^2 -2r + 5}}{2})}{||(1, \frac{1 - r- \sqrt{r^2 -2r + 5}}{2})||} $$ $$ ||(1, \frac{1 - r- \sqrt{r^2 -2r + 5}}{2})|| = \sqrt{1 \cdot 1 + \frac{1 - r- \sqrt{r^2 -2r + 5}}{2}) \cdot \overline{\frac{1 - r- \sqrt{r^2 -2r + 5}}{2})}}$$
In particular, I'm not sure what to do with the conjugate within the square root to simplify the process since there isn't going to necessarily be a complex component if $ r^2 -2r + 5 > 0 $
Thanks!