How to get the eigenvalue of this matrix Let $\theta \in \mathbb R$, and let $T\in\mathcal L(\mathbb C^2)$ have canoncial matrix
$M(T)$ = $$ 
            \left(
            \begin{matrix}
            1 & e^{i\theta}  \\
            e^{-i\theta} & -1  \\
            \end{matrix}
            \right)
         $$
(a) Find the eigenvalues of $T$.
(b) Find an orthonormal basis for $\mathbb C^2$ that consists of eigenvectors for $T$.
I can get the eigenvalues of T and they are $\sqrt 2$ and $-\sqrt 2$. However, I cannot get each eigenvector respect to each eigenvalue. I know how to get eigenvectors by calculating the null space of $(T - \lambda I)$, but it looks like this is not a proper method to solve this problem. So, anyone can help? Thank you! 
 A: The characteristic polynomial of your matrix is
$$
X^2-2
$$
so indeed the eigenvalues are $\sqrt{2}$ and $-\sqrt{2}$.
The eigenvectors relative to $\sqrt{2}$ are the vectors $\begin{pmatrix}x_1\\x_2\end{pmatrix}$ such that
$$
(1-\sqrt{2})x_1+e^{i\theta}x_2=0
$$
so you get one by choosing $x_2=1-\sqrt{2}$ and $x_1=-e^{i\theta}$.

 The eigenspace has dimension $1$ meaning the matrix $M(T)-(1-\sqrt{2})I$ has rank $1$, so any one of the equation can be taken.

Similarly, an eigenvector $\begin{pmatrix}x_1\\x_2\end{pmatrix}$ relative to $-\sqrt{2}$ satisfies
$$
(1+\sqrt{2})x_1+e^{i\theta}x_2=0
$$
so you can choose $x_2=1+\sqrt{2}$ and $x_1=-e^{i\theta}$.
Note that these vectors are orthogonal, because
$$
\begin{pmatrix}
-e^{i\theta}\\
1-\sqrt{2}
\end{pmatrix}^{\!H}
\begin{pmatrix}
-e^{i\theta}\\
1+\sqrt{2}
\end{pmatrix}
=(-e^{-i\theta})(-e^{i\theta})+(1-\sqrt{2})(1+\sqrt{2})=1+1-2=0
$$
Just normalize them, noting that the first has norm
$$
\sqrt{1+(1-\sqrt{2})^2}=\sqrt{4-2\sqrt{2}}
$$
and, similarly, the second one has norm $\sqrt{4+2\sqrt{2}}$.
A: Finding the null space of $A - \lambda I$ always work. Let us find for example the eigenvectors of $T$ corresponding to the eigenvalue $\sqrt{2}$. Denote by $v = (x, y)^T$ such an eigenvector. We must have
$$ Av = \begin{pmatrix} 1 & e^{i \theta} \\ e^{-i \theta} & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + e^{i \theta} y \\ e^{-i \theta} x - y \end{pmatrix} = \sqrt{2} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \sqrt{2}x \\ \sqrt{2}y \end{pmatrix}. $$
Thus, we get a system of two linear equations for $v$:
$$ x + e^{i\theta}y = \sqrt{2}x, \\ e^{-i\theta} x - y = \sqrt{2}y. $$
Written differently, we have:
$$ (1 - \sqrt{2})x + e^{i \theta} y = 0, \\ e^{-i \theta} x - (\sqrt{2} + 1)y = 0. $$
Since this homogeneous system must have a non-zero solution, we know that the equations must be linearly dependent and so we can choose to solve only one of the equations. The first equation implies that $x = \frac{e^{i \theta}}{\sqrt{2} - 1} y$. Choosing $y = 1$ we obtain the eigenvector $v = \left( \frac{e^{i \theta}}{\sqrt{2} - 1}, 1 \right)^T$ and the eigenspace associated with $\sqrt{2}$ is
$$ \mathrm{span} \left \{ \begin{pmatrix} \frac{e^{i \theta}}{\sqrt{2} - 1} \\ 1 \end{pmatrix} \right \}. $$
