Eigenvectors of the following diagonal matrix

Question

Let $x = \frac{2 i \pi k}{n}$ where $1 \leq k \leq \frac{n}{2} - 1$, where $n = 2k$ even. $$ A = \begin{bmatrix} e^{ix} & 0 \\ 0 & e^{-ix} \end{bmatrix}$$ I'm trying to find the eigenvectors of of this matrix. The eigenvalues are clearly $e^{ix}$ and $e^{-ix}$. So, let $ \begin{bmatrix} a \\ b \end{bmatrix}$ denote the eigenvector. Then $$ \begin{bmatrix} e^{ix} & 0 \\ 0 & e^{-ix} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = e^{ix} \begin{bmatrix} a \\ b \end{bmatrix}$$ So, we get that $a = 1$ from the first equation. Solving for $b$ is my problem. I get $e^{-ix}b = e^{ix}b$, so $b = e^{2ix}b$, then $b(1 - e^{2ix}) = 0$, but then I get stuck. The eigenvalues are supposedly $$\begin{bmatrix} 1 \\ i \end{bmatrix}, \begin{bmatrix} 1 \\ -i \end{bmatrix}$$

Answer 1

Hint:

The eigenvalues are $\mathrm e^{ix}$ and $\mathrm e^{-ix}$. This requires the eigenvectors have each one coordinate equal to $0$.

Answer 2

To find the first Eigenvector, you have to solve

$$\begin{bmatrix}e^{ix}-e^{ix}&0\\0&e^{-ix}-e^{ix}\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}0&0\\0&e^{-ix}-e^{ix}\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}.$$

This is immediate.