How is this matrix being transformed?

Question

I have a matrix: \begin{pmatrix}\frac{1}{2}&\frac{1}{2}&-\frac{\sqrt{2}}{2}\\ \frac{1}{2}&\frac{1}{2}&\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}&0\end{pmatrix}

which has eigenvalues $\lambda = 1, i, - i$

My task is to make it diagonal with a block: \begin{pmatrix}\cos(a)& -\sin(a)\\ \sin(a)& \cos(a)\end{pmatrix} Where $a$ is not equal to $k \pi$ for some $k \in \mathbb{Z}$

In my taskbook the solution is: \begin{pmatrix}1&0&0\\ 0&0&1\\ 0&-1&0\end{pmatrix}

How can I do this?

Answer 1

For a general complex number $\lambda =re^{i\theta} =r(\cos\theta +i\sin\theta)$, we have $\bar\lambda =re^{-i\theta} =r(\cos\theta - i\sin\theta)$.
Find a basis transformation (over $\Bbb C$) that proves that the diagonal matrix $\pmatrix{\lambda & 0\\ 0&\bar\lambda}$ is similar to the real matrix $\pmatrix{r\cos\theta & - r\sin\theta \\ r\sin\theta & r\cos\theta}$.

Then just apply this for $\lambda=i$ ($r=1,\,\theta=\frac\pi2$).

Answer 2

Start with a matrix of form $$P=\begin{pmatrix} \cos(a) & -\sin(a) & 0\\\sin(a) & \cos(a) & 0\\0 & 0 & 1\end{pmatrix}$$ and compute $P^{-1}AP$. Then it should be clear which $a$ will work,