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$$A=\begin{pmatrix} 1&i&i\\ i&1&i\\ i&i&1 \end{pmatrix}$$

I have proved that $A$ is normal. Now I want to find a unitary matrix $P$ such that $P^*AP$ is a diagonal matrix ($P^*$ is the conjugate transpose of $P$).

The eigenvalues of $A$ are: $\{1+2i, 1-i\}$ and the eigenvectors: $\{(1,1,1)^t, (-1,0,1)^t,(-1,1,0)^t\}$

How can I get this matrix $P$?

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Hint: for a unitary matrix $P$ one has $P^*=P^{-1}$. So you are looking for an orthonormal (the unitary condition) basis of eigenvectors (these will be the columns of $P$). You're almost there... – Marc van Leeuwen Oct 22 '12 at 8:21
Thank you very much – Hiperion Oct 22 '12 at 13:27
up vote 1 down vote accepted

Choose the eigenvectors corresponding to $1-i$ so they are orthogonal to $(1,1,1)^T$ and each other. For example, replace the third eigenvector by $(1,-2,1)^T$. Then normalize the eigenvectors. This gives: $$P = \pmatrix{ 1 & -1 & 1 \\ 1 & 0 & -2 \\ 1 & 1 & 1 }\pmatrix{\frac{1}{\sqrt{3}} & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0\\0 & 0 & \frac{1}{\sqrt{6}} }$$ Then $P^* A P$ is diagonal.

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When you normalize the vector $(1,−2,1)^T$, you need to change $\frac{1}{2}$ by $\frac{1}{\sqrt{6}}$... but actually $P$(with this change) is unitary and $P^*AP$ is diagonal. Thank you very much! – Hiperion Oct 22 '12 at 15:03
Thanks for catching that! – copper.hat Oct 22 '12 at 15:25

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