# Normal matrix $A= \left( \begin{smallmatrix}2 & i \\ i & 2 \end{smallmatrix} \right)$

Given the normal matrix $$A= \left( \begin{array}{cc} 2 & i \\ i & 2 \end{array} \right)$$ what is the unitary matrix $$P$$ with positive entries in the first column and the second row and for which $$P^* AP$$ is diagonal?

I have proceeded in this way: first I find there are two eigenvalues $$2 \pm i$$ and then I got corresponding eigen vectors as $$\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$$ and $$\left( \begin{array}{c} 1 \\ -1 \end{array} \right)$$ so if I write $$P =\left( \begin{array}{cc} 1& 1 \\ 1 & -1 \end{array} \right)$$ then $$P^* AP$$ gives a diagonal matrix.

I am confused if this process is ok, or if you can say some better way. Also the P which I found is not satisfying the asked P. Please say some hint, where I am getting wrong.

You have found a matrix $P$ of eigenvectors which has orthogonal columns, such that $P^* A P$ is diagonal. Only one requirement has not been satisfied: the columns of $P$ are not unit. That is easy enough to fix by dividing each column by its norm, so you get
$$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}$$
• @radha Well, your problem is asking for $P^*$ in my notation. As written my matrix is actually Hermitian, but you could change the order of the eigenvectors and eigenvalues, and then $P^*$ would be $\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$, which seems to match what you want. – Ian May 3 '15 at 14:53