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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.

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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

$$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}$$

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  • $\begingroup$ but in the question it is asked that first column and second row has all positive entries, so if we change the eigenvectors accordingly, it is not giving a diagonal matrix. There is something wrong in my way, please check it once. $\endgroup$ – Sankha May 3 '15 at 14:49
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    $\begingroup$ @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. $\endgroup$ – Ian May 3 '15 at 14:53
  • $\begingroup$ thanks...fool me, I was taking product without considering transpose. $\endgroup$ – Sankha May 3 '15 at 15:00

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