# Finding a unitary matrix that diagonalizes a given matrix

Let $$T=\begin{pmatrix}5 & 0 & 0 \\ 0 & 2 & i\\ 0 & -i & 2 \end{pmatrix}$$ be a Hermitian matrix.

I found the eigenvalues and eigenvectors already and they are $1,3,5$ and $\begin{pmatrix}0\\-i\\1\end{pmatrix}$,$\begin{pmatrix}0\\i\\1\end{pmatrix}$, and $\begin{pmatrix}1\\0\\0\end{pmatrix}$.

Normalizing each vector, I get $<1,0,0>$, $<0,\frac{-i}{\sqrt{2}},\frac{1}{\sqrt{2}}>$, and $<0,\frac{i}{\sqrt{2}},\frac{1}{\sqrt{2}}>$.

I need to find a matrix $P$ such that $P^*AP$ is diagonal. My first idea was $P=\begin{pmatrix}1 & 0 & 0\\ 0 & -i & i\\ 0 & 1 & 1 \end{pmatrix}$. $P^*=P^{-1}$, but $P^*P\neq I$.

I'm having a some problems trying to find a $P$ that will allow me to diagonalize $T$.

• Well you have to do what you did except that instead of putting the "random" eigenvectors you found as columns for the matrix, you put the normalized ones. May 29 '13 at 17:47
• You had found one using Mathematica in your previous question, hadn't you? May 29 '13 at 17:50
• What you wrote: "$\,P^*=P^{-1}\;$ but $\;P^*P\neq I\;$ " makes no sense... May 29 '13 at 22:42

$$P=\begin{pmatrix}1 & 0 & 0\\ 0 & \!\!-i & i\\ 0 & 1 & 1 \end{pmatrix}\;,\;\;P^*=\begin{pmatrix}1 & 0 & 0\\ 0 & i & 1\\ 0 & \!\!-i & 1 \end{pmatrix}$$

and since

$$PP^*\neq I\;\;\text{then}\;\;P^*\neq P^{-1}$$

The problem is this matrix's columns (rows) aren't orthonormal though they're orthogonal. We must apply Gram-Schmidt (I assume the usual euclidean inner product and let's write all the vector as row ones, for simplicity)):

$$u_1=(1,0,0)$$

$$w_2=(0,-i,1)-\langle\;(1,0,0),(0,-i,1)\;\rangle\,(1,0,0)=(0,-i,1)-0\cdot(1,0,0)=(0,-i,1)$$

and since

$$||w_2||=\sqrt 2\implies u_2=\frac1{\sqrt 2}(0,-i,1)$$

$$w_3:=(0,i,1)-\langle\;(0,i,1),(1,0,0)\;\rangle\,(1,0,0)-\left\langle\;(0,i,1),\frac1{\sqrt 2}(0,-i,1)\;\right\rangle\,\frac1{\sqrt 2}(0,-i,1)=$$

$$=(0,i,1)-\frac13\cdot 0=(0,i,1)$$

and since

$$||w_2||=\sqrt 2\implies u_3=\frac1{\sqrt2}(0,i,1)$$

Thus, our $\,P\,$ is (well, let's call it $\,Q\,$ to avoid confussion):

$$Q=\begin{pmatrix}1&0&0\\ 0&-\frac{i}{\sqrt 2}&\frac1{\sqrt 2}\\ 0&\frac i{\sqrt 2}&\frac1{\sqrt 2}\end{pmatrix}\;\implies\;Q^*=\begin{pmatrix}1&0&0\\ 0&\frac{i}{\sqrt 2}&-\frac i{\sqrt 2}\\ 0&\frac 1{\sqrt 2}&\frac1{\sqrt 2}\end{pmatrix}$$

It is now a simple exercise to verify that indeed $\,QQ^*=UI\,$ and, thus, $\,Q^{-1}=Q^*\,$ ...

Note: The above G-S process was almost trivial because the columns of $\,P\,$ were already orthogonal. The above stuff shows how to carry on the process in general, though in this case it was sufficient to divide each vector by its norm.