Finding diagonal and unitary matrices Let $A=\begin{pmatrix}
1 & 1+i\\ 
1-i & 2 
\end{pmatrix}$ I'm trying to find a diagonal matrix $D$ and a unitary matrix $U$ so that 
$U^\star AU=D$.
(We define $U^*=\overline{U}^t$ ).  
I found the eignvalues: $\lambda_1 = 3$, $\lambda_2 = 0$.
The eignvectors are: $V_1=\begin{pmatrix}
1+i\\2 
\end{pmatrix}$, $V_2=\begin{pmatrix}
-1-i\\1 
\end{pmatrix}$.
What should I do from here?
Thanks in advance.
 A: Since $U$ is unitary, $U^*=U^{-1}$. So, let us first think about how to find a matrix $U$ such that $U^{-1}AU$ is diagonal, without worrying about the condition that $U$ be unitary; we'll come back to fix that detail at the end.
Think about it this way: to diagonalize $A$, you want to find a change of basis: we know that the matrix for this transformation with respect to the basis $\{V_1,V_2\}$ is
$$
\begin{pmatrix}3 & 0\\0 & 0\end{pmatrix},
$$
because it maps $V_1$ to $3V_1$ and $V_2$ to $0V_2$.
If $U^{-1}AU$ is going to give you this, we want it to work like like so: when we plug in the vector $(1,0)^{T}$, we want $U$ to transform that in to $V_1$, so that $AU$ transforms it in to $3V_1$, so that $U^{-1}AU$ transforms it in to $(3,0)^T$. Does that make sense?
If $U$ is going to transform $(1,0)^T$ in to $V_1$, then the first column of $U$ must be exactly $V_1$. Similarly, the second column must be $V_2$. So, by this logic, we should take
$$
U=\begin{pmatrix}1+i & -1-i\\2 & 1\end{pmatrix}.
$$
For this matrix $U$, we certainly have $U^{-1}AU$ being the above matrix. However, there's one problem here: this matrix is not unitary.
So, instead of using $V_1$ and $V_2$ directly, let's use $W_1=\frac{V_1}{\|V_1\|}$ and $W_2=\frac{V_2}{\|V_2\|}$. If you take the matrix $U$ whose first column is $W_1$ and second column is $W_2$, you should get the properties you want.
In this case, we have
$$
\|V_1\|=\sqrt{(1+i)(1-i)+2^2}=\sqrt{6}\qquad \|V_2\|=\sqrt{(-1-i)(-1+i)+1^2}=\sqrt{3}
$$
and so we would take
$$
U=\begin{pmatrix}\frac{1+i}{\sqrt{6}} & \frac{-1-i}{\sqrt{3}}\\\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}}\end{pmatrix}
$$
You can check that in this case, $U^*=U^{-1}$.
