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Let $A \in \mathrm{U}(n) \subset \mathbb{C}^{n \times n}$ a unitary matrix. Show that: $\exists ~ S\in \mathrm{U}(n)$ so that

$\bar{S^t}AS=D:=\begin{pmatrix}\lambda_1&&0\\&\ddots & &\\0&&\lambda_n\end{pmatrix}$

where $\lambda_i\in \mathbb{C}$ are eigenvalues of $A$. Show moreover, that $|\lambda_i|=1$.

My idea is to show that the orthogonal complement $W$ of an eigenvector $v$ of $A$ from $A$ is mapped into itself, meaning $AW \subset W$. But how to start this?

Let $\lambda\in \mathbb{C}$ be an eigenvalue and $v \in \mathbb{C}$ a corresponding eigenvector with $|v|$=1. Let $W\perp v = \{w \in \mathbb{C}^n|\langle w,v\rangle=0\}$ an orthogonal vector subspace to $v$.

${\langle Aw,v\rangle}=\overline{\langle v,Aw\rangle}=\overline{\langle v,\lambda w\rangle}=\lambda \overline{\langle v,w\rangle}=\lambda\cdot 0=0$,
so $Aw\in v^{\perp}=W$.

So now choose a basis of $W$ with normalized vectors $v_2,\ldots,v_n$ perpendicular to each other (Gram–Schmidt process). Be $v_1:=v$ and $S:=(v_1,v_2,\ldots,v_n) \in \mathrm{GL}(n,\mathbb{C})$. Then one has: $\langle v_i,v_i\rangle=1\ \forall i$, $\langle v_i,v_j\rangle=0\ \forall i \neq j $. Let $w_j:=Av_j\in W, j=2,\ldots,n$

$\Rightarrow \overline{S}^tAS=\overline{S}^t(\lambda v_1,w_2,\ldots,w_n)=\begin{pmatrix}\lambda&0&0&0\\0&& & &\\0&&B\\0\end{pmatrix}.$

Is this correct so far? Why is $S$ orthogonal?

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The last part follows like this: you have $I_n=AA^*=SDS^*SD^*S^*=SDD^*S^*$. Now focus on the equality $I_n=SDD^*S^*$ and try to conclude that the eigenvalues are on the unit circle. For the first part, do you know anything about diagonalizability of normal matrices? – Git Gud Jun 22 '14 at 20:24
    
How to start? By assuming that $w\perp v$, and looking at $\langle Aw,v\rangle$. You want to see that that is $0$. You have an inner product in which a matrix appears in one argument. What is a very common operation in such circumstances? – Daniel Fischer♦ Jun 22 '14 at 20:43
    
The property $|\lambda|=1$ follows from the very definition of unitarity $\|Ax\|=\|x\|$. – Peter Franek Jun 22 '14 at 21:43
    
Are you aware of the spectral theorem? – Omnomnomnom Jun 23 '14 at 23:20
    
@Omnomnomnom No, spectral theorem is unknown for us so far – fear.xD Jun 24 '14 at 8:17
up vote 0 down vote

How to show $A(v^\perp) \subset v^\perp$ for an eigenvector $v$:

Consider any $u \in v^\perp$. We have $$ \langle Au,v\rangle = \frac 1 {\bar\lambda} \langle Au,\lambda v\rangle = \frac 1 {\bar\lambda} \langle Au, Av\rangle = \frac 1 {\bar\lambda} \langle u, v\rangle =0 $$ In your proof, you assume that $w$ is an eigenvector, which is an invalid assumption.

For your second question: a matrix $S$ is orthogonal (unitary) if and only if its columns form an orthonormal basis of $\mathbb{R}^n$ ($\mathbb{C}^n$). To see why this is the case, note that $S^*S$ is the matrix of pairwise inner products of columns of $S$. That is, if $s_1,\dots,s_n$ are the columns of $S$, then $$ (S^*S)_{ij} = \langle s_j,s_i \rangle $$

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