About the canonical form of an finite rank operator Let $\mathcal H$ be an seperable Hilbertspace. And let $T\colon \mathcal{H \to H}$ be an finite rank operator, that means $\textbf{dim range } \mathcal H= N<\infty$. In our lecture about operator theory there is the following statement:
There are some orthonormal system $\{v_k\}_k$ and $\{w_k\}_k$ and some $\lambda_k\in \mathbb C$ for $k=1,\dots,N$, such that
$$Tu=\sum_{k=1}^N\lambda_k\langle v_k,u\rangle w_k.$$
I tried to prove this statement. My first attempt was, lets just choose some ONB $\{w_k\}_k$ of the range of $T$. It follows then
$$Tu=\sum_{k=1}^N\langle w_k,Tu\rangle w_k=\sum_{k=1}^N\langle T^*w_k,u\rangle w_k.$$
So we would have $T^*w_k=\overline{\lambda_k}v_k$, but I do not see any reason why $T^*w_k$ should be orthogonal. So this does not work out.
My next guess was to use some prove via induction. So let the rank of $T$ be just 1, then for $w\in \textbf{range }T$ with $||w||=1$ we have
$$Tu=\varphi(u)w$$
for some linear bounded function $\varphi:\mathcal H \to \mathbb C$. With the Riesz representation theorem we have $\varphi(u)=\langle v,u\rangle$ for some $v\in \mathcal H$. So the statement is true for $N=1$.
Now let $N\in \mathbb N$ and assume the statement is true until $N-1$. Ok, we can now choose some $w \in \textbf{range } T$ with $||w||=1$ and moreover we can define the subspaces $U=\textbf{span }w$  and $W= U^\perp\cap \textbf{range }T$, where $\textbf{dim }W=N-1$. We have now $\textbf{range } T=U\oplus W$ and so we have the orthogonal projections $P_U$ and $P_M$. So we have now some $\{v_k\}_k$ and $\{w_k\}_k$ and $\lambda_k\in \mathbb C$ for $k=1,\dots,N-1$ such that
$$Tu=P_UTu+P_WTu=P_UTu+\sum_{k=1}^{N-1}\lambda_k\langle v_k,u\rangle w_k,$$
since $\textbf{rank }P_MT=N-1$. But $P_UTu$ is also uniquly given by some $v$ and $\lambda$ with
$$P_UT_M=\lambda\langle v,u\rangle w.$$
So there is again the same problem, why should $v$ be orthogonal to $v_k$?
What am I missing? I would be glad about some help here.
 A: I'm not sure if there is a straightforward way to do this. The usual way  is to use the Polar Decomposition and the Spectral Theorem, with the added benefit that we get $\lambda_k\geq0$. We can write $T=U|T|$, with $U$ a partial isometry. As $|T|$ is positive (in particular, selfadjoint) we can write
$$
|T|=\sum_{k=1}^n\lambda_k\,P_k,
$$
where $P_k$ are the orthogonal projection onto the eigenspace for $\lambda_k$. If we let $\{v_k\}$ be an orthonormal basis made out of orthonomal bases of each eigenspace, we get (abusing notation a bit, because now some $\lambda_k$ may be repeated and $n$ is possibly different)
$$
|T|x=\sum_{k=1}^n\lambda_k\langle v_k,x\rangle,v_k.
$$
Then
$$
Tx=U|T|x=\sum_{k=1}^n\lambda_k\langle v_k,x\rangle\,Uv_k
=\sum_{k=1}^n\lambda_k\langle v_k,x\rangle\,w_k.
$$
Since $U$ is an isometry on the range of $|T|$,
$$
\langle w_k,w_j\rangle=\langle Uv_k,Uv_j\rangle=\langle U^*Uv_k,v_j\rangle=\langle v_k,v_j\rangle=\delta_{k,j}, 
$$
so $\{w_k\}$ is orthonormal. The expression
$$
Tx=\sum_{k=1}^n\lambda_k\langle v_k,x\rangle\,w_k
$$
is the Singular Value Decomposition of $T$.
