# About the canonical form of an finite rank operator

functional-analysisoperator-theory

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.

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 $$T$$.