# Bounded Linear Operator on Hilbert Space - Finite Dimensional Image Case

Let $$T:H \to H$$ be a Bounded Linear Operator on a Hilbert Space $$H$$ and consider that $$T(H)$$ is finite dimensional. Such a $$T$$ operator have the represensation $$\begin{equation*} Tx= \sum_{i=1}^n \langle x,v_i\rangle w_i\,\, \end{equation*}$$

where $$w_i,v_i \in H$$ and $$n=dim(T(H))$$.

Given that $$T(H)$$ is finite dimensional then exist a basis $$P=\{p_1,...,p_n\} \in H$$ of $$T(H)$$. Since $$T$$ is linear the inverse image $$T^{-1}(P)$$ must be LI. Using Gram-Schmidt Process on $$P$$ we can obtain an orthonormal basis $$V=\{v_1,...,v_n\}$$ such that $$W=T(V)=\{w_1,...,w_n\}$$ still being a basis for $$T(H)$$. Since $$H$$ is Hilbert Space and $$span(V)$$ is complete we can get the decomposition $$H=span(V)\bigoplus span(V)^{\perp}$$. Then, given $$x \in H$$ we have that

$$\begin{equation*} Tx= T(\sum_{i=1}^n \langle x,v_i\rangle v_i + v^{\perp}) = \sum_{i=1}^n \langle x,v_i\rangle w_i + T(v^{\perp}) \end{equation*}$$

It looks like I should get that the Kernel of $$T$$ must be $$span(V^{\perp})$$. I am aware that i have not used the fact that $$T$$ is bounded and a mapping between the same Hilbert Space.

Hints: If $$(w_i=)Tu_i, 1\leq i \leq n$$ be a basis for $$T(H)$$ the we can wriet $$Tx= \sum\limits_{k=1}^{n} c_iTu_i$$. The coefficients are unique. Let $$1\leq i \leq n$$. Then $$Tx \to c_i$$ is a continuous map because any linear map on a finite dimensional space is automaticallty continuous. Combine this with $$T$$ to see that $$x \to c_i$$ is a continuous linear map on $$H$$. By Riesz Represnetation Theorem we can write $$c_i= \langle x, v_i \rangle$$ for some $$v_i$$ so we have $$Tx=\sum\limits_{k=1}^{n} \langle x, v_i \rangle w_i$$