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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.

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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 $Tx=\sum\limits_{k=1}^{n} \langle x, v_i \rangle w_i$

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  • $\begingroup$ Is there another way to get that? I have never studied Closed Graph Theorem and it is not on my course curriculum. $\endgroup$ Aug 27 '21 at 5:55
  • $\begingroup$ @JoanGuastalla Closed Graph Theorem was not required after all!. (Any linear map on a finite dimensional space is continuous). I have edited the answer. $\endgroup$ Aug 27 '21 at 6:29

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