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Let $H$ be a Hilbert space, and let $T \colon H \to H$ be a bounded linear operator. Then how to show the following?

The range of $T$ is finite-dimensional if and only if $T$ can be represented in the form $$Tx = \sum_{j=1}^n \langle x, v_j \rangle w_j \ \ \ [ v_j, w_j \in H].$$

My effort:

Supppose that $\dim \mathscr{R}(T) = n$, and let $\{ e_1, \ldots, e_n \}$ be an orthonormal basis for $\mathscr{R}(T)$. Then, for every $x \in H$, we have $$Tx = \sum_{j=1}^n \langle Tx, e_j \rangle e_j = \sum_{j=1}^n \langle x, T^* e_j \rangle e_j,$$ where $T^*$ denotes the Hilbert adjoint operator of $T$. So we can take $w_j \colon= e_j$ and $v_j \colon= T^* e_j$.

Is this reasoning correct?

If so, then how to show the converse? Doest the above representation mean that the same $v_j$s and $w_j$s will be used for every $x \in H$? If so, then range of $T$ is spanned by the finitely many elements $w_1, \ldots, w_n$ and is clearly finite-dimensional.

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The reasoning is correct. For the second one: yes, the same $v_i$'s and $w_i$'s will be used for any $x$, and as you pointed out this means that $\Im(T)$ is spanned by the $w_i$'s.

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The reasonning seems correct to me. Note that the adjoint $T^*$ is well defined and independant of $x$. Thus the vectors $v_j, w_j$ that you find are well defined and do not depend on the $x$ you choose.

Thus for the second part, it suffices to observe that the range is finite dimensional because it is spanned by a finite number of vectors (that do not depend upon $x$, otherwise the result would be empty).

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