Let $V$ be a finite-dimensional unitary space and let $A\in\mathcal L(V)$ s. t. $\operatorname{rank}(A)=k>0$. Show there exist an orthonormal set $\left\{e_1,e_2,\ldots,e_k\right\}$ and vectors $f_1,f_2,\ldots,f_k\in V$ s. t. $$Ax=\sum_{i=1}^k\langle x, f_i\rangle e_i,\ \forall x\in V$$
My thoughts:
By definition, $\operatorname{rank}(A)=\dim\mathscr Im(A)=k>0$.
If we take some $x\in\mathscr Im(A)$, then $Ax=\sum\limits_{i=k}^k\alpha_i e_i\ne0,\ \alpha_i\in\Bbb F$, and if we keep in mind that $\left\{e_1,e_2,\ldots,e_k\right\}$ is an orthonormal set, then $\left\{e_1,e_2,\ldots,e_k\right\}$ might be an orthonormal basis for $\mathscr Im (A)$, so $$\mathscr Im(A)=\operatorname{span}\left\{e_1,e_2,\ldots,e_k\right\}.$$
The orthonormal set $\left\{e_1,e_2,\ldots,e_k\right\}$ can be extended to an orthonormal basis $\left\{e_1,e_2,\ldots,e_k,e_{k+1},\ldots,e_n\right\}$ for the whole $V$.
If we consider $e_j\in\mathscr Im(A),\ \forall j\in\{1,\ldots,k\}$, which means$Ae_j=\sum\limits_{i=1}^k\left\langle e_j,f_i\right\rangle e_i\ne0$, I thought $A$ could be represented by a block matrix $$[A]_e^e=\begin{bmatrix}E&0\\0&0\end{bmatrix}, E\in M_k(\Bbb F),$$ but I don't think I have proven anything.
The only thing I'm sure of is that $A$ is, indeed, linear, because, $$\begin{aligned}A(\lambda x+\mu y)&=\sum_{i=1}^k\langle \lambda x+\mu y,f_i\rangle e_i\\&=\sum_{i=1}^k\left(\langle \lambda x, f_i\rangle e_i+\langle \mu y,f_i\rangle e_i\right)\\&=\sum_{i=1}^k\lambda\langle x,f_i\rangle e_i+\sum_{i=1}^k\mu\langle y,f_i\rangle e_i\\&=\lambda\sum_{i=1}^k\langle x,f_i\rangle e_i+\mu\sum_{i=1}^k\langle y,f_i\rangle e_i\\&=\lambda Ax+\mu Ay\end{aligned}$$
May I ask for advice on how to solve this task?
Thank you in advance!
Let $\{e_1,...,e_k\}$ be an orthonormal basis of $\mathscr Im(A).$ If $x \in V$, then $Ax \in \mathscr Im(A).$ Hence there are $s_1,...,s_k \in \mathbb F$, such that
$$Ax= \sum_{j=1}^k s_j e_j.$$
Then we get $<Ax,e_j> = s_j.$ Thus $<x, A^*e_j>=s_j.$
$f_j:=A^*e_j, \quad j=1,...k$ will do the job.