Let $\{e_1,\ldots,e_k\}$ be an orthonormal set in a unitary space $V$. Prove:
$$\|x\|^2=\sum\limits_{i=1}^k|\langle x,e_i\rangle|^2\iff x\in\operatorname{span}\{e_1,\ldots,e_k\}$$
My attempt:
My concerns are precision and conciseness of my proof, as well as deduction and order of conclusions.
$\color{brown}{\Rightarrow}$
Let $M=\operatorname{span}\{e_1,\ldots,e_k\}$. Since $\{e_1,\ldots,e_k\}$ is an orthonormal set, $\implies\{e_1,\ldots,e_k\}$ is an orthonormal basis for $M$. Therefore, according to this post regarding the Parseval's identity , the statement simply follows from the proof given in the answer by @MattL. :
$$\|x\|^2=\langle x,x\rangle=\sum_{i=1}^k\langle x, e_i \rangle \overline {\langle x, e_i \rangle}=\sum_{i=1}^k|\langle x,e_i\rangle|^2$$
$\color{brown}{\Leftarrow}$
Having in mind $M^\perp:=\{z\in V\mid\langle y,z\rangle=0,\ \forall y\in M\}, M\oplus M^\perp=V$, we can write: $x=y+z, y\in M,z\in M^\perp$.
Then:
$$\|x\|^2=\langle x, x\rangle=\langle y+z, y+z\rangle=\langle y,y\rangle+\underbrace{\langle z,y\rangle}_{0} +\underbrace{\langle y,z\rangle}_{0} +\langle z,z\rangle=\langle y,y\rangle+\langle z,z\rangle\tag 1$$
Then:
$$y=\sum_{i=1}^k\langle y,e_i\rangle e_i=\sum_{i=1}^k\langle x-z,e_i\rangle e_i=\sum_{i=1}^k(\langle x,e_i\rangle-\langle z,e_i\rangle) e_i=\sum_{i=1}^k\langle x,e_i\rangle e_i\tag 2$$
Now,from $(1)\ \&\ (2)$ we obtain: $\color{Maroon}{\langle z,z\rangle=0}$ and, therfore, $\color{blue}{x=y\in M}$
May I ask if this is legitimate and what I could do to improve my proof?
P. S. I've seen some posts dealing with similar problems, but most of them include Hilbert spaces which we haven't learned yet. Thank you in advance!
Best Answer
You can directly verify that $\|x-\sum\limits_{i=1}^{k} \langle x, e_i \rangle e_i\|^{2}=0$ by just expanding the left side and using definition of inner product.