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So I am reviewing Linear Algebra and looking to understand why this inequality holds.

I have that $V$ is a real vector space (with no precursor included here that it is finite dimensional. I have that $\langle{,}\rangle$ is an inner product and that $\{ e_1,e_2,...,e_k\}$ is an orthonormal set of vectors of $V$. From this information I want to prove that for $v\in V$ $$\sum_{i=1}^k \langle{v,e_i}\rangle^2 \leq ||v||^2$$

I am confused about how I can go about showing this. I now that, as $\{ e_1,e_2,...,e_k\}$ is an orthonormal set of vectors they are linearly independent in $V$, but I don't think this means I know I can extend them to a basis for $V$, as there is always an orthonormal basis for a finite dimensional space $V$ (by Gram-Schmidt) but I do not know this holds for infinite dimensional space.

I was thinking, if I could get an orthonormal basis, I could write $v$ as a linear combination of these orthonormal vectors, and then expand out, cancelling the $\langle{v_i,v_j}\rangle$ vectors when $i\neq j$, but as this doesn't seem possible to me, how do I prove this inequality?

Thanks.

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Note that$$\sum_{i=1}^k\langle v,e_i\rangle e_i\text{ and }v-\sum_{i=1}^k\langle v,e_i\rangle e_i$$are orthogonal and their sum is $v$. Therefore\begin{align}\lVert v\rVert^2&=\left\lVert\sum_{i=1}^k\langle v,e_i\rangle e_i\right\rVert^2+\left\lVert v-\sum_{i=1}^k\langle v,e_i\rangle e_i\right\rVert^2\\&\geqslant\left\lVert\sum_{i=1}^k\langle v,e_i\rangle e_i\right\rVert^2\\&=\sum_{i=1}^k\langle v,e_i\rangle^2.\end{align}

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Hint: Since $e_1,\dots,e_k$ are orthonormal, $$ P(v)=\sum_{i=1}^k \langle v,e_i\rangle e_i $$ is the orthogonal projection $V\to\operatorname{span}\{e_1,\dots,e_k\}$. In particular, $\lVert P(v)\rVert^2\leq\lVert v\rVert^2$ follows.

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