# Linear Algebra Orthogonal set inequality

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.

## 2 Answers

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}

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.