# Proving an inequality in inner product spaces

I'm studying for a upcoming exam and I've came across this problem:

Let $$\{v_1,...,v_k\}$$ be a set of non-zero orthogonal vectors in $$\mathbb{R}^n$$. Show that, for every $$v \in \mathbb{R}^n$$ $$\sum_{i = 1}^{k}\frac{\left \langle v_i,v\right \rangle ^2}{\left \| v_i \right \|^2} \leq \left \| v \right \|^2.$$

My first attempt was to complete the set $$\{v_1,...,v_k\}$$ to a orthogonal basis $$\{v_1,...,v_k,v_{k+1},...,v_n\}$$ of $$\mathbb{R}^n$$, and then write any vector $$v \in \mathbb{R}^n$$ as $$v = \sum_{i = 1}^{n}\left \langle v_i,v\right \rangle v_i$$, but this has lead me to nothing. Can anyone give me any hints (or even straightfoward answers)?

• Your idea of completing to an orthogonal basis is reasonable. If you are getting "lost" in the details, consider using the special case of the standard orthonormal basis in $\mathbb R^n$ bearing in mind that a change of basis is always possible between two orthonormal bases in a norm-preserving way. – hardmath Jul 23 '19 at 16:04

You're close; don't forget that $$v = \sum_{i=1}^n \langle v_i, v\rangle v_i$$ holds when $$\{v_1, \ldots, v_n\}$$ is orthonormal, not just orthogonal. The orthogonal version of this is obtained by normalising the orthogonal vectors: $$v = \sum_{i=1}^n \left\langle \frac{v_i}{\|v_i\|}, v\right\rangle \frac{v_i}{\|v_i\|} = \sum_{i=1}^n \frac{\langle v_i, v\rangle}{\|v_i\|^2} v_i.$$ Now, given the orthogonality, we may use Pythagoras' theorem: $$\|v\|^2 = \sum_{i=1}^n \frac{\langle v_i, v\rangle^2}{\|v_i\|^4} \|v_i\|^2 = \sum_{i=1}^n \frac{\langle v_i, v\rangle^2}{\|v_i\|^2} \ge \sum_{i=1}^k \frac{\langle v_i, v\rangle^2}{\|v_i\|^2}.$$
Actually, your did works, because then $$\left\{\frac{v_1}{\lVert v_1\rVert},\ldots,\frac{v_n}{\lVert v_n\rVert}\right\}$$ will be an orthnormal basis of $$\mathbb R^n$$ and therefore$$v=\sum_{j=1}^n\left\langle v,\frac{v_j}{\lVert v_j\rVert}\right\rangle\frac{v_j}{\lVert v_j\rVert}$$and so$$\lVert v\rVert^2=\sum_{j=1}^n\frac{\langle v,v_j\rangle^2}{\lVert v_j\rVert^2}\geqslant\sum_{j=1}^k\frac{\langle v,v_j\rangle^2}{\lVert v_j\rVert^2}.$$
Hint: With you basis $$\{v_1,\dots,v_n\}$$, define $$u_i = \frac{v_i}{\|v_i\|}$$ for $$1 \leq i \leq n$$. Now, note that $$v = \sum_{i=1}^n \langle u_i,v \rangle u_i$$, so that $$\|v\|^2 = \sum_{i=1}^n \langle u_i,v \rangle^2$$
No need to extend to a basis. Observe that with $$u_i:= \frac{v_i}{\|v_i\|}$$ we have $$v-\langle v,u_i\rangle\ u_i\text{ is orthogonal to } u_i$$ for each $$i$$ and so is their sum. Now use Pythagoras.