Proving an inequality in inner product spaces

Question

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)?

Answer 1

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: $$\|v\|^2 = \sum_{i=1}^n \frac{\langle v_i, v\rangle^2}{\|v_i\|^4} \|v_i\|^2 = \sum_{i=1}^n \frac{\langle v_i, v\rangle^2}{\|v_i\|^2} \ge \sum_{i=1}^k \frac{\langle v_i, v\rangle^2}{\|v_i\|^2}.$$

Answer 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$$\lVert v\rVert^2=\sum_{j=1}^n\frac{\langle v,v_j\rangle^2}{\lVert v_j\rVert^2}\geqslant\sum_{j=1}^k\frac{\langle v,v_j\rangle^2}{\lVert v_j\rVert^2}.$$

Answer 3

Your attempt is on exactly the right track.

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 $\|v\|^2 = \sum_{i=1}^n \langle u_i,v \rangle^2$

Answer 4

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.