Squared norm of sum of vectors

Question

Consider $v_1,\ldots,v_m\in\mathbb{R}^n$ and let $z=v_1+\cdots+v_m$. I am currently trying to show that $$\|z\|^2=\sum_{i=1}^m\|v_i\|^2+2\cdot\sum_{1\leq i<j\leq m}\langle v_i,v_j\rangle$$ I have tried writing out long sums of inner products to find the pattern, and I did, but that is just too cumbersome to write as an argument. Any hints on how to clarify that the identity in fact holds while still keeping the argument formal?

Answer 1

Well it is possible to do it without induction, let $ n\in\mathbb{N}^{*} $, we have :

\begin{aligned} \left\Vert z\right\Vert^{2}=\left\langle z,z\right\rangle=\left\langle\sum_{i=1}^{n}{v_{i}},\sum_{i=1}^{n}{v_{i}}\right\rangle=\sum_{1\leq i,j\leq n}{\left\langle v_{i},v_{j}\right\rangle}&=\sum_{1\leq i=j\leq n}{\left\langle v_{i},v_{j}\right\rangle}+\sum_{1\leq i\neq j\leq n}{\left\langle v_{i},v_{j}\right\rangle}\\ &=\sum_{i=1}^{n}{\left\Vert v_{i}\right\Vert^{2}}+\sum_{1\leq i<j\leq n}{\left\langle v_{i},v_{j}\right\rangle}+\sum_{1\leq j<i\leq n}{\left\langle v_{i},v_{j}\right\rangle} \\ &=\sum_{i=1}^{n}{\left\Vert v_{i}\right\Vert^{2}}+\sum_{1\leq i<j\leq n}{\left(\left\langle v_{i},v_{j}\right\rangle+\left\langle v_{j},v_{i}\right\rangle\right)}\\ \left\Vert z\right\Vert^{2}&=\sum_{i=1}^{n}{\left\Vert v_{i}\right\Vert^{2}}+2\sum_{1\leq i<j\leq n}{\left\langle v_{i},v_{j}\right\rangle}\end{aligned}

Answer 2

I am assuming here that $z$, and each $v_i$, are all real.

$z = \sum_{i=1}^m v_i$ and $\|z\|^2 = \langle z , z \rangle$. Therefore we can write:

\begin{align} \|z\|^2 = \langle z , z \rangle &= \langle \sum_{i=1}^m v_i , \sum_{j=1}^m v_j \rangle \\ & = \sum_{i=1}^m \sum_{j=1}^m \langle v_i , v_j \rangle \quad \quad (\text{by linearity}) \\ & = \sum_{i=1}^m \langle v_i , v_i \rangle + \sum_{i=1}^m \sum_{j\neq i}\langle v_i , v_j \rangle + \sum_{j=1}^m \sum_{i\neq j}\langle v_i , v_j \rangle \\ & = \sum_{i=1}^m \|v_i\|^2 + 2\sum_{i=1}^m \sum_{i<j}\langle v_i , v_j \rangle \quad \quad (\text{by symmetry}) \end{align} The last line is just due to the symmetry of inner product ($\langle v_i , v_j \rangle = \langle v_j , v_i \rangle$). So when I did $i\neq j$, I can just rewrite it as $i<j$ and then double each summation.

I hope this makes some sense. I recommend you try it with a simple example ($m=3$) and you will see why this makes sense.