# Thread: Norm of sum of vectors

1. ## Norm of sum of vectors

Given vectors $x_{1}, x_{2}, ..., x_{k}\in \mathbb{R}^{n}$, what is $|x_{1} + x_{2} + ... + x_{k}|^{2}$?

So $|x_{1} + x_{2}|^{2} = |x_{1}|^{2} + |x_{2}|^{2} + 2x_{1}\cdot x_{2}$ and extending it to the sum of three vectors it seems to be $|x_{1}|^{2} + |x_{2}|^{2} + |x_{3}|^{2} + 2( x_{1}\cdot x_{2} + x_{1}\cdot x_{3} + x_{2}\cdot x_{3})$. So it seems to be that for the norm of $k$ vectors we have $|x_{1} + x_{2} + ... + x_{k}|^{2} = |x_{1}|^{2} + |x_{2}|^{2} + ... + |x_{k}|^{2} + 2(\sum x_{i}x_{j})$ for $i\ne j, 1\le i, j \le k$. Is this correct and is there a better way to formulate this? And how would I go about giving a formal proof for this problem? Thanks.

Edit: I think I found a more simplified formulation: $\sum_{j=1}^{k}\sum_{i=1}^{k} x_{i}\cdot x_{j}$, but how would I go about giving a formal proof? Thanks.

2. The norm of a vector can always be written as follows:

$\|x\|=\sqrt{\langle x|x\rangle},$

where $\langle x|y\rangle$ is the inner product of vector $x$ with vector $y$. The inner product is just a generalization of the dot product. Therefore,

$\|x\|^{2}=\langle x|x\rangle.$ It follows that

$\|x+y\|^{2}=\langle (x+y)|(x+y)\rangle=\langle x|x\rangle+\langle x|y\rangle+\langle y|x\rangle+\langle y|y\rangle.$

I think this line of reasoning should enable you to see how to do your proof. How would you start?