length of sum of orthonormal basis vectors

Question

Let $U\in\mathbb{R}^{D\times D}$ be an orthogonal basis,

$x:=\sum_{i=1}^D u_i$

Is there an easy way to see that

$\|x\|_2 = \sqrt{D}$

?

Answer 1

We have:

$$\|x\|_2^2 = \langle x , x \rangle = \sum_{1\le i, j\le D} \langle u_i,u_j\rangle$$

Since $u_i$ are orthonormal, $\langle u_i, u_j\rangle = \delta_{ij}$. So $\|x\|^2 = \sum_{i=1}^D 1 = D \therefore \|x\|_2 = \sqrt{D}$.