# when is $u = \sum_{i=1}^n \langle u,v_i\rangle v_i$?

I am preparing for an exam, and don't have solutions for this exercise.

What condition has to be met for $v_1,...,v_n \in \mathbb R^n$ so that every $u \in \mathbb R^n$ can be written as
$$u = \sum_{i=1}^n \langle u,v_i\rangle v_i$$

I found that when $u_i$ is an orthonormal basis then $\langle v,w\rangle = \sum^n_{i=1}\langle v,u_i\rangle \langle u_i,w\rangle$.

## 4 Answers

Every $u\in\Bbb{R}^n$ is a linear combination of $v_1,\cdots,v_n$. So that family of $n$ vectors form a basis of $\Bbb{R}^n$. Now one has

$$\forall i,\,v_i=\sum_{j=1}^n\langle v_i,v_j\rangle v_j$$

$v_1,\cdots v_n$ being a basis we have

$$\langle v_i,v_j\rangle=\delta_{ij}$$

$\delta_{ij}=1$ if $i=j$ and $0$ otherwise and so the basis is orthonormal relative to the inner product $\langle \cdot,\cdot\rangle$

All $v_i$'s must be orthonormal i.e. $||v_i|| =1 \forall i \in[1,n] and <v_i,v_j>=0$ $i \neq j$.
If $v_i$'s are orthogonal, they will be linearly independent and as the dimension of $R^n$ is $n$. They will form a basis for $R^n$.

I am not sure how helpful will this be but basically you can say that the invariant measure of $v$ must be 1, that is $|\vec{v}|^2=1$. We see this taking a component of $u$, say $u_j$. Then $$u_j = u_j v_i v^i$$ thus we must have that $v_iv^i=1$ but $v_iv^i = |\vec{v}|^2$ the L2-norm squared, and since $|\vec{v}|^2 = 1$ we conclude that $\vec{v}$ is a radial unit vector in $S^n$.

Precisely if $\{v_1, \ldots, v_n\}$ is an orthonormal basis. Indeed, given $u \in \Bbb R^n$, we may write $u = \sum_{i=1}^n c_i v_i$. Now for each $j$, $\langle u, v_j \rangle = \sum_{i=1}^n c_i \langle v_i, v_j \rangle = c_j \langle v_j,v_j \rangle = c_j$ and it follows that $u = \sum_{i=1}^n \langle u,v_i \rangle v_i$.

Conversely, if for each $u$ it holds that $u = \sum_{i=1}^n \langle u, v_i \rangle v_i$, then given $j$, we have for $u = v_j$:

$$v_j = \sum_{i=1}^n \langle v_j , v_i \rangle v_i \tag1$$

It follows that:

$$\|v_j\|^2 = \langle v_j, v_j \rangle = \langle \sum_{i=1}^n \langle v_j , v_i \rangle v_i, v_j\rangle =\sum_{i=1}^n \langle v_i ,v_j \rangle ^2 = \|v_j\|^2 + \sum_{i \neq j} \langle v_i ,v_j \rangle ^2$$

So for each $i \neq j$, $\langle v_i, v_j \rangle = 0$, and returning to $(1)$, we see that $v_j = \|v_j\|^2 v_j$, so $v_j = 0$ or $\|v_j\|^2 =1$.

By the assumption that each $u$ can be written as $\sum_{i=1}^n \langle u,v_i \rangle v_i$, $\{v_1, \ldots, v_n\}$ spans $\Bbb R^n$, so it must be a basis, hence each $v_i$ is non-zero, and it follows that for all $j$, $\|v_j\|^2 = 1$ i.e. $\|v_j\| = 1$.