Cfrgtkky

I want to transpose my vector $v$ to an arbitrary orthonormal basis $U = {u_1,u_2, u_3}$.

Which would be,

$v = sum_i langle u_i cdot v rangle u_i =sum_i u_i^Tvu_i$

How do I prove the above decomposition is correct?

You could have specified the coordinates of the vector in the new base to be $c_i$, with $c_i = <u_i, v>$.

That being said, you only need to prove one thing:

$$
sum u_i c_i = v
$$

Which is already done by definition.

You want to show that
$$
v=sum_i langle v,u_irangle u_i
$$
If you do
$$
leftlangle v-sum_i langle v,u_irangle u_i,u_jrightrangle=
langle v,u_jrangle-langle v,u_jranglelangle u_j,u_jrangle=0
$$
A vector $w$ is zero if and only if $langle w,u_jrangle=0$ for every $j$.

To specify the vector $mathbf{v} in mathbb{R}^n$ in a general different basis $U = left [ mathbf{u}_1 dots mathbf{u}_nright ]$, where $mathbf{u}_i in mathbb{R}^n, forall i;$, you need to find a $mathbf{v}'$ such that:

$$U mathbf{v}' = mathbf{v}$$

For the general basis the coordinates of $mathbf{v}$ in the basis $U$ is given by (multiply by the inverse in both sides):
$$mathbf{v}' = U^{-1}mathbf{v}$$

In the case where $U$ is an orthonormal basis(means that $U$ is orthogonal matrix), we know that:

$$U^TU = I = UU^T$$

Hence $U^{-1} = U^T$, therefore $mathbf{v}'$ becomes:

$$mathbf{v}' = U^T mathbf{v}$$

Now I can prove your decomposition in a simple way:
$$mathbf{v} = U mathbf{v}' = Uleft( U^T mathbf{v}right) = UU^T mathbf{v} = mathbf{v}$$

Note that this is exactly your formula:
$$mathbf{v} = UU^Tmathbf{v} = sum_{i} mathbf{u}_i left< mathbf{u}_i, mathbf{v}right>$$

Hope this answers your question.