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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?

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  • $\begingroup$ @Masacroso hey but sorry, i cannot still connect your suggestion to the solution for above. $\endgroup$ – hadi k Nov 12 '18 at 17:15
  • $\begingroup$ To clarify: You want to prove mathematically? Or are you looking for some counter-validation (i.e. maybe you are programming a function, and is required to unit test your code). The result you provide (given that the basis is orthonormal) is almost the definition of the decomposition, so a bit more context on allowed assumptions would be needed if you are looking for a formal proof of sorts. $\endgroup$ – Mefitico Nov 12 '18 at 18:09
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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.

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You want to show that $$ v=\sum_i \langle v,u_i\rangle u_i $$ If you do $$ \left\langle v-\sum_i \langle v,u_i\rangle u_i,u_j\right\rangle= \langle v,u_j\rangle-\langle v,u_j\rangle\langle u_j,u_j\rangle=0 $$ A vector $w$ is zero if and only if $\langle w,u_j\rangle=0$ for every $j$.

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To specify the vector $\mathbf{v} \in \mathbb{R}^n$ in a general different basis $U = \left [ \mathbf{u}_1 \dots \mathbf{u}_n\right ]$, 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: $$\mathbf{v} = U \mathbf{v}' = U\left( U^T \mathbf{v}\right) = UU^T \mathbf{v} = \mathbf{v}$$

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

Hope this answers your question.

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