# Linear Algebra: proving a decomposition of vector to orthonormal basis

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?

• @Masacroso hey but sorry, i cannot still connect your suggestion to the solution for above. – hadi k Nov 12 '18 at 17:15
• 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. – Mefitico Nov 12 '18 at 18:09

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

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 $$v=\sum_i \langle v,u_i\rangle u_i$$ 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 $$j$$.

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: $$\mathbf{v} = UU^T\mathbf{v} = \sum_{i} \mathbf{u}_i \left< \mathbf{u}_i, \mathbf{v}\right>$$