Question about Orthonormal Bases

Question

I solved this question but I am not quite sure if it is a valid solution. The question is:

Let $u$ be a vector in an inner product space $V$ and let $\{v_1, v_2, v_3, \ldots, v_n\}$ be an orthonormal basis for $V$. Show that if $a_i$ is the angle between $u$ and $v_i$ we have*

$$\cos^2(a_1) + \cos^2(a_2) + \cdots + \cos^2(a_n) = 1$$

My approach was that:

The vector zero is included in both of the bases. Other than the zero vector, all vectors have an angle of $90^\circ$ between them. However, since the vector zero is in both of them there is an angle of $0^\circ$. Hence, the equation is true.

Is this a valid solution? Does any vector exist that opposes this? What would be a valid solution? Thank you for your help.

Answer 1

Because $\{v_{1},...,v_{n}\}$ is an orthonormal basis, you know that:

(1) $||v_{k}||=1$ for every $k=1,...,n$.

(2) $u=\sum_{k=1}^{n}\alpha_{k}v_{k}$, with $\alpha_{1},...,\alpha_{n}$ scalars and $||u||^{2}=\sum_{k=1}^{n}\alpha_{k}^{2}$.

(3) $\langle u, v_{k}\rangle = \alpha_{k}$

Now, for each $k \in \{1,...,n\}$ we have (using (1)): $$\cos a_{k} = \frac{\langle u, v_{k}\rangle}{||u||||v_{k}||} = \frac{\langle u, v_{k}\rangle}{||u||}.$$ Thus, using (3) and (2) we get: $$\cos^{2}a_{1}+\cdots+\cos^{2}a_{n} = \sum_{k=1}^{n} \frac{\langle u, v_{k}\rangle}{||u||} = \sum_{k=1}^{n} \frac{\alpha_{k}^{2}}{||u||^{2}} = \frac{1}{||u||^{2}}\sum_{k=1}^{n}\alpha_{k}^{2} = \frac{1}{||u||^{2}}||u||^{2} = 1$$

Answer 2

Sorry, but I think your solution for $u=\sum\limits_k c_k v_k$ is totally wrong.
But here's the one possible approach, IMHO:
We recall $\cos a_k=\frac{v_k.u}{|v_k||u|}=\frac{v_k.u}{|u|}$ so $c_k=|u|\cos a_k$ (why?)
Let's say, for simplicity, $|u|=1$, so we need to show $\sum\limits_k c_k^2=1$.
But from where can we get $\sum\limits_k c_k^2$? Maybe considering something like $\left(\sum\limits_k c_k\right)^2=\sum\limits_k c_k^2+ 2\sum\limits_{i\ne j} c_ic_j$ where like all $c_ic_j$ somehow cancel to $0$?
And indeed, as we have $v_iv_j=0$ for $i\ne j$ we consider $\left(\sum\limits_k c_k v_k\right)^2=\sum\limits_k c_k^2 v_k^2+2\sum\limits_{i\ne j} c_ic_jv_i.j_j=\sum\limits_k c_k^2$.
The LHS $=u^2$ and the RHS $=|u|^2\sum\limits_k\cos^2 a_k$, hence the desired result.