Proof of the following equality with vectors

Question

Let $\{v_1,v_2,\dots,v_n\}$ be an orthogonal set in $V$, and let $a_1,a_2,\dots,a_n$ be scalars. Prove that $$\left\Vert \sum_{i=1}^na_iv_i \right\Vert^2=\sum_{i=1}^n|a_i|^2\Vert v_i\Vert^2$$

Here's what I've tried, but I don't know if it is correct: $$\left\Vert \sum_{i=1}^na_iv_i \right\Vert^2=\left<\sum_{i=1}^n a_iv_i, \sum_{i=1}^n a_iv_i\right>=\sum_{i=1}^na_i\left<v_i, \sum_{i=1}^na_iv_i \right>=\sum_{i=1}^na_i\left(\sum_{i=1}^n\overline{a_i} <v_i,v_i>\right)=\sum_{i=1}^na_i\left(\sum_{i=1}^n\overline{a_i}\ \Vert v_i\Vert^2 \right)$$ I think that when proving this, I should also relate to the fact that the vectors are orthogonal.

Answer 1

I think the only mistake in your work is the fact that the two summations are linked.

Instead, you should have \begin{align} \left\| \sum^n_{j=1} a_j v_j\right\|^2 =&\ \left\langle \sum^n_{j=1} a_j v_j, \sum^n_{i=1} a_i v_i\right\rangle\\ =&\ \sum^N_{i=1}\sum^N_{j=1} \bar a_ja_i\langle v_j, v_i\rangle \\ =&\ \sum^N_{i=1}\sum^N_{j=1} \bar a_ja_i \delta_{i, j}\|v_i\|^2 = \sum^N_{j=1} |a_j|^2\|v_i\|^2. \end{align}

Answer 2

As an alternative, in matrix form we have

$$M=\begin{bmatrix}v_1&v_2&\ldots&v_n\end{bmatrix},\:a^T=\begin{bmatrix}a_1&a_2&\ldots&a_n\end{bmatrix}$$

then

$$ \sum_{i=1}^na_iv_i =Ma \implies \left\Vert \sum_{i=1}^na_iv_i \right\Vert^2=\left<Ma, Ma\right>=a^TM^TMa=a^T\begin{bmatrix}v_1^Tv_1&v_1^Tv_2&\ldots&v_1^Tv_n\\v_2^Tv_1&v_2^Tv_2&\ldots&v_2^Tv_n\\\ldots&\ldots&\ldots&\ldots\\v_n^Tv_1&v_n^Tv_2&\ldots&v_n^Tv_n\end{bmatrix}a=$$

$$=a^T\begin{bmatrix}v_1^Tv_1&0&\ldots&0\\0&v_2^Tv_2&\ldots&0\\\ldots&\ldots&\ldots&\ldots\\0&0&\ldots&v_n^Tv_n\end{bmatrix}a=\sum_{i=1}^na_i^2(v_i^Tv_i)=\sum_{i=1}^na_i^2\|v_i\|^2$$