I am required to prove the following Claim.
Let $\{v_1,v_2,...,v_k\}$ be an orthogonal set in $V$, and let $a_1,a_2,...,a_k$ be scalars. Prove that $$\left|\left|\sum_{i=1}^{k}a_iv_i\right|\right|^2 = \sum_{i = 1}^{k}|a_i|\cdot ||v_i||^2$$
The following is my attempt is it correct?
Proof. We construct the proof by recourse to Mathematical Induction. For $k=1$ by considering the properties of norm we have $||\sum_{i=1}^{k}a_iv_i||^2 = ||a_1v_1||^2 = |a_1|^2\cdot||v_1||^2 = \sum_{i = 1}^{k}|a_i|\cdot ||v_i||^2$.
Assume now for an arbitrary $j\in\{1,2,...,k-1\}$ that $||\sum_{i=1}^{j}a_iv_i||^2 = \sum_{i = 1}^{j}|a_i|\cdot ||v_i||^2$. then $||\sum_{i=1}^{j+1}a_iv_i||^2 = ||\sum_{i=1}^{j}a_iv_i+a_{j+1}v_{j+1}||^2$. We now show that $\langle\sum_{i=1}^{j}a_iv_i,a_{j+1}v_{j+1}\rangle = 0$.
By considering the conjugate homogeneity of $\langle\cdot,\cdot\rangle$ in second component followed by linearity in first component we may simplify the inner product $\langle\sum_{i=1}^{j}a_iv_i,a_{j+1}v_{j+1}\rangle$ to yield $\sum_{i=1}^{j}\overline{a_{j+1}}a_i\langle v_i,v_{j+1}\rangle$ furthermore since the set $\{v_1,v_2,...,v_k\}$ is orthogonal $\langle v_i,v_{j+1}\rangle = 0$ consequently the inner product $\langle\sum_{i=1}^{j}a_iv_i,a_{j+1}v_{j+1}\rangle = 0$.
Having shown that $\sum_{i=1}^{j}a_iv_i$ and $a_{j+1}v_{j+1}$ are orthogonal it follows by $\textbf{(10)}$ that $||\sum_{i=1}^{j}a_iv_i+a_{j+1}v_{j+1}||^2 = ||\sum_{i=1}^{j}a_iv_i||^2 +||a_{j+1}v_{j+1}||^2$. From the inductive hypothesis $ ||\sum_{i=1}^{j}a_iv_i||^2 = \sum_{i = 1}^{j}|a_i|^2\cdot ||v_i||^2$ and by considering the properties of norm as in the base case we have $||a_{j+1}v_{j+1}||^2 = |a_{j+1}|^2\cdot||v_{j+1}||^2$ consequently we have $\||\sum_{i=1}^{j+1}a_iv_i||^2 = \sum_{i = 1}^{j}|a_i|^2\cdot ||v_i||^2+||a_{j+1}v_{j+1}||^2 = \sum_{i = 1}^{j+1}|a_i|^2\cdot ||v_i||^2$
$\blacksquare$
Note: $(10)$ here is simply the result that $||x+y||^2 = ||x||^2+||y||^2$ whenever $x$ and $y$ are orthogonal.