Prove the following set of unit vectors is orthogonal.

Question

Suppose ${v_1 , v_2 , ……..v_n}$ are unit vectors in $\mathbb R^n$ such that $ || v||^2 = \sum_{n=1}^{\infty} | <v_i , v>|^2 $ for all $v \in\mathbb R^n $ Then I have to prove that the set of unit vectors is orthogonal set of vectors .

I have proved it. Can anyone please check if it is okay or not.

My Attempt::

EDIT : $[v_1 , v_2 , ..... ,v_n]$ are unit vectors in $\mathbb R^n$. Let's say $A$ bea matrix whose rows are the vectors $[v_1 , v_2 , ..... ,v_n]$. Now $||v||^2 = v^tv$ where $v\in \mathbb R^n$ and $\sum |<v_i , v>|^2 = ||Av||^2 = v^tA^tAv$.

$v^tv $ and $ v^tA^tAv$ are non negative so $v^tv = v^tA^tAv$ which implies $v^t(A^tA - I)v = 0 $

So $(A^tA - I)$ is skew symmetric matrix as well as a symmetric matrix so $(A^tA - I)$is a zero matrix. So $A$ is an orthogonal MatrIX. So the rows are mutually orthogonal and $[v_1 , v_2 , ..... ,v_n]$ is a basis of $\mathbb R^n$.

I have deleted the photo of my attempt I have uploaded here. Instead I wrote my attempt in MathJax.

Answer 1

Your proof is right except a minor mistake. In line 6, you should have written$$||Av||^2=v^TA^TAv$$not $$||Av||^2=(v^TA^TAv)^2$$any way, Your proof is correct and elegant.

Answer 2

I'm not going to try and decipher that picture. But there is a very easy proof:

Choose one of the unit vectors $v_k$, and set $v=v_k$ in the equation $|| v||^2 = \sum_{i=1}^n \left| \left<v_i , v\right>\right|^2$ (assuming that this is what you meant to write). This gives \begin{align} ||v_k||^2 & = \sum_{i=1}^n \left| \left<v_i , v_k\right>\right|^2\\ & = ||v_k||^2 + \sum_{i\ne k} \left| \left<v_i , v_k\right>\right|^2 \end{align}

Hence $\sum_{i\ne k} \left| \left<v_i , v_k\right>\right|^2=0$, which means that $v_k$ is orthogonal to every other $v_i$. So the set is orthogonal.