Show that:
if $Cond_2(A)=1$ then $A=\alpha Q$; where $Q$ is an orthonormal matrix and $\alpha$ is a real number$\neq 0$
The condition number is defined as theorem as
$Cond_2(A)= ||A|| \hspace{0.1cm}||A^{-1}||$
In this case, the operator norm is $L_2$.
The condition number theorem is extracted from the book: Numerical Linear Algebra, Lloyd N. Trefethen, David Bau, First Edition, page 95.
My work first idea
If the condition number is is equal to 1 then
$$||A||=\frac{1}{||A^{-1}||} \Leftrightarrow |\sigma_{max}|=|\sigma_{min}|$$
Where $\sigma_{max}$ and $\sigma_{min }$ is the eigenvalue with the biggest and lowest norm respectively.
From here we can conclude that. If the condition number is one then all the eigenvalues have the same norm $|\sigma_{i}|=1$.
Main problem
How to conclude from this result that the matrix A has the form of $\alpha Q$
My work second idea
Using SVD we obtain that.
$$||Q\Sigma V^*||||(Q\Sigma V^*)^{-1}||=||\Sigma|||\Sigma^{-1}|||= |\sigma_{max}|\frac{1}{|\sigma_{min}|}=1$$
From the last equation, we obtain the same results of My work first idea
The norms of all eigenvalues are equal to one.
General overview
How to conclude that the structure of A is $A=\alpha Q$