Given an $n \times n$ matrix $A$ is diagonalizable where, $A$ is non-symmetric i.e. $A^T\not= A$
And let $S = \{ \vec{v}_1 \,,\, \ldots \,,\, \vec{v}_n\}$ the set of the eigenvectors of $A$ , ($S$ is not orthogonal)
I understand that $A$ isn't orthogonally diagonalizable since it's not symmetric.
But another equivalent statement about the conditions of orthogonal diagonalization, is that the set of the eigenvectors of $A$ has to be orthonormal.
Hence, $$ A \text{ is symmetric } \Longleftrightarrow S \text{ is orthonormal} $$
But what if we apply Gram-Schmidt to $S$ to get a new orthonormal set $S'$ doesn't this allow $A$ to be orthogonally diagonalizable, even though $A$ was not symmetric, to begin with?
I tried to understand the flaw in this line of reasoning and, the conclusion that I landed on is that, it may have to do with the fact that the Gram-Schmidt process knocks every vector from $S$ off their span -except the initial vector $\vec{v}_k = \vec{u}_1$- to the span of the orthonormal component of its projection over the vectors before it.
Meaning, the Gram-Schmidt preserved the total span of $S$ but, not necessarily the span of each vector in $S$,
then $S'$ is a different set in this sense and, no longer represents the eigenvectors of $A$.
Hence, $A$ is still not orthogonally diagonalizable.
I'd love to know if I'm close to the right answer or missing something.