Why can't all diagonalizable matrices be decomposed into $Q\Lambda Q^T$?

Question

I am a beginner Linear Algebra student and I have the following question:

If a matrix is diagonalizable matrix can be decomposed into $A = S\Lambda S^{-1}$, where $S$ is a matrix of $A$'s eigenvectors, can I simply orthogonalize via Gram-Schmidt and then normalize those vectors to change $S$ into $Q$, and then use $A = Q\Lambda Q^T$?

Answer 1

Because if $A$ can be diagonalized with an orthogonal matrix, eigenvectors relative to different eigenvalues are orthogonal. This may not happen for a diagonalizable matrix.

Indeed, suppose $Av=\lambda v$ and $Aw=\mu w$, with $\lambda\ne\mu$. Then $$ \lambda\langle v,w\rangle= \langle \lambda v,w\rangle= \langle Av,w\rangle= \langle Q\Lambda Q^Tv,w\rangle= \langle v,Q\Lambda^TQ^Tw\rangle= \langle v,Aw\rangle= \mu\langle v,w\rangle $$ and so $(\lambda-\mu)\langle v,w\rangle=0$, hence $\langle v,w\rangle=0$ (since $\Lambda$ is diagonal, $\Lambda=\Lambda^T$).

If you apply Gram-Schmidt to the columns of $S$ there's no reason for the matrix $Q$ you get to have eigenvectors of $A$ as columns.

By the way, this shows that if a real matrix $A$ is diagonalizable with an orthogonal matrix, then it is symmetric.