I am learning about diagonalizing a matrix, one of the theorem is that given $P$ some nonsingular matrix, then we can find a $\Sigma$ such that $P^{-1} \Sigma P$ = $A$, where $\Sigma$ is pure diagonal, and $A$ non-diagonal.
But then this would mean that (multiplying $P$ to the other side) $P^{-1} A P$ = $\Sigma$.
But we know that A is not a diagonal matrix, but $\Sigma$ is. Wouldn't the equation above indicate that $A$ is a diagonal matrix as well?
No. Let $A=\begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}$.
and $P=\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$
and $D=\begin{pmatrix} 3 & 0\\ 0 & 1 \end{pmatrix}$
Then $AP=PD$, i.e. $P^{-1}AP=D$ but $A$ is not diagonal.