Is the diagonalization of A Invertable? There is a theorem which says that given a diagonalizable matrix $A$ such that $P^{-1}AP=D$ if $D$ is invertible then A is invertible.
I suspect that the other direction isn't true, but I can't think of a counter example.
 A: The other direction is true. Suppose $A$ is invertible. We have that $D = P^{-1}AP$, a product of invertible matrices, and thus invertible. In fact, $D^{-1} = P^{-1}A^{-1}P$: we can verify that $D(P^{-1}A^{1}P) = (P^{-1}A^{1}P)D = I$.
More generally, if $A \cong B$ ($A$ and $B$ are similar matrices), $A \text{ is invertible} \Longleftrightarrow B \text{ is invertible}$. To prove this, suppose $A \cong B$ and $A$ is invertible. Then, there exists an invertible matrix $P$ such that $B = P^{-1}AP$, and $B$ has inverse $B^{-1} = P^{-1}A^{-1}P$. Conversely, if $A \cong B$ are similar and $B$ is invertible, since similarity is an equivalence relation, $B \cong A$ with $B$ invertible and we're back in the implication proved, so $A$ is invertible.
A: You know that your matrix $P$ is invertible. Now working with determinants you don't only know that $A$ is invertible, but that it even has the same determinant as $D$:
$$\det(D)=\det(P^{-1}AP)=\det(P^{-1})\det(A)\det(P)=\det(P^{-1})\det(P)\det(A)$$
$$\det(P^{-1})\det(P)\det(A)=\det(P^{-1}P)\det(A)=1\cdot\det(A)=\det(A)$$
