Let $A$ be a diagonalizable matrix, show that $A^{-1} = A$ This is the question in full:

Let $A$ be a diagonalizable matrix whose eigenvalues are all either -1
  or 1. Show that $A^{-1} = A$.

Since it's diagonalizable then I think we should use $A = PDP^{-1}$ but I'm not sure how to prove it, because I also know that it might involve $|A -\lambda I | = 0 $ or $Ax=\lambda x$
 A: If $A$ is diagonalizable then you can write
$$A = PDP^{-1}$$
then
$$A^{-1} = (PDP^{-1})^{-1}=PD^{-1}P^{-1}$$
but $D$ is diagonal with $\pm 1$ then $D^{-1}$ is also diagonal with $(\pm 1)^{-1}=\pm 1$, so $D^{-1}=D$ and then
$$A^{-1} = PD^{-1}P^{-1}= PDP^{-1}=A$$
A: You're on the right track. A diagonalisation contains most of the $|A-\lambda I| = 0$ and $Ax = \lambda x$ in it already, so you don't have to worry much about that. With that said, if $A = PDP^{-1}$ is a diagonalisation, what are the values in $D$? Then what is $(PDP^{-1})^2$?
A: As $A$ is invertible, the equality $A^{-1}=A$ is equivalent to $A^2 = I$.
And if $A= PDP^{-1}$, you have $A^2=PD^2P^{-1}=I$.
A: As $A$ is diagonalisable, $\mathbb{R}^n$ (where $n$ is the dimension of the matrix) has a basis of eigenvectors $\{v_1, \ldots, v_n\}$.
The corresponding eigenvalues are either $+1$ or $-1$, so $A(A(v_i)) = A(v_i) = v_i$ in the former case, and $A(A(v_i)) = A(-v_i) = -A(v_i) = --v_i = v_i$ in the latter case. So always $A^2(v_i) = v_i$. It follows that for any $x \in \mathbb{R}^n$ (that we can write as a linear combination of $v_i$), we also have $A^2(x) =x$. As $A$ is invertible (no $0$-eigenvalue), it follows that $A^{-1} = A$.
