Does invertible diagonalizable matrix have its inverted matrix diagonalizable?

Question

From invertibility, $(PAP^{-1})^{-1}=PA^{-1}P^{-1}=D^{-1}$, the thing is how do you make sure the $P$ doesn't change?

Answer 1

You've essentially proved that $P$ doesn't change. I'm assuming that your $A$ matrix is the original, invertible, diagonalizable matrix, and $D$ is the diagonal matrix. So, writing what you did in a different way,

$$ A^{-1} = (PDP^{-1})^{-1} = (P^{-1})^{-1}D^{-1}P^{-1} = PD^{-1}P^{-1} $$

Hence you have expressed the inverse matrix in the form

$$ A^{-1} = P\tilde{D}P^{-1} $$ where $\tilde{D}$ is a diagonal matrix and $P$ is the same matrix of eigenvectors as before. This is the beauty of diagonalization - all you have to do is invert a diagonal matrix $D$, which is easy - you don't have to touch the $P$ matrix.