$A$ is diagonalizable $\iff A^{-1}$ is diagonalizable.

Question

I used that proposition while doing an exercise and later realized it hadn't been demonstrated in class so i proceeded trying to prove it myself as it follows: Let $A\in M_n(\Bbb{K})$ be invertible, than exists $A^{-1}\in M_n(\Bbb{K})$; Let then V be a vector space and $L_A=:f\in End(V)$. Now if $A$ is diagonalizable there exist a base $\mathscr{B}=\left( v_1,\dots,v_n \right)$ for V such that $$M_{\mathscr{B}}\left(f\right)= \begin{pmatrix} \lambda_1 & 0 & \dots & \dots & 0 \\ 0 & \lambda_2 & 0 & \dots & 0 \\ \vdots & 0 & \ddots & & \vdots \\ \vdots & \vdots & & \ddots & 0\\ 0 & 0 & \dots & 0 & \lambda_n \\ \end{pmatrix} $$ Now since $f\circ f^{-1}=Id_V$, it must be true that $M_{\mathscr{B}}\left(f\right) \cdot M_{\mathscr{B}}\left(f^{-1}\right)=Id_n$; therefore $$ M_{\mathscr{B}}\left(f^{-1}\right)= \begin{pmatrix} 1\over\lambda_1 & 0 & \dots & \dots & 0 \\ 0 & 1\over\lambda_2 & 0 & \dots & 0 \\ \vdots & 0 & \ddots & & \vdots \\ \vdots & \vdots & & \ddots & 0\\ 0 & 0 & \dots & 0 & 1\over\lambda_n \\ \end{pmatrix} $$ and so we found that $A$ being diagonalizable implies $A^{-1}$ is diagonalizable. Now, since the inverse of $A^{-1}$ is actually $A$, for the same argument we can conclude that $A^{-1}$ being diagonalizable implies $A$ is diagonalizable and therefore the thesis. Is my argument correct? Could I prove it quicker? Because I believe my proof is correct as much as I believe it could be done in a smarter way.

Answer 1

Your argument is correct. A quicker way could be the following:
$A$ is diagonalizable $\iff$ $D=BAB^{-1}$ with $D$ diagonal (and invertible since so is $A$) and $B$ invertible $\iff$ $BA^{-1}B^{-1}=(BAB^{-1})^{-1}=(D)^{-1}=D^{-1}$ $\iff$ $A^{-1}$ is diagonalizable ($D^{-1}$ is clearly diagonal as is the inverse of a diagonal matrix).