I have a complex square matrix with the only eigenvalues being $1$ and $-1$. Show that $A$ and the inverse of this matrix $A^{-1}$ are similar

Question

I was thinking I use the formula for Jordan blocks I have $J=P^{-1} AP$ where

$$J= \left[\begin{array}{rr} 1 & 0 \\0 &-1 \end{array}\right]$$

Then $J^{-1}=P A^{-1}P^{-1}$

I can note $J=J^{-1}$

Then could I set the two sides equal? As in

$P^{-1} AP=P A^{-1}P^{-1}$ when simplifying $A=A^{-1}$

Thus proving similarity? I'm not entirely sure if my logic here is sound however.

Answer 1

If you regard your basis of eigenvectors $(v_i)_{i=1}^n$ then you know \begin{align*} Av_i = \pm v_i\quad \text{and} \quad AAv_i = v_i \end{align*} Now you know every $x\in \mathbb{C}^n$ can be written as $\sum_{i=1}^n x_i v_i$ for certain $x_i \in \mathbb{C}$. By using the linearity of the matrix multiplication you receive \begin{align*} AAx = AA\sum_{i=1}^n x_i v_i = \sum_{i=1}^n x_i AAv_i = \sum_{i=1}^n x_i v_i = x \end{align*} So $A$ is its own inverse.