Show that if A is diagonalizable and 1 and −1 are the only eigenvalues of $A$, then $A^{-1 }= A$. I'm not quite sure how to approach this, I can see how the eigenvalues will be on the diagonal and finding the inverse would result in $A^{-1}=A$. However, please give a formal explanation. 
 A: Since $A$ is diagonalizable and $-1$ and $1$ are only eigenvalues so minimal polynomial must contain distinct linear factors i.e. $m(t)=(t-1)(t+1)$ which implies $t^2-1=0$ or $A^2-I=0\implies A^2=I$. Multilply by $A^{-1}$ to get $A^{-1}=A$.
A: One approach for this is:
$ A = P[ Diag( \pm 1, \dots, \pm 1 ) ] P^{-1}$ 
Thus,
$$A^2 = P[ Diag( \pm 1, \dots, \pm 1 ) ] P^{-1} P[ Diag( \pm 1, \dots, \pm 1 ) ] P^{-1} = P[ Diag( (\pm 1)^2, \dots, (\pm 1)^2 ) ] P^{-1} $$
A: Here is a sketch of a proof:


*

*$A$ is diagonalizable, so there exists invertible matrix $P$ such that $P^{-1}AP = \Lambda$, a diagonal matrix. Since similar matrices have same eigenvalues, the eigenvalues of $\Lambda$ are 1’s and -1’s, that is, the entries of the diagonal of $\Lambda$ are 1’s and -1’s.

*$P^{-1}A^2P = P^{-1}APP^{-1}AP =\Lambda^2 = I$, the identity matrix. Therefore, multiplying $P$ on the left and $P^{-1}$ on the right, you get $A^2=I$. Now you are done.
You may also argue something like any polynomial with a square matrix plugged in is with the eigenvalues that you get from plugging in the eigenvalues of the original matrix into the polynomial, and in your question the polynomial is the square function.
