Each eigenvalue of $A$ is equal to $\pm 1$. Why is $A$ similar to $A^{-1}$? $A$ is a non-singular matrix ($n \times n$) and each eigenvalue of $A$ is equal to $\pm 1$. Why is $A$ similar to $A^{-1}$? (by Jordan form)
 A: Note that if 
$$
A = \pmatrix{A_1\\&\ddots\\&&A_n}
$$
then 
$$
A^{-1} = \pmatrix{A_1^{-1}\\&\ddots\\&&A_n^{-1}}
$$
So, it suffices to show that a Jordan block on $1$ or $-1$ is similar to its own inverse.
For now, take
$$
J = \pmatrix{1&1\\&1&1&\\&&&\ddots &1\\ &&&&1}
$$
Following the explanation here, we find
$$
J^{-1} = 
\pmatrix{
1&-1&1&-1&\cdots\\
&1&-1&1 & \cdots\\
&& \ddots\\
&&&&1
}
$$
It then suffices to note that $J^{-1}$ has $1$ as its only eigenvalue and that the matrix $J^{-1} - I$ has rank $n-1$.
A similar argument can be applied to the Jordan block on $-1$.
A: Suppose $Ax=\lambda x$. Then $A^{-1} (Ax) = x = A^{-1} \lambda x = \lambda A^{-1} x.$ So $1/\lambda$ is an eigenvalue of $A^{-1}$ with the same eigenvector $x$. But if $\lambda = 1$ then $1/\lambda = 1$, or if $\lambda = -1$ then $1/\lambda = -1$. So $A^{-1}$ has the same eigenvalues as $A$ with the same geometric multiplicities. Can we conclude that they are similar from here? (We certainly can in the diagonalizable case.)
