$ \operatorname{rank} \left( A - A^{-1} \right) = n $ for every invertible $n \times n$ matrix $A$? Given invertible matrix $A$ of size $n \times n$. Is it true that:
$$ \operatorname{rank} \left( A - A^{-1} \right) = n $$
if $A$ has no eigenvalue of magnitude 1, i.e., $|\lambda_i| \neq 1$ for all $i$. It's clear for diagonalizable matrices and it seems to hold for Jordan normal form as well.
 A: Yes, it is true. Note that
$$
\operatorname{rank}(A - A^{-1}) = 
\operatorname{rank}(A^{-1}[A^2 - I]) = 
\operatorname{rank}(A^2 - I).
$$
From there, it suffices to note that $A^2$ has full rank if and only if $A^2$ does not have $1$ as an eigenvalue, which holds if and only if $A$ has neither $1$ nor $-1$ as an eigenvalue. So, if $A$ does not have any eigenvalues of magnitude $1$, then $A^2 - I$ has full rank, which means that $A - A^{-1}$ has full rank.
A: Suppose that $A-A^{-1}$ is not invertible, there exists $x$ such that $(A-A^{-1})(x)=0$, this implies that $A(x)=A^{-1}(x)$ and $A^2(x)=x$. If $A(x)=cx$, $A^2(x)=c^2x=x$ implies that $|c|=1$. If $x,A(x)$ are independent,$A(x+A(x))=x+A(x)$. Contradiction.
A: If $J$ is Jordan matrix for $A$ then $A= P^{-1}JP$ for some invertibile $P$.
Then $$A-I = P^{-1}(J-I)P$$ and $$A+I = P^{-1}(J+I)P$$
Let $X= A-A^{-1}$ then $$XA = A^2-I =  P^{-1}(J-I)(J+I)P$$
so $$\det X \det A = \det (J-I)\cdot \det (J+I) \ne 0$$ and thus $\det X \ne 0$ so the claim is true if all $\lambda _i \neq \pm 1$.
