Suppose $V$ is an infinite-dimensional vector space (dimension $N$ over the field $k$), and suppose $A$ is an $(N \times N)$-matrix over $k$ which defines a bijective linear transformation $f$ on $V$. (We assume the Axiom of Choice and work relative to a fixed base.)
My first naive question is: since $f$ has a unique inverse, does this mean that $A$ also has a unique inverse ?
Now for my real question.
Suppose $A, B, C$ are $(N \times N)$ $(0,1)$-matrices such that $AB = CA$ (where we assume that the products are well defined), and such that $B$ and $C$ are permutation matrices. (Note that $N$ is still supposed to be infinite.)
When can we conclude that $\mathrm{trace}(B) \ne 0$ if and only if $\mathrm{trace}(C) \ne 0$ ?
Can even more be concluded about the traces of $B$ and $C$ (such as in the finite-dimensional case) ?
EDIT: in a first version, I assumed $A$ to be a permutation matrix instead of $B$ and $C$, and then the answer appears to be that $B$ and $C$ have the same trace. But when $A$ is not necessarily a permutation matrix while $B$ and $C$ are, in general this property does not seem to be true anymore ...