The only detail I know is that $f^2+2f+I=0$, with $I$ being the identity matrix and 0 being the trivial transformation (takes every vector to zero). Also, we're talking about linear transformations and $V$ is finite-dimensional. I don't know how to approach this problem, but my guess is this:
- The expression contains an $f^2$, so the matrix $A$ associated with $f$ must be a squared matrix.
So that points in the direction of $f$ being an automorphism, because squared, regular matrices are all associated with a particular, bijective linear transformation. But I have no idea how to find out if the associated matrix is regular, so I can just stop now and conclude that there's no way for me to assure that $f$ is an automorphism. Is there a way to go the extra step and find a definitive answer?
From the equation it follows that $f(-f-2I)=I$ and $(-f-2I)f=I$ so that $-f-2I$ is an inverse for $f$ (and this works independently of whether $V$ is finite dimensional or not)