Eigenvalues and eigenspaces with isomorphisms I have the following question about eigenvalues and eigenspaces in the case of an isomorphism. I am not sure how to prove those 2 statements. My intuition tells me it has something to do with diagonalizability.
Suppose that $f(v)=\lambda v$ for some non-zero vector $v$, then $g(f(v))=v=\lambda g(v)$. Hence $g(v)=\frac{1}{\lambda}v$. This is sufficient. It has nothing to do with diagonalizability.
Since $f$ is an isomorphism, you know it's 1-1 and onto, and since you know the inverse $g$ exists, you know none of the elements of $V$ can have zero eigenvalue, otherwise they're non-invertible.  (b) follows from (a) since all $v$ in $V$ are invertible, and a matrix and its inverse have the same eigenvectors.