Show that is $\upsilon$ is an eigenvector of the matrices A and AB (assume invertibility) Show that is $\upsilon$ is an eigenvector of the matrices A and AB with corresponding eigenvalues $\lambda \neq 0$, $\mu$ respectively, then $\upsilon$ is also a corresponding eigenvector of B with corresponding eigenvalue $\frac{\mu}{\lambda}$. What I have so far: $$A\upsilon=\lambda\upsilon$$ $$A^{-1}A\upsilon=A^{-1}\lambda\upsilon$$ $$\upsilon=\lambda A^{-1}\upsilon$$ $$\frac{1}{\lambda}\upsilon=A^{-1}\upsilon$$ $$AB\upsilon=\mu\upsilon$$ $$A^{-1}AB\upsilon=A^{-1}\mu\upsilon$$ $$B\upsilon=\mu A^{-1}\upsilon$$ $$B\upsilon=\frac{\mu}{\lambda}\upsilon$$ Is this enough to prove the statement or am I missing anything?

Your reasoning is right, since you seem to assume invertibility of $A$.

Without invertibility of $A$, the statement is wrong: Take $$ A=\pmatrix{1&0\\\0&0}, \ B = \pmatrix{0&0\\\1&0}, \ v = \pmatrix{1\\\0}, $$ then $Av = v$, $AB=0$, but $v$ is not an eigenvector of $B$.