I am trying to solve the problem below but without luck:
Let V be a finite-dimensional vector space, A $\in $End(V), and U $\subseteq$ V an invariant subspace. Let $A_r \in End(U)$ denote the restriction of A to U, regarded as a map into U, and let $A_q \in End(V/U)$ denote the quotient map given by $A_q(x+U)=Ax+U$.
Show that A is invertible $\iff$ $A_r$ and $A_q$ are both invertible.
It seems like "$\implies$" must be true, since any restriction on A invertible must also be invertible on some restriction, but i have trouble with how to prove this. For the "iff" part i am totally lost.
Suppose $A$ is invertible. We need to show that both $A_r$ and $A_q$ are injective (because of finite dimension). For $A_r$ it is obvious. Suppose $A_q(x+U)=0+U$: then $A(x)\in U$; but $A_r$ is surjective, so…
Suppose both $A_r$ and $A_q$ are invertible. By finite dimensionality, we need to show $A$ is surjective. Take $y\in V$; then $y+U=A_q(x+U)$, for some $x$. This means $y-A(x)\in U$, so…