# A invertible iff both restrictions are invertible

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…
• @MathBro For a linear map $A\colon V\to V$, with $V$ finite dimensional, bijectivity, injectivity and surjectivity are equivalent. Oct 18, 2020 at 8:19