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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.

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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…

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  • $\begingroup$ I have looked at your answer for a while now, and i can not quite finish the proof. I see that we try to show that the functions are bijective, since this implies that they are invertible. $\endgroup$
    – MathBro
    Oct 18, 2020 at 7:22
  • $\begingroup$ @MathBro For a linear map $A\colon V\to V$, with $V$ finite dimensional, bijectivity, injectivity and surjectivity are equivalent. $\endgroup$
    – egreg
    Oct 18, 2020 at 8:19

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