# Show endomorphisms commute

Problem: Consider a finitely generated vector space $V$ over $\mathbb{Q}$. Let $a,b$ be element of $End(V)$ satisfying

$$3a^3+7a^2-2ab+4a=id_V$$

where multiplication represents function composition. Show that $ab=ba$.

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I like Potato's answer, but here's another way to approach it.

We have $I=a(3a^2+7a-2b+4I)$, which, since $V$ is finite dimensional, implies that $a$ is invertible and $a^{-1}=3a^2+7a-2b+4I$. Thus $b=-\frac{1}{2}a^{-1}+2I+\frac{7}{2}a+\frac{3}{2}a^2$, and right-hand side of the last equation clearly commutes with $a$.

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Composing on the left and right by $a$ gives

$$3a^4+7a^3-2a^2b+4a=a=3a^4+7a^3-2aba+4a$$

from which we see $a(ab)=aba\Rightarrow a(ab-ba)=0$. So for elements $v$ not in $\ker(a)$, $ab(v)=ba(v)$. We want to prove that $ker(a)$ is just 0. Assume $w\neq 0$ lies in $\ker(a)$. Then applying the original sums of compositions to $w$, we see $ab(w)=w$, which shows $ab$ is the identity on $\ker a$. Then $a$ is surjective, and because $V$ is finitely generated, injective, so the kernel is just 0 and we have the desired result.

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