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I have a question regarding the proof of this corollary 2.11, which I have encountered in a course on Linear Algebra. As this is a corollary, I have a feeling that the proof should be somewhat obvious and self-explanatory but I fail to see it.

Theorem 2.10

Assume that both $U$ and $V$ both have finite dimension $n$. If $A$ is surjective or injective, then $A$ is bijective.

Corollary 2.11

Suppose that $dim(U) = dim (V)<\infty$ and let $T\in Hom(U,V)$. If $AT=\iota_V$ or $TA=\iota_U$, then $A$ is an isomorphism and $A^{-1}=T$.

Proof. If $AT=\iota_V$, then $A$ is surjective, and if $TA=\iota_U$ then $A$ is injective. In both cases Theorem 2.10 implies $A$ is bijective. Then $A^{-1}=T$ follows.

What I do not understand is why $A$ is surjective when we have $A\circ A^{-1}$ but it is injective when $A^{-1}\circ A$.

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1 Answer 1

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If $AT=\iota_V$, then, for each $u\in U$, $A(T(u))=u$. Therefore, $A$ is surjective.

And, if $TA=\iota_U$, then, if $v_1,v_2\in V$ are such that $A(v_1)=A(v_2)$, then $T(A(v_1))=T(A(v_2))$, which means that $v_1=v_2$. So, $A$ is injective.

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