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I want to prove the following statement.

Let $V$ and $W$ be $n$-dimensional vector space and $T \in \mathcal{L}(V,W)$. Prove that $S \in \mathcal{L}(W,V)$ is a left inverse if and only if it is a right inverse.

My first thought was that I can use the following facts:

(1) A function $f: A \to B$ is injective if and only if it possesses a left inverse and is surjective if and only if it possesses a right inverse.

(2) A linear map $T \in \mathcal{L}(V)$ is injective if and only if it surjective if and only if it is invertible.

The first statement seems valid, but the 'link' between surjectivity and injectivity breaks down, I believe, because $V$ and $W$ only have the same dimension, but are not necessarily the same vector space. Finite-dimensional doesn't imply finite, so I cannot use the usual fact for functions that if $f: A \to B$ where $|A| = |B| = n$ is injective, then it is surjective (and vice-versa). I also can't use the fact that a function is bijective if and only if it is invertible because having a left or a right inverse says nothing about invertibility (with the key word being 'or').

Any help would be appreciated. I'd like to think I'm on the right track.

Tom Baker
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2 Answers2

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If $S$ is a left-inverse, then $T$ must be injective.

You say that you know that a linear map $L : V \rightarrow V$ from a finite-dimensional vector space to itself is injective iff it is surjective. But $V, W$ are both $n$-dimensional. So they are in fact the same space; more precisely, they're isomorphic. We can (by sending a basis of $V$ to a basis of $W$) find a bijective linear map $\phi : V \rightarrow W.$ Then $\phi^{-1} \circ T$ is still an injective linear map, but now it is from $V$ to $V.$ Thus $\phi^{-1} \circ T$ is surjective... but this means $T$ is surjective. Now concluding is straightforward: $T$ has a right inverse. But if a function has a left- and right-inverse, those inverses must be equal.

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  • If $S$ is a left inverse of $T$, then $ST = \operatorname{id}_V$, which means that $\ker T = \{0\}$ (since $Tv = 0$ implies that $v = S(Tv) = S(0) = 0$). Thus, by the rank-nullity theorem we have $$\dim W = \dim V = \underbrace{\dim (\ker T)}_0 + \dim(\operatorname{im} T)$$ and then $T$ is surjective, that is, $T$ is invertible. Hence $TS = \operatorname{id}_W$, since the inverse is unique.
  • Now suppose that $S$ is a right inverse of $T$. Reversing the roles of $S$ and $T$ in the previous dot shows that $S$ is also a left inverse of $T$.
azif00
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