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Problem: Let $V$ be a vector space over a field $F$ which is not finitely generated, and fix $\alpha\in End(V)$ which is not the 0-endomorphism. Consider the set $A=\{\beta\in End(V)\ |\ \alpha\circ \beta=id\}$. Show that if $A$ has more than one element, it is infinite.

I have posted a solution attempt in the answers.

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up vote 5 down vote accepted

This is actually a result that holds for rings: if an element has more than one right inverse, then it has infinitely many right inverses.

Suppose that there are at least two elements in $A$, so that $\alpha$ cannot have a left inverse. Let $\beta_1$ be any element of $A$. Consider the map $f\colon A\to A$ given by $f(\beta) =\beta\alpha - \mathrm{id} + \beta_1$.

Note that this is well-defined: the image of $\beta$ is an endomorphism of $V$ (composition and sum of linear transformations is linear), and $$\alpha\Bigl(\beta\alpha - \mathrm{id} + \beta_1\Bigr) = \alpha\beta\alpha -\alpha + \alpha\beta_1 = \alpha - \alpha +\mathrm{id} = \mathrm{id}$$ so if $\beta$ is a right inverse for $\alpha$, then so is $f(\beta)$.

Next: $f$ is one-to-one: if $f(\beta)=f(\beta')$, then $$\beta\alpha - \mathrm{id} + \beta_1 = \beta'\alpha - \mathrm{id} + \beta_1$$ which yields $\beta\alpha = \beta'\alpha$. Now composing with $\beta_1$ on the right we obtain $\beta=\beta'$, since $\alpha\beta_1=\mathrm{id}$.

Finally, $f$ is not onto: we can never get $\beta_1$ in the image. For if $f(\beta)=\beta_1$, then $\beta\alpha - \mathrm{id} + \beta_1 = \beta_1$, hence $\beta\alpha=\mathrm{id}$. But that would mean that $\beta$ is both a left and a right inverse for $\alpha$, which is impossible since $\alpha$ has at least two right inverses.

Thus, $f\colon A\to A$ is one-to-one and not onto, which proves that $A$ is infinite.

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This is very nice. – Potato Jun 2 '12 at 4:36
@Arturo Seems like $\,\beta_2\,$ in the third line in your post is superfluous – DonAntonio Jun 2 '12 at 12:42
@DonAntonio: Yes; it's only used implicitly in arguing that $\beta_1$ cannot be in the image. – Arturo Magidin Jun 2 '12 at 21:07

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