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Let $ L, K: V \to V $ be linear maps that satisfy $L\circ K=1_V$. Show that

  • (a) If $\ V $ is finite dimensional, then $ K\circ L=1_V$.
  • (b) If $\ V $ is infinite dimensional give an example where $K\circ L \neq 1_V $.

I am having trouble seeing why finite or infinite dimensions apply in these cases.

MarianD
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Jerome Turner
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  • The problem is that a linear injective (surjective) endomorphism of an infinite dimensional vector spaces needs not to be surjective (injective), on the contrary in the finite dimensional case a linear injective (surjective) endomorphism is always surjective (injective). – Bargabbiati Mar 16 '19 at 21:02
  • The statement regarding finite dimensional spaces is addressed here (as well as in the linked and related problems on the right) – Ben Grossmann Mar 16 '19 at 21:12

2 Answers2

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I will show a counterexample for the infinite dimensional case. Let $V$ be $\mathcal l_2 (\mathbb N)$, $K$ the right-shift operator, defined as $K(a_1, a_2, \ldots)=(0, a_1, a_2, \ldots)$ and $L$ the left-shift operator, defined as $L(a_1, a_2, \ldots)=(a_2, a_3, \ldots)$; they are both linear, and of course $LK=1_V$. It is also clear that $KL$ is not the identity.

The statement is true in the finite dimensional case, because a surjective/injective endomorphism is also injective/surjective (for example, because it sends basis in basis), so it is invertible, and moreover the inverse is unique.

Bargabbiati
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Hint: There's an analogous statement for sets: let $f,g:S\to S$ be functions with $f\circ g=\mathrm{id}_S$.
Then, $f$ must be surjective and $g$ injective.
Moreover, for any $t=g(s)$ we have $g(f(t)) =g(f(g(s)))=g(s)=t$.

But if $S$ is finite, simply by counting elements, both $f$ and $g$ will be bijective, so by the above, $g\circ f=\mathrm{id}_S$,
while if $S$ is infinite, this conclusion does not hold.

Berci
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