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Let $X$ be a normed space. I am working on prove that if $U, V$ are both linear operator on $X$. Then operator $UV$ is invertible if and only if $U$ and $V$ are both invertible.

I can prove one direction: assume that $UV$ is invertible, then there is a operator $A$ so that $$ UVA=AUV=I $$

Hermi
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  • Only square matrices are invertible. Therefore this is not a counter example. For a linear operator $B$ to be the inverse of an operator $A$, $AB=BA=I$ must be satisfied, and then we write $A^{-1}=B$. So you have not yet shown that $UV$ is invertible. – bayes2021 Dec 02 '23 at 21:19

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This is not true. Take $R$ and $L$ to be the right- and left-shift operator on $l^2$. Then $LR = Id$ is invertible, but $R$ is not surjective and $L$ is not injective.

If $X$ is finite-dimensional, then the claim is true (take determinants).

Martin Brandenburg
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daw
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    We don't need determinants in the finite-dimensional case: https://math.stackexchange.com/questions/3852 – Martin Brandenburg Dec 02 '23 at 21:22
  • @MartinBrandenburg I am confused that. I mean here we apply this 'wrong' Lemma there https://math.stackexchange.com/questions/2708101/if-lambda-is-in-spectrum-of-bounded-linear-operator-a-then-lambdan-is: if $\lambda^n I-A^n$ is invertible, then $\lambda I-A$ is invertible. If our result is wrong, why can we say $\lambda I-A$ is invertible? – Hermi Dec 02 '23 at 22:58
  • $\lambda I - A$ is a two-sided divisor of $\lambda^n I - A^n$, hence invertible. PS: please create follow-up question if further explanation is required. Your original question is answered. – Martin Brandenburg Dec 02 '23 at 23:12
  • @MartinBrandenburg How to modify the wrong statement based on that question? – Hermi Dec 02 '23 at 23:19
  • @MartinBrandenburg See this question:https://math.stackexchange.com/questions/4818986/if-lambdan-i-an-has-an-inverse-then-so-does-lambda-i-a – Hermi Dec 02 '23 at 23:22