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$A$ is an invertible matrix and $B$ is a non-invertible matrix. Can $AB$ be invertible? I have the following idea:

Sup. $AB$ is invertible, then:

$B=IB=(A^{-1}A)B=A^-1(AB)$, then apply inverse both sides:

$B^{-1}=(AB)^{-1}(A^{-1})^{-1}=(AB)^{-1}A$, but $B$ is non-invertible (hip). This leads to a contradiction, as we supposed $AB$ is invertible.

Therefore $AB$ is non-invertible.

I'm not sure if the step where I apply "inverse both sides" is right. Otherwise I'm not sure how to prove this.

Note 1: I CAN'T use $(AB)^{-1}=B^{-1}A^{-1}$ since the hypothesis for that theorem is $A, B$ invertible matrices and this is not the case.

Note 2: I CAN'T use determinants yet.

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    $\begingroup$ A product of invertible matrices is invertible. $\endgroup$
    – Jakobian
    Oct 9, 2020 at 13:59
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    $\begingroup$ If $A$ is invertible and $B$ is not, then $AB$ is not invertible. $\endgroup$
    – JMoravitz
    Oct 9, 2020 at 13:59
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    $\begingroup$ Can you use $\det AB=\det A\det B$? $\endgroup$ Oct 9, 2020 at 13:59
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    $\begingroup$ Refer to the invertible matrix theorem which gives a long list of equivalent conditions for a matrix to be considered invertible. You should have seen at least some of these proven before. It may be easiest to use a specific one of these, be it related to determinants, or related to nullspaces or injectivity, or what have you. $\endgroup$
    – JMoravitz
    Oct 9, 2020 at 14:02
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    $\begingroup$ @FabrizioGambelín you can use what I said to show, in a very simple way, that $B$ must be invertible if $A$ and $AB$ are $\endgroup$
    – Jakobian
    Oct 9, 2020 at 14:02

1 Answer 1

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Your argument is correct, though depending on the level, you may want to explain why $$B^{-1}=\big(A^{-1}(AB)\big)^{-1}=(AB)^{-1}(A^{-1})^{-1}.$$ Another approach would be to suppose that $AB$ is invertible with inverse $C$.Then $$(CA)B=C(AB)=I,$$ so $CA$ is the inverse of $B$, a contradiction.

Alternatively, you could note that if $B$ is non-invertible, then there exists some nonzero vector $x$ such that $Bx=0$. Then also $ABx=0$, and so $AB$ is non-invertible.

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  • $\begingroup$ Seems good. However, I can't clearly see why is Bx=0 then ABx=0 also has a nonzero vector :/ $\endgroup$
    – Fabrizio G
    Oct 9, 2020 at 14:02
  • $\begingroup$ Oh it was trivial! I'm a bit sleepy sorry, thank you very much for your help :) $\endgroup$
    – Fabrizio G
    Oct 9, 2020 at 14:06

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