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I am having trouble with a proof for linear algebra. Could somebody explain to me how to prove that if $A$ and $B$ are both $n\times n$ non singular matrices, that their product $AB$ is also non singular.

A place to start would be helpful. Thank you for your time.

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  • $\begingroup$ You should check out the following article. Well I mean other's searching for the proof: yutsumura.com/… $\endgroup$ Oct 12, 2021 at 12:10

3 Answers 3

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There's different manners to prove this result for example:

  • Using the determinant: $$\det(AB)=\det A\det B$$ and the fact that $C$ is singular iff $\det C=0$.
  • Using the fact that $AB$ is invertible then $A$ is surjective and $B$ is injective and that in finite dimensional space: $C$ is injective iff $C$ is surjective iff $C$ is bijective.
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  • $\begingroup$ How'd this go without an upvote? $\endgroup$
    – amWhy
    Feb 14, 2014 at 13:22
  • $\begingroup$ @amWhy can you please elaborate more on second approach? Didnt get anything...Any web links for the relevant basics? $\endgroup$
    – Mahesha999
    Oct 5, 2017 at 15:02
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Note that a matrix is non-singular if and only if it has an inverse.

Suppose $A$ and $B$ have inverses $A^{-1} B^{-1}$. What do you get when you multiply $$ (AB)(B^{-1}A^{-1}) $$ and why can we now conclude that $AB$ is non-singular?

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  • $\begingroup$ Ok thanks I think this helps me out a lot. $\endgroup$
    – cogle
    Feb 12, 2014 at 22:09
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Depends how far into linear algebra you are and what you can use. One possible and very short solution: a square matrix is nonsingular iff its determinant is nonzero. Now use the property for $\det(AB)$.

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  • $\begingroup$ I haven't gotten this far yet, but thank you. $\endgroup$
    – cogle
    Feb 12, 2014 at 22:34

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