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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.

k170
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cogle
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  • You should check out the following article. Well I mean other's searching for the proof: https://yutsumura.com/two-matrices-are-nonsingular-if-and-only-if-the-product-is-nonsingular/ – Shaheer ziya Oct 12 '21 at 12:10

3 Answers3

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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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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?

Ben Grossmann
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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 $\det(AB)$.

zipirovich
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