# Proving the product of two non singular matrices is also non singular.

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.

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

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.
• How'd this go without an upvote? Feb 14, 2014 at 13:22
• @amWhy can you please elaborate more on second approach? Didnt get anything...Any web links for the relevant basics? Oct 5, 2017 at 15:02

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?

• Ok thanks I think this helps me out a lot. Feb 12, 2014 at 22:09

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 $\det(AB)$.

• I haven't gotten this far yet, but thank you. Feb 12, 2014 at 22:34