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.

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. – cogle Feb 12 '14 at 22:09

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? – Namaste Feb 14 '14 at 13:22
• @amWhy can you please elaborate more on second approach? Didnt get anything...Any web links for the relevant basics? – anir123 Oct 5 '17 at 15:02

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. – cogle Feb 12 '14 at 22:34