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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$\begingroup$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.
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?
$\endgroup$ 1 $\begingroup$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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