# Problem with proof

Im stuck on this proof.

Let A and B be nxn matricies such that AB is singular. Prove that either A or B is singular.

Sooooo, here we go.

Let M = AB where is M is the given singular matrix.

Becuase M is singular then

Mx=0 has an infinite amount of solutions.

Let J be one of the non zero solutions

Mj=0

ABj=0

this is where I get stuck.
If knew that B was singular I think I could prove M is singular but Im having trouble from this way around.
Any ideas?

matt grime
Homework Helper
Stop proving this and prove the converse - if A and B are invertible, then so is AB.

matt grime
Homework Helper
this is where I get stuck.
If knew that B was singular I think I could prove M

You aren't asked to prove this (though it is clearly true). You are told that M is singular, and asked to show that either A or B is too (or both).

HallsofIvy
Homework Helper
You could just use determinants: det(AB)= det(A)det(B)= det(M). Since M is singular, det(M)= what? What does that tell you about det(A) or det(B)?

so

Let A and B be invertible then M must be inverble also.

But since M is singular then A or B must be singular.

You could just use determinants: det(AB)= det(A)det(B)= det(M). Since M is singular, det(M)= what? What does that tell you about det(A) or det(B)?

so det(M)=0 so det(A)=0 or det(B)=0

nonsingular matrix cannot equal zero so A or B must be singular.

?

You've got to use determinants. That way you just map your matrices to set of real numbers R, and everything is just much easier ther!

Homework Helper
so det(M)=0 so det(A)=0 or det(B)=0

nonsingular matrix cannot equal zero so A or B must be singular.

?

Something like that, yes. You know that M = AB is singular, so, det(AB) = detA detB = 0 implies det A = 0 or det B = 0 ("or", of course, includes the case where both equal zero, too).

You've got to use determinants. That way you just map your matrices to set of real numbers R, and everything is just much easier ther!

Nonsense. It's equally easy to simply find the inverse of AB directly.

Nonsense. It's equally easy to simply find the inverse of AB directly.

Yes, but we're saying here that AB is singular here, so it doesn't have an inverse. Plus, finding inverse involves finding the determinant first...

Using determinant gives the easiest proof.

AKG