Prove Either A or B is Singular Given AB is Singular

[robierob12]

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 I am having trouble from this way around.
Any ideas?

 

[matt grime]

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

 

[matt grime]

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]

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)?

 

[robierob12]

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.

Is that it? By contradiction?

 

[robierob12]

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.

?

 

[sab47]

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!

 

[radou]

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).

 

[DeadWolfe]

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.

 

[sab47]

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...

 

[leon1127]

Using determinant gives the easiest proof.

 

[AKG]

There are (at least) two easy methods:

1. Use determinants. det(AB) = det(A)det(B). If AB is singular, det(AB) is zero. det(A) and det(B) are just real numbers, so if their product is zero, what do you know?
2. Prove the contrapositive (not the converse), that is, prove that if A and B are non-singular, then AB is non-singular. So if A and B have inverses, A^-1 and B^-1, find a matrix that should be the inverse of AB, and prove that the matrix you found really is the inverse.