# Problem with proof

1. Mar 13, 2007

### 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 Im having trouble from this way around.
Any ideas?

2. Mar 13, 2007

### matt grime

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

3. Mar 13, 2007

### matt grime

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).

4. Mar 13, 2007

### 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)?

5. Mar 13, 2007

### 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.

6. Mar 13, 2007

### robierob12

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

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

?

7. Mar 14, 2007

### 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!

8. Mar 14, 2007

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).

9. Mar 14, 2007

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

10. Mar 14, 2007

### sab47

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...

11. Mar 14, 2007

### leon1127

Using determinant gives the easiest proof.

12. Mar 14, 2007

### 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.