Invertibility of the product of matrices

[Mr Davis 97]

Homework Statement

Let A and B be n by n matrices such that A is invertible and B is not invertible.
Then, AB is not invertible.

Homework Equations

The Attempt at a Solution

We know that A is invertible, so there exists a matrix C such that CA = I. Then we can right -multiply by B so that CAB = IB = I. Then by the associative property C(AB) = I. By the same argument, we can show that there is a C such that (AB)C = I. So AB has an inverse.

Obviously this is wrong, because in order for AB to have an inverse, both A and B must have an inverse. So what am I doing wrong?

 

[cragar]

CAB=IB but you can't conclude IB=I, IB=B

 

[fresh_42]

Maybe you should concentrate on the non invertible part. What does it mean to B, not being invertible? Is there a positive property, i.e. without the use of non, not or no?

 

[Ray Vickson]

Homework Statement

Let A and B be n by n matrices such that A is invertible and B is not invertible.
Then, AB is not invertible.

Homework Equations

The Attempt at a Solution

We know that A is invertible, so there exists a matrix C such that CA = I. Then we can right -multiply by B so that CAB = IB = I. Then by the associative property C(AB) = I. By the same argument, we can show that there is a C such that (AB)C = I. So AB has an inverse.

Obviously this is wrong, because in order for AB to have an inverse, both A and B must have an inverse. So what am I doing wrong?

What tools/results are you allowed to use? Do you know about determinants? Do you know how determinants relate to the invertability/non-invertability of a matrix?

 

[micromass]

Assume that $AB$ is invertible. This means that there is a $C$ such that $CAB = I$ and $ABC = I$. Can you prove now that $B$ is invertible? (and thus deriving a contradiction).