Inverses

[Gear300]

For any two elements A and B that form AB, neither A nor B have to be invertible for (AB)-1 to exist, right?

[defunc]

Both have to be invertible. Inv(AB) = inv(B)inv(A).

[Mark44]

Right. For example..
\left[
\begin{array}{c c c}
1&0&0\\
0&1&0
\end{array}
\right]
\left[
\begin{array}{c c}
1&0\\
0&1\\
0&0
\end{array}
\right]~=~\left[
\begin{array}{c c}
1&0\\
0&1\\
\end{array}
\right]


The matrix on the right is clearly invertible, while the two matrices in the product aren't event square, let alone invertible.

[Mark44]

Both have to be invertible. Inv(AB) = inv(B)inv(A).
See my counterexample.

[Gear300]

Thanks for the reply. So that would mean that Inv(AB) = inv(B)inv(A) iff inv(B) and inv(A) exist (whereas Inv(AB) may exist without a defined inv(A) or inv(B)), right?

[pbandjay]

Looks correct to me. The condition that I am familiar with is:

If A and B are invertible matrices of the same size, then AB is invertible and (AB)-1 = B-1A-1.

This is easy to prove by showing that (AB)(B-1A-1) = A(BB-1)A-1 = AIA-1 = AA-1 = I.

As far as the other way of your "iff", if (AB)-1 = B-1A-1 is given then it seems to me that the existence of B-1 and A-1 would directly follow since they are used in the initial condition.

[Gear300]

I see. Thanks again for the replies. I have an additional question:

For a matrix C that is not a square matrix, there is no defined inverse; however, it is possible that there is a left inverse A and a right inverse B, in which A =/= B, for the matrix C, right?

[Mark44]

Yes. See the wikipedia article here (http://en.wikipedia.org/wiki/Inverse_element#Matrices), under the section titled Matrices.

[Gear300]

Thanks for the link.

So in the case of linear systems Ax = b, I suppose it wouldn't always be possible to use the left inverse of A to isolate x as a general method since the left inverse of A does not necessarily exist. How would one isolate x in these linear systems (other than parametrization of x)?