Here's an idea:
Try computing
is a square matrix such that =0. Prove that for any real number the matrix is invertible
Attempt: if =0 then is a singular matrix. A singular matrix times is still a singular matrix. so the idea is to prove that is an invertible matrix when A is singular.
I'm not sure where to go from here but this is something interesting I found:
C = D = C and D are both singular matrices.
1) C + = = invertible matrix
however,
2) D + = = singluar matrix
So I believe the idea is to figure out why D is not part of the matrix we are looking for, even though it is singular, and it probably has to do with the fact that
Thanks!
I'm not sure if it's common. I haven't seen it before, but it's a fun result.
As for my approach, I was first inspired by the condition to try and square to see what that'd get me. In the result of , the term of course vanishes, but the cross-term doesn't, which is pretty useless. Then I just remembered that to get the cross-term out of the way, you can change the sign on in one of the factors. That's it.