Here's an idea:
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
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
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.