Proving Invertibility of AB-I & BA-I When A & B Are Both nxn Matrices

[daniel_i_l]

Homework Statement

Q: If A and B are both nxn matrices and AB-I is invertable then prove that BA-I is also invertable.

Homework Equations

if A is invertible iff |A|<>0

The Attempt at a Solution

I've been thinking about this for over an hour I've only managed to prove it if either A or B are invertable. because if let's say A is invertable then:
|AB-I|<>0 => |AB-I||A|<>0 => |ABA-A|<>0 => |A||BA-I|<>0 => |BA-I|<>0 and so it's invertable. if B is invertable then you do pretty much the same thing on starting on the left side.
But what if they're both singular?
Thanks.

 

[matt grime]

X is not invertible if and only if there is a v=/=0 with Xv=0.

Suppose (AB-I)v=0, and see what you can deduce. (I don't promise this works, but is the first thing that springs to mind.)

 

[Dick]

You are too hung up on determinants. Try a proof by contradiction. Assume (BA-I) is NOT invertible. Then there is a nonzero vector x such that (BA-I)x=0. Now tell me what is (AB-I)Ax=?. (By playing the same game you did with the determinants).

 

[daniel_i_l]

Thanks a lot! now i can go to sleep...