If AB^2-A Invertible Prove that BA-A Invertible

[ThankYou]

Homework Statement

As in the tile
If AB^2-A is a Invertible matrix Prove that BA-A is also a Invertible matrix

Homework Equations

liner algebra 1 , only the start...

The Attempt at a Solution

Made many things noting work

Thank you

 

[ThankYou]

I've solve it..
Thank you..
AB^2-A = A(B^2-I)
Means that A and (B^2-I) are Invertible
(B-I)(B+I) are Invertible and so
A(B-I) Is Invertible

 

[Mark44]

I've solve it..
Thank you..
AB^2-A = A(B^2-I)
Means that A and (B^2-I) are Invertible
(B-I)(B+I) are Invertible and so
A(B-I) Is Invertible

You're supposed to show that BA - A is invertible.

A(B - I) = AB - A, which is not necessarily equal to BA - A.

 

[losiu99]

Nevertheless, we can save this reasoning provided you are allowed to use det(AB)=det(A)det(B).

 

[ThankYou]

I am allowed

 

[losiu99]

Then we can just say det(AB²-A)=det(A)det(B-I)det(B+I) is nonzero, so det(BA-A)=det(B-I)det(A) is nonzero.