Linear Algebra Proof involving idempotency

[PsychonautQQ]

Homework Statement

I = Identity matrix
Suppose that A^2 = A. Prove that I - 2A = (I - 2A)^-1

Homework Equations

ahh don't know what to put here

The Attempt at a Solution

So I have to prove this thing is it's own identity... interesting..

I - 2A = I - 2A^2

(I - 2A^2)*(I - 2A)^-1 = I

Distributive law?
Idk honestly this is all I have gotten.. And it's probably not the right direction, just trying to put all the information I know into one line I guess. Any Mathamavericks out there want to help a noob out?

 

[Mark44]

Homework Statement

I = Identity matrix
Suppose that A^2 = A. Prove that I - 2A = (I - 2A)^-1

Homework Equations

ahh don't know what to put here

The Attempt at a Solution

So I have to prove this thing is it's own identity... interesting..

I - 2A = I - 2A^2

(I - 2A^2)*(I - 2A)^-1 = I

Distributive law?
Idk honestly this is all I have gotten.. And it's probably not the right direction, just trying to put all the information I know into one line I guess. Any Mathamavericks out there want to help a noob out?

To show that A and B are inverses -- IOW, that B = A^-1 -- show that AB = I.

 

[PsychonautQQ]

Isn't that what I set up? I don't know how to solve it

 

[Dick]

Isn't that what I set up? I don't know how to solve it

You want to show (1-2A) is its own inverse. I.e. (1-2A)*(1-2A)=I.

 

[Mark44]

Isn't that what I set up? I don't know how to solve it

You want to show (1-2A) is its own inverse. I.e. (1-2A)*(1-2A)=I.

What you wrote, Psychonaut, was (1-2A)*(1-2A)^-1=I. Do you see the difference?