Linear Algebra Proof involving idempotency

PsychonautQQ
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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?
 
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PsychonautQQ said:

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.
 
Isn't that what I set up? I don't know how to solve it
 
PsychonautQQ said:
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.
 
PsychonautQQ said:
Isn't that what I set up? I don't know how to solve it
Dick said:
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[/color]=I. Do you see the difference?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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