Linear Algebra Proof involving idempotency

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Homework Help Overview

The discussion revolves around proving that the expression I - 2A is its own inverse, given that A is an idempotent matrix (A^2 = A). Participants are exploring the implications of this property in the context of linear algebra.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to manipulate the expression I - 2A and relate it to its inverse. There are questions about the validity of their setups and whether they are on the right track. Some participants express uncertainty about the steps taken and seek clarification on proving the identity.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning each other's reasoning. There is a focus on confirming the setup for proving that I - 2A is its own inverse, but no consensus has been reached on the correct approach yet.

Contextual Notes

Participants are working under the assumption that A is an idempotent matrix, which may influence their reasoning and the steps they take in the proof. There is also a mention of confusion regarding the definitions and properties of inverses in this context.

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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