Idempotent matrix problem - HELP

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Homework Help Overview

The discussion revolves around the properties of idempotent matrices, specifically examining the invertibility of the matrix expression E - nA, where E is the identity matrix and n is a real number. The original poster expresses confusion regarding the implications of idempotency on this expression's invertibility.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the conditions under which the matrix E - nA might be invertible, questioning the existence of a non-zero vector that satisfies the equation (E - nA)v = 0. They discuss the implications of this condition on the kernel of the matrix and its relation to invertibility.

Discussion Status

Participants are actively engaging with the concepts of linear algebra, particularly focusing on the invertible matrix theorem and its implications. Some have provided guidance on revisiting foundational concepts, while others are attempting to clarify the reasoning behind the conditions for invertibility.

Contextual Notes

There is a noted uncertainty regarding the application of linear algebra principles, particularly for participants who are newer to the subject. The discussion reflects an exploration of definitions and properties related to idempotent matrices and their implications for invertibility.

chefobg57
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idempotent matrix problem - HELP :(

Hi, I have the following problem with an idempotent matrix and I am stuck...

If A is an idempotent (A^2 = A), is E - nA invertible /E is the identity matrix and n is a real number/ and why?

I've tried with setting (E - nA)(E - nA) but it doesn't get me anywhere..
My instinct is telling me there it should be invertible in a lot of the cases, but... :/
 
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If (E-nA) were NOT invertible, then there would be a nonzero vector v such that (E-nA)v=0. Where does that lead you?
 


I am sorry, I am quite new to Linear Algebra I am not really sure what the answer to your question should be... Can you explain a little bit more why there should be a non-zero vector such that (E-nA)v=0?

Thank you for helping me out! You're a life saver!
 


If M is a square matrix, then M being invertible is equivalent to the kernel or null space of M being {0}. If M is NOT invertible, then the kernel of M is not {0}. Meaning there is nonzero vector v in the kernel, hence Mv=0. You must have covered this. Can you look back in your book or notes for that kind of thing? It's pretty important.
 


I checked my notes and as far as I understand you're talking about the invertible matrix theoremy...
Can we then say that:
Assuming that E-nA is not invertible means that there will be a non-zero vector x, such that (E-nA)x=0. However, this suggests that E-nA is not row equivalent to E which then implies that E-nA is not a product of elementary matrices. This implies that E-nA is not invertible?

I'm sorry I'm such a pain but I am really lost...

Thanks again..
 


chefobg57 said:
I checked my notes and as far as I understand you're talking about the invertible matrix theoremy...
Can we then say that:
Assuming that E-nA is not invertible means that there will be a non-zero vector x, such that (E-nA)x=0. However, this suggests that E-nA is not row equivalent to E which then implies that E-nA is not a product of elementary matrices. This implies that E-nA is not invertible?

I'm sorry I'm such a pain but I am really lost...

Thanks again..

You are making it much too hard. If (E-nA)v=0 then Ev-nAv=0. What is Ev?
 


Hm..

Well, Ev = nAv => v = nAv .. Doesn't this mean that nA=1?
 


chefobg57 said:
Hm..

Well, Ev = nAv => v = nAv .. Doesn't this mean that nA=1?

NO! You can't divide both sides by a vector! But it does mean Av=v/n. Now operate on both sides with A again remembering A is idempotent.
 

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