Linear Algebra -- Projection matrix question

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Homework Help Overview

The discussion revolves around the properties of a projection matrix, specifically an n×n matrix A that satisfies the equation A^2 = A. The original poster expresses confusion regarding the nature of the problem, particularly in writing down the most general polynomial in A.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of A being a projection matrix and discuss the form of the polynomial in A. The original poster questions the initial requirements of the problem, while others suggest that the polynomial can be simplified due to the properties of projection matrices.

Discussion Status

There is an ongoing exploration of the polynomial's structure, with some participants providing insights into the implications of A^n = A. The conversation reflects a mix of understanding and uncertainty, with participants building on each other's ideas without reaching a definitive conclusion.

Contextual Notes

Participants note the importance of including the identity matrix in the polynomial representation, indicating a consideration of the polynomial's degree and terms.

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Homework Statement


Let A be an n×n matrix which has the property that A^2 =A.
(i) Write down the most general polynomial in A

Homework Equations

The Attempt at a Solution


My biggest problem is that I don't even understand what the question is asking

Is it just sum (alphaA^n)=0

but A^n=A

sum(alpha A)=0 ?

I know its not the equation to find the eigenvalues as that follows, and I'm fine with that

Av=pv where p are the eigenvalues, and v the corresponding eigenvectors
A^2v=Apv
Av=p(Av)
pv=p^2v
v(p)(p-1)=0 and hence p= 0 or 1

But I just don't understand the first bit

Many thanks in advance :)
 
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The absolute most general polynomial, degree N, in some arbitrary A would be ##P_N(A)=\sum_{n=0}^N a_nA^n## wouldn't it?
What is special about this polynomial?
 
Simon Bridge said:
The absolute most general polynomial, degree N, in some arbitrary A would be ##P_N(A)=\sum_{n=0}^N a_nA^n## wouldn't it?
What is special about this polynomial?
Well since A is a projection matrix, surely that would imply than ##A^n=A## and hence ##P_N(A)=\sum_{n=0}^N a_nA^n## -> ##P_N(A)=\sum_{n=0}^N a_nA##
 
i.e. the most general polynomial is 1st order. Well done.
 
Wouldn't you want to include ##A^0 = I##?
 
LCKurtz said:
Wouldn't you want to include ##A^0 = I##?
Yes, yes I would ;) Thank you
 

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