Eigenvalues of Matrix Function

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Homework Help Overview

The discussion revolves around the properties of matrix functions, specifically focusing on the eigenvalues of a matrix function defined by a Taylor series expansion. The original poster seeks to understand how the eigenvalues of a matrix T relate to the eigenvalues of the function f(T).

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore direct calculations involving eigenvectors and eigenvalues, questioning how powers of the matrix T affect these eigenvectors. There is an attempt to derive the relationship between the eigenvalues of T and those of f(T) through substitution into the function's series expansion.

Discussion Status

The discussion is progressing with participants providing insights into the relationship between eigenvalues and the matrix function. Some calculations have been proposed, and there is a focus on verifying the implications of these calculations on the eigenvalues of f(T). However, no consensus has been reached yet.

Contextual Notes

Participants are working under the constraints of the problem statement, which requires a proof without additional context or external resources. The original poster expresses uncertainty about how to begin the proof, indicating a potential gap in foundational understanding.

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Homework Statement


Define a matrix function f(T) of an nxn matrix T by its Taylor series f(T)=f0 +f1T +f2T2+...
Show that if matrix T has the eigenvalues t1,t2...tn, then f(T) has eigenvalues f(t1), f(t2)...f(tn)


Homework Equations





The Attempt at a Solution


I am at a loss of how to prove this, could someone help me with this problem? I have no idea where to start.
 
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Well, how about a direct calculation? Suppose v is an eigenvector of matrix T with eigenvalue t_1. That is, Tv= t_1v. Okay, so what is T^2v? T^3v, etc?
 
So you'd have T2v=t2v ... Tnv=tnv
 
Then f(T)v=(f0+f1T +f2T2...)v = (f0+f1t1 +f2t2...)v =f(t1)v
 
Then for all eigenvalues... λ=tn f(T)v=f(tn)v , therefore f(B) has eigenvalues f(t1), f(t2),... f(tn)
 

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