Eigenvalues of Matrix Function

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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)
 
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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