Characteristic and minimal polynomials

corey2014
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Homework Statement


Let V be a finite dimensional complex vector space and T be the linear operator of V. Prove that the following are equivalent

a V has a basis consisting of eigenvectors of T.
b T can be represented by a diagonal matrix.
c all the eigenvalues of T have multiplicity one.
d. the Characteristic polynomial of T equals the minimal polynomial of T.


Homework Equations



Not really applicable

The Attempt at a Solution



Ok so I proved A implies B, However, I feel that B does not imply C.

I just want to see if my argument is valid. because the identity matrix, is definitely diaganol however, it does not have a multiplicity of One. Can I assume wlog that T has distinct eigenvalues along the main diaganol?
 
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You are quite right. But I wouldn't assume wlog. That looses generality. B doesn't imply C period. I think whoever wrote the problem up had a memory lapse or forgot to state an assumption. You could assume they forgot that assumption and proceed from there.
 
Last edited:
corey2014 said:

Homework Statement


Let V be a finite dimensional complex vector space and T be the linear operator of V. Prove that the following are equivalent

a V has a basis consisting of eigenvectors of T.
b T can be represented by a diagonal matrix.
c all the eigenvalues of T have multiplicity one.
? There are two "multiplicities", the geometric multiplicity (number of independent eigenvectors corresponding to the eigenvalue) and algebraic multiplicity (multiplicity as a root of the characteristic polynomial). But neither of those must be one in order that T can be represented by a diagonal matrix, only that, for each eigenvalue, the two multiplicities be the same.

d. the Characteristic polynomial of T equals the minimal polynomial of T.


Homework Equations



Not really applicable

The Attempt at a Solution



Ok so I proved A implies B, However, I feel that B does not imply C.

I just want to see if my argument is valid. because the identity matrix, is definitely diaganol however, it does not have a multiplicity of One. Can I assume wlog that T has distinct eigenvalues along the main diaganol?
 
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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