# Characteristic and minimal polynomials

1. Nov 27, 2011

### corey2014

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

Not really applicable

3. 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?

2. Nov 27, 2011

### Dick

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: Nov 27, 2011
3. Nov 28, 2011

### HallsofIvy

Staff Emeritus
??? 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.