Characteristic and minimal polynomials

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SUMMARY

The discussion centers on the equivalence of conditions for a linear operator T on a finite-dimensional complex vector space V. It establishes that having a basis of eigenvectors (condition a) implies T can be represented by a diagonal matrix (condition b). However, it clarifies that condition b does not guarantee that all eigenvalues have multiplicity one (condition c). The participants emphasize the distinction between geometric and algebraic multiplicities, noting that both must be equal for T to be diagonalizable, which is essential for the characteristic polynomial of T to equal its minimal polynomial (condition d).

PREREQUISITES
  • Understanding of linear operators in finite-dimensional vector spaces
  • Knowledge of eigenvalues and eigenvectors
  • Familiarity with diagonalization of matrices
  • Concepts of characteristic and minimal polynomials
NEXT STEPS
  • Study the relationship between geometric and algebraic multiplicities in linear algebra
  • Learn about diagonalization criteria for matrices
  • Explore the implications of the Cayley-Hamilton theorem
  • Investigate examples of linear operators with distinct eigenvalues and their properties
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Students and educators in linear algebra, mathematicians focusing on operator theory, and anyone interested in the properties of eigenvalues and polynomials in vector spaces.

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?
 

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