Minimal vs Characteristic Polynomials

[cadillacclaire]

Let T:V to V be a linear operator on an n-dimensional vector space V. Let T have n distinct eigenvalues. Prove that the minimal polynomial and the characteristic polynomial are identical up to a factor of +/- 1

I'm probably over thinking this, but it seems that if you have n distinct eigenvalues then the minimal polynomial would have to be EXACTLY the same as the characteristic. How could it be different?

Thanks in advance!

 

[HallsofIvy]

The characteristic polynomial for a matrix is the polynomial det(A- \lambda I). The minimal polynomial is the polynomial of least degree, p, such that p(A)= 0.

Yes, if an n by n matrix has n distinct eigevalues, \{\lambda_i\}, then, since eigenvectors corresponding to distinct eigenvalues are independent, there exist a basis for the space, \{\vec{v}_i\}, consisting of eigenvectors. In order that p(A)= 0 we must have p(A)(\vec{v}_i)= 0 for every such eigenvector and that means p(A) must have a factor of the form (x- \lambda_i) which, in turn, means that the minimal polynomial is the characteristic polynomial.

In fact, it is not necessary that all eigenvalues be distinct. If an n by n matrix has n independent eigenvectors (if it is diagonalizable) then the characteristic polynomial is the same as the minimal polynomial. If an eigenvalue, \lambda, has "algebraic multiplicity" (the number of factors of the form (x- \lambda) in the characteristic polynomial) n, then, by definition, the charactaristice polynomial has a factor of (x- \lambda)^n. If an eigenvalue, \lambda has "geometric multiplicity" (the dimension of the subspace of eigenvectors corresponding to the eigenvalue) m, then the minimal polynomial contains the factor (x-\lambda)^m.

So the characteristic polynomial is the same as the minimal polynomial if and only if the geometric multiplicity of every eigenvalue is the same as it algebraic multiplicity.

 

[cadillacclaire]

Thanks for the quick reply. I think the problem itself was not worded very well and that's what threw me off. Good to know that I'm on the right track!