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 [itex]det(A- \lambda I)[/itex]. 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, [itex]\{\lambda_i\}[/itex], then, since eigenvectors corresponding to distinct eigenvalues are independent, there exist a basis for the space, [itex]\{\vec{v}_i\}[/itex], consisting of eigenvectors. In order that p(A)= 0 we must have [itex]p(A)(\vec{v}_i)= 0[/itex] for every such eigenvector and that means p(A) must have a factor of the form [itex](x- \lambda_i)[/itex] 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, [itex]\lambda[/itex], has "algebraic multiplicity" (the number of factors of the form [itex](x- \lambda)[/itex] in the characteristic polynomial) n, then, by definition, the charactaristice polynomial has a factor of [itex](x- \lambda)^n[/itex]. If an eigenvalue, [itex]\lambda[/itex] has "geometric multiplicity" (the dimension of the subspace of eigenvectors corresponding to the eigenvalue) m, then the minimal polynomial contains the factor [itex](x-\lambda)^m[/itex].

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!