Question

Is every eigen value satisfies its annihilating polynomial?

Answer 1

Step-by-step explanation:

The theorem of Cayley-Hamilton says that a matrix satisfies it's characteristic polynomial. But can we also make a statement about the eigenvalues if a matrix satisfies a monic polynomial in general?

So let's say that we have a matrix A∈Cm,mA∈Cm,m and a monic polynomial pp of degree nn (with n<mn<m) for which p(A)=0p(A)=0. What can be said about the eigenvalues of AA with this information?

Answer 2

Step-by-step explanation:

By applying the polynomial p[A] in A term by term to an eigenvector of eigenvalue λ, one sees that it acts on it as multiplication by p[λ]. But on the other hand p[A]=0, so one must have p[λ]=0, and λ is a root of p. Every eigenvalue of Amust be a root of p (but not every root of p needs to be eigenvalue).

This is elementary, and much easier than Cayley-Hamilton.