# Minimal Polynomial A nxn Matrix

1. ### wurth_skidder_23

39
Let A be an n x n matrix; denote its distinct eigenvalues by a_1,...,a_k and denote the index of a_i by d_i. How do I prove that the minimal polynomial is then:

m_A(s) = (s-a_1)^d_1*...*(s-a_k)^d_k

?

The characterstic polynomial is defined as:

p_A(s) = (s-a_1)*...*(s-a_n);

Last edited: Oct 26, 2006
2. ### Hurkyl

16,089
Staff Emeritus
What do you mean by index? If you mean "multiplicity", then you can't prove it. For example, consider the minimum polynomials of the zero and identity matrices. (1 and x-1, respectively)

3. ### wurth_skidder_23

39
The book defines the index as:

"N_m = N_m(a) the nullspace of (A-a*I)^m. The subspaces N_m consist of generalized eigenvectors; they are indexed increasingly, that is N_1 is included in N_2 is included in...

Since these are subspaces of a finite-dimensional space, they must be equal from a certain index on. We denote by d = d(a) the smallest such index, that is, N_d = N_(d+1) = ... but N_(d-1) /= N_d_j

d(a) is called the index of the eigenvalue a."

This is from Peter D. Lax's book on linear algebra.

Last edited: Oct 26, 2006
4. ### Hurkyl

16,089
Staff Emeritus
If f is the minimal polynomial of A, then f(A) is the zero matrix. So f(A)v = 0 for any vector. Does that help?

5. ### wurth_skidder_23

39
That is also true for the characteristic polynomial. I know that the characteristic polynomial can be divided by a minimal polynomial and I want to show that the minimal polynomial is equal to (s-a_1)^(d_1)*...*(s-a_k)^(d_k)

6. ### Hurkyl

16,089
Staff Emeritus
Right. But you want the smallest polynomial that satisfies what I wrote. So you first want to prove:

(s-a_1)^(d_1)*...*(s-a_k)^(d_k)

satisfies what I wrote, and then you want to prove that any proper factor of it does not. I think good choices of f(A)v will do that.