Minimal Polynomial A nxn Matrix


by wurth_skidder_23
Tags: matrix, minimal, polynomial
wurth_skidder_23
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#1
Oct26-06, 03:41 PM
P: 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);
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Hurkyl
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#2
Oct26-06, 06:01 PM
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the index of a_i by d_i
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)
wurth_skidder_23
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#3
Oct26-06, 06:27 PM
P: 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.

Hurkyl
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#4
Oct26-06, 07:08 PM
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Minimal Polynomial A nxn Matrix


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?
wurth_skidder_23
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#5
Oct26-06, 07:31 PM
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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)
Hurkyl
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#6
Oct26-06, 07:35 PM
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That is also true for the characteristic polynomial.
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.


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