
#1
Oct2606, 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) = (sa_1)^d_1*...*(sa_k)^d_k ? The characterstic polynomial is defined as: p_A(s) = (sa_1)*...*(sa_n); 



#2
Oct2606, 06:01 PM

Emeritus
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PF Gold
P: 16,101





#3
Oct2606, 06:27 PM

P: 39

The book defines the index as:
"N_m = N_m(a) the nullspace of (Aa*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 finitedimensional 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_(d1) /= N_d_j d(a) is called the index of the eigenvalue a." This is from Peter D. Lax's book on linear algebra. 



#4
Oct2606, 07:08 PM

Emeritus
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PF Gold
P: 16,101

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?




#5
Oct2606, 07:31 PM

P: 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 (sa_1)^(d_1)*...*(sa_k)^(d_k)




#6
Oct2606, 07:35 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

(sa_1)^(d_1)*...*(sa_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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