Calculating the Minimal Polynomial for a Given Matrix A

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SUMMARY

The minimal polynomial for the given matrix A, with a characteristic polynomial of (1-λ)5, is determined to be (1-λ)2. This conclusion is based on the definition that the minimal polynomial is the monic polynomial of lowest degree satisfying M(A)=0, where all irreducible factors must divide the characteristic polynomial. The calculation involves checking powers of (I-A) until the result equals zero, confirming that (I-A)2=0 while (I-A)≠0.

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  • Understanding of characteristic and minimal polynomials in linear algebra
  • Familiarity with matrix operations, specifically the identity matrix and matrix subtraction
  • Knowledge of monic polynomials and their properties
  • Basic concepts of eigenvalues and eigenvectors
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Students and professionals in mathematics, particularly those studying linear algebra, matrix theory, and polynomial functions. This discussion is beneficial for anyone looking to deepen their understanding of minimal polynomials and their applications in matrix analysis.

gottfried
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I've been given a matrix A and calculated the characteristic polynomial. Which is (1-λ)5. Given this how does one calculate the minimal polynomial?

Also just to check, is it correct that the minimal polynomial is the monic polynomial with lowest degree that satisfies M(A)=0 and that all the irreducible factors of the minimal polynomial divide the characteristic polynomial?

Given this I think the minimal polynomial is (1-λ)2 since (I-A)≠0 and (I-A)2=0 but this method to figure it out seems a little ad hoc.

A=
[1 1 0 0 0]
[0 1 0 0 0]
[0 0 1 1 0]
[0 0 0 1 0]
[0 0 0 0 1]
 
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No, that's perfecty valid. The characteristic polynomial is (1- \lambda)^5 and the minimal polynomial is the polynomial, p, of lowest degree that is a factor of the characteristic polynomial and such that p(A)= 0. The obvious thing to do is to start with the factor of lowest degree, 1- x, that is a factor of that and see if I- A= 0. Since it does not, try (I- A)^2. I presume that did gives 0 but if it had not, you would then try (I- A)^3 and so on. There is nothing "Ad hoc" about using the definition of something.
 
Thanks for clearing it up:)
 

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