Finding the minimal polynomial of a matrix?

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SUMMARY

The discussion centers on finding the minimal polynomial of a 3x3 matrix A with a characteristic polynomial defined by an irreducible cubic polynomial f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are rational coefficients. The user identifies the companion matrix corresponding to f(x) and concludes that the minimal polynomial m(x) is derived from the normalized companion matrix, resulting in m(x) = x^3 + (b/a)x^2 + (c/a)x + (d/a). This conclusion is reached by recognizing that the minimal polynomial coincides with the characteristic polynomial when the matrix is in companion form.

PREREQUISITES
  • Understanding of polynomial functions, specifically cubic polynomials.
  • Familiarity with matrix theory, particularly companion matrices.
  • Knowledge of characteristic and minimal polynomials in linear algebra.
  • Proficiency in rational numbers and their properties.
NEXT STEPS
  • Study the properties of companion matrices in linear algebra.
  • Learn about the relationship between characteristic and minimal polynomials.
  • Explore irreducible polynomials over the rationals and their implications in matrix theory.
  • Investigate generalizations of minimal polynomials for matrices of higher dimensions.
USEFUL FOR

Students and educators in linear algebra, mathematicians focusing on polynomial theory, and anyone interested in the properties of matrices and their associated polynomials.

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Homework Statement



Let f(x) be an irreducible polynomial cubic in Q. For example

f(x) = ax^3 + bx^2 + cx + d

Let A be a 3 x 3 matrix with entries in Q such that char(A,x) = f(x). Find the minimal polynomial m(x) of A. Can you generalize to a degree n polynomial?


Homework Equations





The Attempt at a Solution



If the char(A,x) = f(x) then the companion matrix is...

[0, 0, -d]
[1, 0, -c]
[0, 1, -b]

Since the companion matrix's characteristic polynomial = its minimal polynomial, does this mean the minimal polynomial is just f(x). I'm missing something, aren't I...
 
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What about the a?
 
Office_Shredder said:
What about the a?

Oh. The companion matrix would be...

[0 0 -d/a]
[1 0 -c/a]
[0 1 -b/a]

so m(x) = x^3 + (b/a)x^2 + (c/a)x + (d/a)?
 

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