Linear algebra - matrices and polynomials

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SUMMARY

The discussion centers on proving that for every square matrix B, there exists a non-zero real polynomial p(x) such that p(B) = 0. The key insight involves the rank-nullity theorem and the dimensionality of square matrices, specifically that the set {B, B^2, B^3, ..., B^{n^2}} is linearly dependent in an n^2-dimensional vector space. This leads to the conclusion that coefficients a0, a1, a2, ..., a(n^2) can be found such that their polynomial representation results in the zero matrix.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly matrix theory.
  • Familiarity with the rank-nullity theorem.
  • Knowledge of polynomial functions and their properties.
  • Experience with linear independence and dependence in vector spaces.
NEXT STEPS
  • Study the rank-nullity theorem in detail to understand its implications for linear transformations.
  • Explore the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic polynomial.
  • Learn about the concepts of linear independence and dependence in vector spaces.
  • Investigate polynomial interpolation and its applications in linear algebra.
USEFUL FOR

This discussion is beneficial for students of linear algebra, mathematicians focusing on matrix theory, and educators teaching polynomial functions in relation to matrices.

Kate2010
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Homework Statement


Prove for each square matrix B there is a real polynomial p(x) (not the zero polynomial) so p(B)=0


Homework Equations


Rank-nullity? dimv = r(T) + n(T)


The Attempt at a Solution


I've found the dimension for nxn square matrices (n²) and a basis (1 in one place and zero in the rest for each place in the matrix) but I don't really know where to go from here. Any ideas would be great :) thanks.
 
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Consider,
1,B,B^2,B^3,...,B^{n^2}
is this set linearly dependent? (can n^2+1 many vectors be linearly independent in a n^2 dimensional vector space?) If they are linearly dependent you can find coefficients a0,a1,a2,...,a(n^2) such that,
\sum_{i=0}^{n^2} a_i B^i = 0
so consider,
p(x) = \sum_{i=0}^{n^2} a_i x^i
 
Thank you, that helped a lot.
 
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