Nilpotent Matrices: Show Jordan Form w/Linear Independence

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SUMMARY

The discussion centers on nilpotent matrices, specifically demonstrating that a nilpotent mxm matrix N, where N^m = 0 and N^{m'} ≠ 0 for m' < m, can be represented in Jordan form as a single Jordan block with vanishing diagonal elements. The proof involves constructing a basis from vectors annihilated by N and subsequently by N^2, ensuring linear independence throughout the process. This method confirms that the basis spans the entire vector space while maintaining the required properties of nilpotent matrices.

PREREQUISITES
  • Understanding of nilpotent matrices and their properties
  • Familiarity with Jordan canonical form
  • Knowledge of linear independence and basis in vector spaces
  • Experience with matrix exponentiation and annihilation of vectors
NEXT STEPS
  • Study the construction of Jordan blocks for nilpotent matrices
  • Learn about the implications of the Cayley-Hamilton theorem on nilpotent matrices
  • Explore the relationship between nilpotent operators and their eigenvalues
  • Investigate the process of finding a basis for generalized eigenspaces
USEFUL FOR

Mathematicians, graduate students in linear algebra, and anyone studying matrix theory or advanced topics in linear transformations will benefit from this discussion.

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Homework Statement


Suppose that N is a nilpotent mxm matrix, N^{m}=0, but N^{m&#039;}\neq0 for m'<m. Show that there exists a basis in which it takes the form of a single Jordan block with vanishing diagonal elements. Prove that your basis set is linearly independent.


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The Attempt at a Solution


So I've recognized the fact that N^{m&#039;}\neq0 for m'<m means that N^{m&#039;} does not annihilate every vector in V. I'm just not really sure where go from here...
 
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Pick a basis for the vectors that are annihilated by N. Then add a basis for the vectors that are annihilated by N^2 but not by N. Continue. Eventually you'll get a basis for the whole space, right? What does N look like in that basis?
 

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