Nilpotent Matrices: Show Jordan Form w/Linear Independence

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In summary, if N is a nilpotent mxm matrix with N^{m}=0 and N^{m'}\neq0 for m'<m, then there exists a basis in which it is represented as a single Jordan block with zero diagonal elements. This basis is also proven to be linearly independent by constructing it through a combination of vectors that are annihilated by N^m and those that are not annihilated by N^m, but are by N^{m+1}.
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Homework Statement


Suppose that N is a nilpotent mxm matrix, N[itex]^{m}[/itex]=0, but N[itex]^{m'}[/itex][itex]\neq[/itex]0 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[itex]^{m'}[/itex][itex]\neq[/itex]0 for m'<m means that N[itex]^{m'}[/itex] 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?
 

FAQ: Nilpotent Matrices: Show Jordan Form w/Linear Independence

What is a nilpotent matrix?

A nilpotent matrix is a square matrix where the powers of the matrix eventually become all zeros. In other words, there exists a positive integer k such that A^k = 0, where A is the nilpotent matrix.

What is the Jordan form of a nilpotent matrix?

The Jordan form of a nilpotent matrix is a diagonal matrix with all zeros except for the superdiagonal, which contains ones. This form can also be written as a block diagonal matrix, where each block is a Jordan block with eigenvalue 0.

How do you show a nilpotent matrix in Jordan form?

To show a nilpotent matrix in Jordan form, you need to find its eigenvalues and corresponding Jordan blocks. The number of Jordan blocks for each eigenvalue is determined by the algebraic multiplicity of the eigenvalue. Then, you can arrange the Jordan blocks in a diagonal or block diagonal matrix to obtain the Jordan form.

What is the significance of linear independence in relation to nilpotent matrices in Jordan form?

The linear independence of the columns of a nilpotent matrix is important in determining the number of Jordan blocks for each eigenvalue. If the columns are linearly independent, then there will be only one Jordan block for each eigenvalue. However, if the columns are linearly dependent, then there will be multiple Jordan blocks for each eigenvalue.

Can a nilpotent matrix have repeated eigenvalues?

Yes, a nilpotent matrix can have repeated eigenvalues. In this case, the number of Jordan blocks for each eigenvalue will be greater than one, and the size of each block will correspond to the geometric multiplicity of the eigenvalue.

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