Sorting an n x n matrix in the Jordan form.

Question

Mohammad Ayoub 2017-3-23

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编辑： Martin Seltmann 2018-11-28

Assume I have a matrix J (n x n dimension), the matrix is originally obtained from MATLAB using the 'jordan' function which returns the matrix in the Jordan canonical form. MATLAB always returns the matrix J sorting the diagonal from lowest to highest, until it encounters repeated eigenvalue(s), which are sorted in Jordan blocks in the lower right corner of the matrix. Here's an example:

A = [0 1 0 0 0;
     0 0 1 0 0;
     0 0 0 1 0;
     0 0 0 0 1;
     -18 -57 -68 -38 -10];
J = jordan(A)
J = [-2 0 0 0 0;
     0 -3 1 0 0;
     0 0 -3 0 0;
     0 0 0 -1 1;
     0 0 0 0 -1]

The question is, is it possible to sort Matrix J's diagonal from highest to lowest (e.g. here [-1;-1;-2;-3;-3]) while keeping the Jordan blocks associated with each repeated eigenvalue?

In my example I want J to look like

[-1 1 0 0 0;
-1 0 0 0;
0 -2 0 0;
0 0 -3 1;
0 0 0 -3],

I want to make this transformation to any matrix of order n (n x n).

Thank you in advance.

Regards, M Ayoub.

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Jan 2017-3-23

编辑：Jan 2017-3-23

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What do you want for:

 J = [-2 0 0 0 0;
-3 1 0 0;
0 -4 2 0;  % <- 2 beside the diagonal
0 0 -1 1;
0 0 0 -1]

If the -3 is moved to the last row, there is no place for the 2. Can this happen?

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Answer 1

David Goodmanson 2017-3-25

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Hello Mohammad, I believe the following code works for a given Jordan matrix J.

a = sort(diag(J),'descend');
d = diff(a)==0;              % look for repeated elements
Jnew = diag(a) + diag(d,1);

This code does not permute the elements of the original Jordan matrix but rather sorts the diagonal elements, so it is sensitive to any very small unintended numerical differences in the diagonal elements. The Jordan decomposition itself has the same issues so I don't think that this process adds new complications.

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Martin Seltmann 2018-11-28

编辑：Martin Seltmann 2018-11-28

The proposal seems to assume that identical eigenvalues belong to the same Jordan block. But this does not have to be the case, there could be several Jordan blocks for the same eigenvalue.

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