Not showing the right set of linearly independent eigenvectors

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G0pie 2017-7-21

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评论： Matt J 2017-7-22

采纳的回答： Ari

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I am using matlab R2013a.

Consider the matrix

A = [0 1 1 1 0;
0 0 0 1;
0 0 0 0;
0 0 0 0;
0 0 0 0];

Clearly, it's rank is 2; so nulity is 3. But while computing all its eigenvectors, it's showing as if it has only one linearly independent eigenvector. Theoretically, it has [1;0;0;0;0],[0;1;-1;0;0],[0;1;0;-1;0] as three linearly independent eigenvectors corresponding to the 0 eigenvalue.

So why is it so with the command [vA,d]=eig(A)?

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Matt J 2017-7-21

编辑：Matt J 2017-7-21

Most intriguing!

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Answer 1

Ari 2017-7-21

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编辑：Ari 2017-7-21

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To compute eigenvalues, the eig function tries to minimize A*V - V*D to numerical precision of double. Since eig uses floating point computation this can only approach zero, and not exactly zero (try printing A*V-V*D). In case of defective matrices this behavior is expected. You can try using Symbolic Math as a workaround as below

B = sym(A);
[V,D] = eig(B);

This will give you correct eigenvalues and eigenvectors but it will come with a computational penalty for large matrices. MATLAB documentation on eigenvalues of defective matrices can be found here.

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Answer 2

John D'Errico 2017-7-21

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编辑：Walter Roberson 2017-7-22

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Commonly known as a...

https://en.wikipedia.org/wiki/Defective_matrix

Your matrix lacks a complete set of eigenvectors.

The classic example is:

[v,d] = eig([1 1;0 1])
v =
                       1                        -1
                       0      2.22044604925031e-16
d =
     1     0
     0     1

The good news is with these defective matrices, you can send it back to the person you bought it from for a full refund, if this is done within 90 days, and you have a valid sales receipt. Did you get the extended warranty? :)

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John D'Errico 2017-7-22

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Lets see. It is clerly a defective matrix.

syms lambda
det(A-eye(5)*lambda)
ans =
-lambda^5

So det tells us the matrix has only zero eigenvalues, with apparent multiplicity 5, even though there are only 3 eigenvectors. null will give us the correct eigenvectors.

null(A - 0*eye(5))
ans =
            0            0            1
      0.57735     -0.57735            0
     -0.78868     -0.21132            0
      0.21132      0.78868            0
            0            0            0

The eig algorithm gets hung up though.

Matt J 2017-7-22

It is strange that eig() gets hung up when, in other cases, it handles eigenvalue multiplicity very gracefully. For a symmetric matrix with multiple eigenvalues, for example, all the independent eigenvectors are always found.

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