linearly independent eigen vectors

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Shannu 2020-9-18

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评论： John D'Errico 2020-9-20

set of linearly independent eigenvectors

A=[2 0 0 0 0 0
2 0 0 0 0
0 2 0 0 0
0 0 2 0 0
0 0 0 2 0
0 0 0 0 2];

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John D'Errico 2020-9-18

NO. It is NOT done using rank. You want to compute the eigenvectors. I wonder what function in MATLAB computes eigenvectors? Duh, I wonder. Could eig be useful?

Shannu 2020-9-18

no i want to comute linearly independent eigen vectors.

eig is help for all vector bt i want linearly independent

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Answer 1

John D'Errico 2020-9-18

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https://ww2.mathworks.cn/matlabcentral/answers/595945-linearly-independent-eigen-vectors#answer_496774

编辑：John D'Errico 2020-9-18

在 MATLAB Online 中打开

And for some strange reason, you think that eig as applied to a diagonal matrix does not produce a set of independent eigenvectors?

Have you tried using eig?

What does eig produce when applied to 2*eye(6)?

[V,D] = eig(2*eye(6))
V =
   0     0     0     0     0
   1     0     0     0     0
   0     1     0     0     0
   0     0     1     0     0
   0     0     0     1     0
   0     0     0     0     1
D =
   0     0     0     0     0
   2     0     0     0     0
   0     2     0     0     0
   0     0     2     0     0
   0     0     0     2     0
   0     0     0     0     2

The columns of V are a set of linearly independent eigenvectors. Do you dispute that fact? The diagonal elements of D are the eigenvalues.

Just for kicks,

V*V'
ans =
   0     0     0     0     0
   1     0     0     0     0
   0     1     0     0     0
   0     0     1     0     0
   0     0     0     1     0
   0     0     0     0     1

It looks good to me.

(In some cases, when the matrix is defective, it will not have a complete set of eigenvectors, but that is not the fault of eig but of mathematics. No complete set will exist in some cases.) But a diagonal matrix is not even remotely a problem. So feel free to explain why the columns of V do NOT form a set of linearly independent basis vectors for the vector space R^6 in this case?

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Shannu 2020-9-18

Iam not sure if they really form linearly independent are not! I just want to confirm this thing And Thanks a lot!!!

John D'Errico 2020-9-20

在 MATLAB Online 中打开

Note that eig can fail to produce a set of linearly depending eigenvectors when your matrix is defective. The classic example is:

>> [V,D] = eig(triu(ones(3)))
V =
            1           -1            1
            0   2.2204e-16  -2.2204e-16
            0            0   4.9304e-32
D =
     1     0     0
     0     1     0
     0     0     1