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linearly independent eigen vectors

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Shannu
Shannu on 18 Sep 2020
Commented: John D'Errico on 20 Sep 2020
set of linearly independent eigenvectors
A=[2 0 0 0 0 0
0 2 0 0 0 0
0 0 2 0 0 0
0 0 0 2 0 0
0 0 0 0 2 0
0 0 0 0 0 2];

  3 Comments

John D'Errico
John D'Errico on 18 Sep 2020
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
Shannu on 18 Sep 2020
no i want to comute linearly independent eigen vectors.
eig is help for all vector bt i want linearly independent

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Accepted Answer

John D'Errico
John D'Errico on 18 Sep 2020
Edited: John D'Errico on 18 Sep 2020
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 =
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
D =
2 0 0 0 0 0
0 2 0 0 0 0
0 0 2 0 0 0
0 0 0 2 0 0
0 0 0 0 2 0
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 =
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
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?

  2 Comments

Shannu
Shannu on 18 Sep 2020
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
John D'Errico on 20 Sep 2020
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
As you can see, the columns of V are not independent. But that is due to the defective matrix.

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More Answers (1)

BOB MATHEW SYJI
BOB MATHEW SYJI on 18 Sep 2020
i is the eigen vector and j is the eigen values
[i j]=eig(A)

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