linearly independent eigen vectors

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];

[Shannu]

is it through rank?

[John D'Errico]

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]

no i want to comute linearly independent eigen vectors.

eig is help for all vector bt i want linearly independent

Answer 1

[John D'Errico]

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?

[Shannu]

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]

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.

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

Answer 2

[BOB MATHEW SYJI]

i is the eigen vector and j is the eigen values

[i j]=eig(A)