Also If I have 1000 of matrices how can I separate those on the basis of number of linearly independent eigenvectors, e.g I want to separate those matrices of order 4 by 4 having linearly independent eigen vectors 2.
How to find number of linearly independent eigenvectors in a matrix?
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I have a matrix
0 -1 -2 -1
-1 0 -1 0
-2 -1 0 -1
-1 0 -1 0
Each column represent an eigen vector, In the above case I have three linearly independent eigenvectors , How can I find this in matlab ?
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possibility
2018 年 2 月 10 日
In the context of Linear Algebra, one finds an eigenvalue of a matrix and then finds the right or the left eigenvector associated to that eigenvalue. Assume you have the matrix (your matrix)
A =
0 -1 -2 -1
-1 0 -1 0
-2 -1 0 -1
-1 0 -1 0
In order to have an idea of how many linearly independent columns (or rows) that matrix has, which is equivalent to finding the rank of the matrix, you find the eigenvalues first. And then you can talk about the eigenvectors of those eigenvalues.
Hence, if I understand it correctly, you're trying to find the rank of the matrix.
Following gives the number of linearly independent columns (or rows) of matrix A
>> rank(A)
ans =
3
And following outputs the eigenvalues (and their right eigenvectors) of that matrix
>> [eigenvec,eigenval]=eig(A)
eigenvec =
0.6015 -0.0000 0.3717 -0.7071
0.3717 0.7071 -0.6015 0.0000
0.6015 0.0000 0.3717 0.7071
0.3717 -0.7071 -0.6015 0.0000
eigenval =
-3.2361 0 0 0
0 0.0000 0 0
0 0 1.2361 0
0 0 0 2.0000
Hope that helps.
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