Null space vs eigenvectors
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Below is a code I ran to compare the null space & the eigenvectors of matrix A. Please correct me if I am wrong, but I thought that the eigenvectors are the same as the null space for the matrix [A-D(n,n)*I]. Unfortunately, my results do not seem to support that premise. What do I have wrong?
A=[[14 8 -19];[-40 -25 52];[-5 -4 6]];
[V,D]=eig(A);
Vnull=null(A-D(1,1)*eye(3));
Vnull=[null(A-D(1,1)*eye(3)) null(A-D(2,2)*eye(3)) null(A-D(3,3)*eye(3))];
Vchek=[V Vnull];
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David Goodmanson
2017-6-10
编辑：David Goodmanson
2017-6-10
Hi Jeff, Since your eigenvalues are all distinct, what you have is basically correct. It's just that the eigenvector and the null vector don't have to be identical, merely proportional. Taking the first column of both Vnull and V and dividing element by element shows proportionality
>> V(:,1)./Vnull(:,1)
ans =
0.7071 - 0.7071i
0.7071 - 0.7071i
0.7071 - 0.7071i
and the same is true for the other two columns.
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