Null space vs eigenvectors

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Jeff 2017-6-10

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评论： Jeff 2017-6-10

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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Answer 1

David Goodmanson 2017-6-10

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编辑：David Goodmanson 2017-6-10

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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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Jeff 2017-6-10

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Thanks David, I guess I was working a bit too late. The proportionality completely escaped me, especially when you take Vnull(3,1)/V(3,1). I was expecting an output of

real(Vnull(3,1))/real(V(3,1))+imag(Vnull(3,1))/imag(V(3,1))*i

ans =

0.0000 + 1.4142i

I completely forgot how to divide complex numbers appropriately. Thanks for setting me straight!!!

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