Null space vs eigenvectors

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];

Answer 1

[David Goodmanson]

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.

[Jeff]

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!!!