# Can I use SVD to find complex eigenvectors from complex eigenvalues?

Assume that we have a real square matrix $$A$$ and we know its eigenvalues.

Let's assume that matrix $$A$$ is

   A = [0.018142,   0.968856,   0.151740,   0.757174,
0.017829,   0.474323,   0.358832,   0.970854,
0.184523,   0.063063,   0.680511,   0.191901,
0.806877,   0.830208,   0.977169,   0.222291];


And the eigenvalues are

eigs = [1.87922 + 0.00000i
-0.45009 + 0.11680i
-0.45009 - 0.11680i
0.41623 + 0.00000i]


First of all, I know the eigenvectors. But how can I find them with SVD or another method?

I know that eigenvectors can be found by finding the null space of

$$N(A - \lambda_i I) = W_i$$ Where $$\lambda_i$$ is the $$i:th$$ eigenvalue.

Do you have any suggestion? I tried to use SVD without any success.

   A = [0.018142,   0.968856,   0.151740,   0.757174,
0.017829,   0.474323,   0.358832,   0.970854,
0.184523,   0.063063,   0.680511,   0.191901,
0.806877,   0.830208,   0.977169,   0.222291];

t = eig(A)

eigenvalue = real(t(2))
B = A - eigenvalue*eye(4);
[u, s, v] = svd(B);
v

[W, ~] = eig(A)


I did a test as Robert Israel said.

A = [0.018142,   0.968856,   0.151740,   0.757174,
0.017829,   0.474323,   0.358832,   0.970854,
0.184523,   0.063063,   0.680511,   0.191901,
0.806877,   0.830208,   0.977169,   0.222291];

t = eig(A)

eigenvalue = t(2) % Complex
B = (A - eigenvalue*eye(4))*(A - eigenvalue*eye(4));
[u, s, v] = svd(B);
v % This is V^T
s
[W, ~] = eig(A)


Output:

t =

1.87922 + 0.00000i
-0.45009 + 0.11680i
-0.45009 - 0.11680i
0.41623 + 0.00000i

eigenvalue = -0.45009 + 0.11680i
v =

-0.28716 - 0.00000i  -0.04672 - 0.00000i   0.89589 - 0.00000i   0.33575 + 0.00000i
-0.56631 + 0.00284i   0.34482 + 0.33887i   0.03732 - 0.12258i  -0.53597 + 0.37665i
-0.55413 + 0.01685i  -0.58743 - 0.55128i  -0.16543 - 0.03894i  -0.11425 + 0.04162i
-0.53799 + 0.00597i   0.25403 + 0.22487i  -0.34698 + 0.17797i   0.50108 - 0.43850i

s =

Diagonal Matrix

6.0213e+00            0            0            0
0   9.8366e-01            0            0
0            0   1.2468e-01            0
0            0            0   8.0735e-17

W =

-0.53992 + 0.00000i   0.25267 + 0.22111i   0.25267 - 0.22111i   0.66764 + 0.00000i
-0.50351 + 0.00000i  -0.65138 - 0.06952i  -0.65138 + 0.06952i   0.20360 + 0.00000i
-0.21211 + 0.00000i  -0.11338 - 0.04392i  -0.11338 + 0.04392i  -0.67929 + 0.00000i
-0.64030 + 0.00000i   0.66585 + 0.00000i   0.66585 - 0.00000i   0.22663 + 0.00000i


No signs of eigenvalues in last column of $$V^T$$

The null space of $$A - \lambda I$$ is the null space of $$B(\lambda) = (A - \lambda I)^*(A - \lambda I)$$. If you take the svd of $$B(\lambda)$$: $$B(\lambda) = U \Sigma V^T$$, a basis for the null space of $$A -\lambda$$ consists of columns of $$V^T$$ for the singular value $$0$$.