# Confusion between eigen value decomposition and singular value decomposition

The Singular Value Decomposition of matrix $H$ gives $$H = U \Sigma V^H$$

The Eigen value decomposition of $$HH^H= U \Sigma \Sigma^t U^H$$

I took an example in matlab and performed EID and SVD respectively

H= [0.1, 0.3, .4; 0.5 , 0.5, 0.9; 0.1, 0.4, 0.5]
[U,e]= eigs(H*(conj(transpose(H))))
[U,D,Vh]= svd(H)

H =

0.1000    0.3000    0.4000
0.5000    0.5000    0.9000
0.1000    0.4000    0.5000

U =

0.3598    0.4271   -0.8295
0.8160   -0.5751    0.0578
0.4524    0.6977    0.5555

e =

1.9450         0         0
0    0.0450         0
0         0    0.0000

U =

-0.3598    0.4271   -0.8295
-0.8160   -0.5751    0.0578
-0.4524    0.6977    0.5555

D =

1.3946         0         0
0    0.2121         0
0         0    0.0034

Vh =

-0.3508   -0.8256    0.4419
-0.4997    0.5641    0.6573
-0.7920    0.0097   -0.6105


1) As a sanity check if I square the singular values of $H$ obtained from SVD, I obtain the eigen values of $HH^H$ obtained from the EID. We need to square the elements in D to obtain e...

BUT

2) But, shouldn't the matrix U from the SVD be equal to the matrix U from the eigen value decomposition ??? MATLAB is not giving me that as you can see above.. In particular the first columns of U matrix dont match.

If any more details or explanation is needed I can provide. Looking forward for your help

If $\lambda$ is an eigenvalue of $A$ with multiplicity $1$, then there exist two real vectors with norm $1$ which satisfy $Ax=\lambda x$. This is because you can always multiply $x$ with $-1$, which does not change its norm.
• @Henry No, they can be different. $U$ contains the eigenvectors, and it can contain either $x$ or $-x$. Both are correct
• One contains left eigenvectors of $HH^H$ vectors and one contains right ones..