0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ so the above U matrices are correct? is my intuiton correct should they be equal althought one is from EID and the other from SVD $\endgroup$ – Henry Sep 2 '15 at 5:16
  • $\begingroup$ @Henry No, they can be different. $U$ contains the eigenvectors, and it can contain either $x$ or $-x$. Both are correct $\endgroup$ – 5xum Sep 2 '15 at 5:34
  • $\begingroup$ thank you very much. can i say that both U and V from the SVD contain the eigen vector or just U? @5xum $\endgroup$ – Henry Sep 2 '15 at 5:35
  • $\begingroup$ if i rephrase my question. i mean in general when we perform svd does it give us the matrices of eigen vector or singular vectors? $\endgroup$ – Henry Sep 2 '15 at 5:39
  • $\begingroup$ One contains left eigenvectors of $HH^H$ vectors and one contains right ones.. $\endgroup$ – 5xum Sep 2 '15 at 5:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.