following @Sivaram Ambikasaran's answer
for SVD, I get the computing (using MATLAB)of:
A = 2 1 3 1 2 5 3 5 4
[U,S,V] = svd(A)
U = -0.3793 -0.2964 -0.8765 -0.5571 -0.6832 0.4721 -0.7387 0.6674 0.0941 S = 9.3111 0 0 0 2.4506 0 0 0 1.1395 V =-0.3793 0.2964 -0.8765 -0.5571 0.6832 0.4721 -0.7387 -0.6674 0.0941
for eigenvalues I get:
[eigVector,lambda] = eig(A)
eigVector = 0.2964 0.8765 0.3793 0.6832 -0.4721 0.5571 -0.6674 -0.0941 0.7387 lambda = -2.4506 0 0 0 1.1395 0 0 0 9.3111
I know understand that the singular values are the magnitudes of the eigen values so:
lambda's diagonal magnitudes are equal to S, in SVD,
|-2.4506| = 2.4506 |1.1395 | = 1.1395 |9.3111 | = 9.3111
The question here is, as the other values
How to get U from eigenvector in that example
what criteria is used to change sign?
- Could you tell me if computing eigenvalue decomposition takes less time than computing SVD (it's order)?? -