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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)??
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1 Answers 1

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Arrange the eigen values in decreasing order of magnitude. Arrange the eigen vectors also based on decreasing values of magnitude of eigenvalues. Now do you see what $U$ and $V$ are. Notice that $U$ and $V$ are almost same except for a change in sign of the column corresponding to the negative eigenvalue.

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    I actually don't remember the computational cost for these. I will look it up somewhere online and get back to you. You can also try to google and find out which is cheaper. Another way out is to compare the time taken for the two different algorithms and check which one takes lesser time.2011-02-22