# Eigenvalue decomposition Singular value decomposition

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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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.