egr 515 singular value decomposition
TRANSCRIPT
SINGULA
R VALU
E
DECOMPOSIT
ION
EGR
51 5 ,
OC
TOB
E R 2
2 , 2 0 1 4
DR .
T I M L
I N
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SINGULAR VALUE DECOMPOSITION
• Example 91. (page 128)
• Let A = . Then A’A = ,
• AA’ =
2
[2 1 11 2 11 1 2]
SINGULAR VALUE DECOMPOSITION
• The eigenvalues of A’A are 0, 1, 1, and 4; and the eigenvalues of AA’ are 1, 1, and 4.
• For l = 4, the eigenvectors of AA’ satisfy
• -2x + y + z = 0
• x – 2y + z = 0
• x + y -2z = 0
• A normalized solution (column eigenvector) is x = y = z =
• For l = 1, the eigenvectors of AA’ satisfy x + y + z = 0.
• There are two orthogonal normalized solutions (why two?) which are also orthogonal to the eigenvector of l = 4.
• x = , y = - , z = 0; and x = , y = , and z = .
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SINGULAR VALUE DECOMPOSITION
• We have S’ = . Note the three columns of S’ were computed earlier as eigenvectors.
• From proof of theorem 9.2 (on page 126), U = A’S’L-1/2 = .=
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SINGULAR VALUE DECOMPOSITION
• A SVD of A = .
•
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