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ds science forum beginner
Joined: 22 Jun 2006
Posts: 1
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Posted: Thu Jun 22, 2006 11:42 am Post subject:
svd question
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Hi all,
I hope this is the right place to ask and sorry for cross-posting. I am
facing the following
problem:
1) vector ai belongs to some R^n subspace W.
2) matrix A=[a1 a2 a3 ... an]^T is a mxn matrix of vectors
3) A=USV^T according to the SVD where V is a nxn matrix of
eigenvectors.
now, the singular vectors vi, do they belong to W? Or, in other words,
what are the conditions that this proposition holds? For example one
might write:
vi=(1/si)*A^T*ui. Therefore, if W=span(ai)then the proposition will
hold. But let's say D is a real positive definite and symmetric matrix
with eigenvalues E and W=ran(E) and ai belong to W. But also span(ai)
belongs to ran(E). As a conclusion, if all this holds then the right
singular vectors are linear combinations of the eigenvectors if all of
the above hold. Does anyone like any of this or am I a complete idiot?
Any help will be greatly appreciated.
TIA and cheers
dps |
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Peter Spellucci science forum Guru
Joined: 29 Apr 2005
Posts: 702
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Posted: Thu Jun 22, 2006 4:00 pm Post subject:
Re: svd question
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In article <e7dvnm$gpn$1@dizzy.math.ohio-state.edu>,
"ds" <junkmailavoid@yahoo.com> writes:
| Quote: |
Hi all,
I hope this is the right place to ask and sorry for cross-posting. I am
facing the following
problem:
1) vector ai belongs to some R^n subspace W.
2) matrix A=[a1 a2 a3 ... an]^T is a mxn matrix of vectors
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you mean am instead of an ? ai is the i-th row of A
| Quote: | 3) A=USV^T according to the SVD where V is a nxn matrix of
eigenvectors.
of A^TA |
| Quote: |
now, the singular vectors vi, do they belong to W? Or, in other words,
what are the conditions that this proposition holds? For example one
might write:
vi=(1/si)*A^T*ui. Therefore, if W=span(ai)then the proposition will
hold. But let's say D is a real positive definite and symmetric matrix
with eigenvalues E and W=ran(E) and ai belong to W. But also span(ai)
belongs to ran(E). As a conclusion, if all this holds then the right
singular vectors are linear combinations of the eigenvectors if all of
the above hold. Does anyone like any of this or am I a complete idiot?
Any help will be greatly appreciated.
TIA and cheers
dps
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vi=(1/si) sum_{k=1 to m} ak*u_{ki}
and since by your assumption each ak is in W also vi in W for any i
hth
peter |
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ds science forum beginner
Joined: 23 Jun 2006
Posts: 1
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Posted: Fri Jun 23, 2006 12:30 am Post subject:
Re: svd question
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Hello Peter,
| Quote: | 2) matrix A=[a1 a2 a3 ... an]^T is a mxn matrix of vectors
you mean am instead of an ? ai is the i-th row of A
Correct, matrix A should have m rows.
vi=(1/si) sum_{k=1 to m} ak*u_{ki}
and since by your assumption each ak is in W also vi in W for any i
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Yes, that's the idea. And since W is the space of the linear
combinations of the eigenvalues then every vi is a linear combination
of the eigenvalues. I haven't found this in any reference and this is
very interesting for me. |
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