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SVD of a reduced rank matrix still has non-zero U and V`?

  1. Apr 29, 2015 #1
    In a given matrix A, the singular value decomposition (SVD), yields A=USV`. Now lets make dimension reduction of the matrix by keeping only one column vector from U, one singular value from S and one row vector from V`. Then do another SVD of the resulted rank reduced matrix Ar.

    Now, if Ar is the result of multiplication of Ur , Sr and V`r, then why the result, shown in the right picture in the attached doc, still has non-vanishing columns of Ur and non-vanishing rows of V`r? in other words, where do Ur1, Ur2, V`r1 and V`r2 come from as long as other values of S, namely S2 and S3 are zero?
     

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  3. Apr 29, 2015 #2

    Stephen Tashi

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    It's better to say "a singular value decomposition" since singular value decompositions are not unique.

    In the right hand side of your page, you could set columns 2 and 3 of [itex] U [/itex] equal to zeroes and rows 2 and 3 of [itex] V '[/itex] equal to zeroes and you'd still have a singular value decomposition.


    It isn't clear what you mean by "keeping" only one of the singular values.

    One way to visualize the singular value decomposition of [itex] M = USV' [/itex] is to say that the entries of [itex] M [/itex] are a table of data of some sort and the the singular value decomposition of [itex] M [/itex] expresses it as a linear combination of "simple" data tables where the singular values are the coefficients in the linear combination. The simple data table are imagined to have a list of "row headings" on their left margin and a list of "column headings" across the top and each entry in the simple table is the product of the corresponding row heading and column heading for that entry. A given simple table has as the jth column of [itex] U [/itex] as its row headings and the jth row of [itex] V' [/itex] as its column headings.

    According to that way of looking at things, the way to "keep only one" a singular value would be to keep only the corresponding term in the linear combination. This amounts to setting the other singular values equal to zero.

    A linear combination where zeroes are allowed lets us write vector equations like (1,2,-4) = (1)(1,2,-4) = (1)(1,2,-4) + (0)(5,6,7) + (0)(9,3,14) where arbitrary vectors can appear as long as their coefficients are zero. A similar statement applies to linear combinations of simple data tables. This implies that some columns of [itex] U [/itex] and some rows of [itex] V' [/itex] can be chosen arbitrarily.
     
  4. Apr 29, 2015 #3
    So I have two concerns here:
    1) does that mean we can construct infinite number of matrices U and V' by arbitrarily choosing orthonormal columns vectors of U and row vectors of V'?
    2) back to my question, U And V' are constructed from AA'=USSU' and A'A=VSSV', but if this is applied for the reduced form of A, where we only have one non-zero singular value, then where do other U and V' vectors apart from the first ones arise as long as A has only rank1 so do AA' and A'A?
     
    Last edited: Apr 29, 2015
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