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

1. Apr 29, 2015

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 Vr, 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 Vr? in other words, where do Ur1, Ur2, Vr1 and V`r2 come from as long as other values of S, namely S2 and S3 are zero?

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• ###### SVD.pdf
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2. Apr 29, 2015

### Stephen Tashi

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 $U$ equal to zeroes and rows 2 and 3 of $V '$ 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 $M = USV'$ is to say that the entries of $M$ are a table of data of some sort and the the singular value decomposition of $M$ 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 $U$ as its row headings and the jth row of $V'$ 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 $U$ and some rows of $V'$ can be chosen arbitrarily.

3. Apr 29, 2015