Page 425 - Matrix Analysis & Applied Linear Algebra
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5.12 Singular Value Decomposition                                                  421

                                    It’s proven on p. 435 that 9 q = P R(A) q is the vector in R (A) (the document
                                    space) that is closest to q, so using 9 q in place of q has the effect of using the
                                    best approximation to q that is a linear combination of the document vectors
                                                      T
                                              T
                                    d i . Since 9 q A = q A and  9 q  ≤ q  , it follows that cos θ j ≥ cos θ j , so
                                                                                            9
                                                                 2      2
                                    more documents are deemed relevant when the projected query is used. Just as
                                    in the unprojected query matching strategy, the noise is filtered out by replacing
                                                                             k       T
                                    A in (5.12.14) with a truncated SVD A k =   σ i u i v . The result is
                                                                             i=1     i
                                                                        T
                                                                       q U k s j
                                                               9
                                                            cos φ j =
                                                                     U q     s j
                                                                    
  T
                                                                       k   2   2
                                    and, just as in (5.12.13), cos φ j is easily and quickly computed for each new
                                                              9
                                    query q because U k and s j need only be computed once.
                                        The next example shows why singular values are the primary mechanism
                                    for numerically determining the rank of a matrix.
                   Example 5.12.5
                                                                                  m×n
                                    Perturbations and Numerical Rank. For A ∈          with p = min{m, n},
                                    let {σ 1 ,σ 2 ,...,σ p } and {β 1 ,β 2 ,...,β p } be all singular values (nonzero as well
                                    as any zero ones) for A and A + E, respectively.
                                    Problem: Prove that

                                               |σ k − β k |≤ E  2  for each  k =1, 2,...,p.      (5.12.15)


                                    Solution: If the SVD for A given in (5.12.2) is written in the form
                                                    p                               k−1
                                                           T                                 T
                                              A =     σ i u i v ,  and if we set  A k−1 =  σ i u i v ,
                                                           i                                 i
                                                   i=1                              i=1
                                    then
                                             σ k =  A − A k−1   =  A + E − A k−1 − E
                                                             2                     2
                                                ≥ A + E − A k−1   − E      (recall (5.1.6) on p. 273)
                                                                 2     2
                                                ≥ β k − E    by (5.12.10).
                                                          2
                                    Couple this with the observation that
                                              σ k = min     A − B  2 = min     A + E − B − E  2
                                                rank(B)=k−1         rank(B)=k−1

                                                 ≤ min      A + E − B  2 +  E  2 = β k +  E  2
                                                rank(B)=k−1
                                    to conclude that |σ k − β k |≤ E  2 .
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