Mathematical Methods for Physics and Engineering - Matematica.NET

MATRICES AND VECTOR SPACESUsing the unitarity of the matrices U and V, we find thatAˆx − b = USSU † b − b = U(SS − I)U † b. (8.142)The matrix (SS − I) is diagonal and the jth element on its leading diagonal isnon-zero (and equal to −1) only when s j = 0. However, the jth element of thevector U † b is given by the scalar product (u j ) † b;ifb lies in the range of A, thisscalar product can be non-zero only if s j ≠ 0. Thus the RHS of (8.142) mustequal zero, and so ˆx given by (8.141) is a solution to the equations Ax = b. Wemay, however, add to this solution any linear combination of the N − r vectors v i ,i = r+1,r+2,...,N, that form an orthonormal basis for the null space of A; thus,in general, there exists an infinity of solutions (although it is straightforward toshow that (8.141) is the solution vector of shortest length). The only way in whichthe solution (8.141) can be unique is if the rank r equals N, so that the matrix Adoes not possess a null space; this only occurs if A is square and non-singular.If b does not lie in the range of A then the set of equations Ax = b doesnot have a solution. Nevertheless, the vector (8.141) provides the closest possible‘solution’ in a least-squares sense. In other words, although the vector (8.141)does not exactly solve Ax = b, it is the vector that minimises the residualɛ = |Ax − b|,where here the vertical lines denote the absolute value of the quantity theycontain, not the determinant. This is proved as follows.Suppose we were to add some arbitrary vector x ′ to the vector ˆx in (8.141).This would result in the addition of the vector b ′ = Ax ′ to Aˆx − b; b ′ is clearly inthe range of A since any part of x ′ belonging to the null space of A contributesnothing to Ax ′ . We would then have|Aˆx − b + b ′ | = |(USSU † − I)b + b ′ |= |U[(SS − I)U † b + U † b ′ ]|= |(SS − I)U † b + U † b ′ |; (8.143)in the last line we have made use of the fact that the length of a vector is leftunchanged by the action of the unitary matrix U. Now, the jth component of thevector (SS − I)U † b will only be non-zero when s j = 0. However, the jth elementof the vector U † b ′ is given by the scalar product (u j ) † b ′ , which is non-zero only ifs j ≠0,sinceb ′ lies in the range of A. Thus, as these two terms only contribute to(8.143) for two disjoint sets of j-values, its minimum value, as x ′ is varied, occurswhen b ′ = 0; this requires x ′ = 0.◮Find the solution(s) to the set of simultaneous linear equations Ax = b, whereA is givenby (8.137) and b = (1 0 0) T .To solve the set of equations, we begin by calculating the vector given in (8.141),x = VSU † b,306