8.18 SIMULTANEOUS LINEAR EQUATIONS
We already know from the above discussion, however, that the non-zero eigenvalues of
this matrix areequalto those ofAA†found above, and that the remaining eigenvalue is
zero. The corresponding normalised eigenvectors are easily found:
λ 1 =16 ⇒ v^1 =^12 (1 1 1 1)T
λ 2 =9 ⇒ v^2 =^12 (1 1 − 1 −1)T
λ 3 =4 ⇒ v^3 =^12 (− 111 −1)T
λ 4 =0 ⇒ v^4 =^12 (1 − 11 −1)T
andsothematrixVis given by
V=
1
2
11 − 11
11 1− 1
1 −11 1
1 − 1 − 1 − 1
. (8.140)
Alternatively, we could have found the first three columns ofVby using the relation
(8.135) to obtain
vi=
1
si
A†ui fori=1, 2 , 3.
The fourth eigenvector could then be found using the Gram–Schmidt orthogonalisation
procedure. We note that if there were more than one eigenvector corresponding to a zero
eigenvalue then we would need to use this procedure to orthogonalise these eigenvectors
before constructing the matrixV.
Collecting our results together, we find the SVD of the matrixA:
A=USV†=
10 0
0 35 −^45
(^04535)
4000
0300
0020
1
2
1
2
1
2
1
2
1
2
1
2 −
1
2 −
1
2
−^121212 −^12
1
2 −
1
2
1
2 −
1
2
;
this can be verified by direct multiplication.
Let us now consider the use of SVD in solving a set ofMsimultaneous linear
equations inNunknowns, which we write again asAx=b. Firstly, consider
the solution of a homogeneous set of equations, for whichb= 0. As mentioned
previously, ifAis square and non-singular (and so possesses no zero singular
values) then the equations have the unique trivial solutionx= 0 .Otherwise,any
of the vectorsvi,i=r+1,r+2,...,N, or any linear combination of them, will
be a solution.
In the inhomogeneous case, wherebis not a zero vector, the set of equations
will possess solutions ifblies in the range ofA. To investigate these solutions, it
is convenient to introduce theN×MmatrixS, which is constructed by taking
the transpose ofSin (8.131) and replacing each non-zero singular valuesion
the diagonal by 1/si. It is clear that, with this construction,SSis anM×M
diagonal matrix with diagonal entries that equal unity for those values ofjfor
whichsj= 0, and zero otherwise.
Now consider the vector
xˆ=VSU†b. (8.141)