10.3 Some linear algebra
It will be useful to analyze in a little more detail the procedure for solving equation (10.19)
above, as the generalization of this method will enter when we deal with degenerate pertur-
bation theory. It is best to illustrate the issues for anN-dimensional Hilbert spaceH, whose
vectors are represented by column vectors, and whose inner product is given by (u,v) =u†v
for allu,v∈ H. In this case, we have a Hermitian linear operatorA, represented by an
N×N Hermitean matrix, and a vectorbin Hilbert space. The relevant linear equation,
analogous to (10.19) is then,
Ax=b (10.25)
wherex∈His the unknown vector. WhenAis invertible, we have simplyx=A−^1 b. When
Ais not invertible, the above equation may still admit solution, under certain conditions.
WhenAis not invertible, it has a non-trivial null-space orkernel, and a non-trivialrange,
defined by
KerA = {v∈Hsuch thatAv= 0}
RangeA = AH (10.26)
Equation (10.25) has at least one solution if and only ifb∈RangeA. This rather abstract
condition may be translated into a simple concrete criterion. To test whether a givenb
belongs to RangeA, we proceed as follows. Ifbbelongs to RangeA, then it follows that
(v,b) = (v,Ax) = (Av,x) = 0 for allv∈KerA. For any vector subspaceV ofH, we define
it orthogonal complement as follows,
V⊥={v∈Hsuch that (u,v) = 0 for allu∈V} (10.27)
For any vector space we clearly have
V ⊕V⊥=H (10.28)
The sum is direct since a vector belonging to bothVandV⊥must necessarily vanish. From
this definition, we conclude that RangeA ⊂ (KerA)⊥. Conversely, if v ∈ (RangeA)⊥,
then we have (v,Ax) = (Av,x) = 0 for allx∈ H, so thatv∈KerA, and we also have
(RangeA)⊥⊂KerA. As a result, we must have^8
RangeA= (KerA)⊥ (10.29)
(^8) For non-selfadjoint operators, the corresponding equation is analogously given by RangeA†= (KerA)⊥,
whereA†is the adjoint operator toA.