69.4. THE IMPLICIT CASE 2375

because ⟨Bxk,xk⟩ is zero by what was just shown.Thus assume there is such a first index. Let

e1 ≡gn1

⟨Bgn1 ,gn1⟩1/2

Then ⟨Be1,e1⟩= 1. Now if you have constructed e j for j ≤ k,

e j ∈ span(gn1 , · · · ,gnk

),⟨Bei,e j

⟩= δ i j,

gn j+1 being the first for which⟨Bgn j+1 −

j

∑i=1

⟨Bgn j+1 ,ei

⟩Bei,gn j+1 −

j

∑i=1

⟨Bgn j,ei

⟩ei

⟩̸= 0,

andspan

(gn1 , · · · ,gnk

)= span(e1, · · · ,ek) ,

let gnk+1 be such that gnk+1 is the first in the list{

gnk

}such that⟨

Bgnk+1 −k

∑i=1

⟨Bgnk+1 ,ei

⟩Bei,gnk+1 −

k

∑i=1

⟨Bgnk+1 ,ei

⟩ei

⟩̸= 0

Note the difference between this and the Gram Schmidt process. Here you don’t necessarilyuse all of the gk due to the possible degeneracy of B.

Claim: If there is no such first gnk+1 , then B(span(ei, · · · ,ek)) = BW so in this case,{Bei}k

i=1 is actually a basis for BW .Proof: Let x ∈W . Let xr ∈ span(g1, · · · ,gr) ,r > nk such that limr→∞ xr = x in W . Then

xr =k

∑i=1

cri ei +

r

∑i/∈{n1,··· ,nk}

dri gi ≡ yr + zr (69.4.17)

If l /∈ {n1, · · · ,nk} , then by the construction and the above assumption, for some j ≤ k⟨Bgl−

j

∑i=1⟨Bgl ,ei⟩Bei,gl−

j

∑i=1⟨Bgl ,ei⟩ei

⟩= 0 (69.4.18)

If l < nk, this follows from the construction. If the above is nonzero all j ≤ k, then l wouldhave been chosen but it wasn’t. Thus

Bgl =j

∑i=1⟨Bgl ,ei⟩Bei

If l > nk, then by assumption, 69.4.18 holds for j = k. Thus, in any case, it follows that foreach l /∈ {n1, · · · ,nk} ,

Bgl ∈ B(span(ei, · · · ,ek)) .