178 CHAPTER 7. IMPLICIT FUNCTION THEOREM

equation f (x,λ) = 0 defines x= x(λ) for small λ and x is Ck. Let {y1, ...,ym}be a basis for L(Rn) and enlarge to get {y1, ...,ym,wm+1, ...,wn} as a basis forRn. Letting Lxk = yk use the above problem to have a basis for X which is of theform {x1, ...,xm,zm+1, ...,zn} with {zm+1, ...,zn} a basis for ker(L) . Thus, fromthe above problem L is one to one on X1 ≡ span(x1, ...,xm) . For x̂ ∈ X1, showDx̂f (0,0) is the restriction of L to X1 and so Dx̂f (0,0) is one to one on X1. Nowdefine the linear map Q : Rn → Rn by Q

(∑

mk=1 akyk +∑

nk=m+1 bkwk

)≡ ∑

mk=1 akyk.

Thus Q2 = Q. We can write the original equations f (x,λ) = 0 as

Qf (x̂, x̃,λ) = Qf (x,λ) = 0, x̃ ∈ ker(L)(I−Q)f (x̂, x̃,λ) = 0

Thus Qf (x,λ)∈ span(y1, ...,ym)≡Y1. Now show that for x̂ the variable in X1, andif v ∈ X1, and Dx̂Qf (0,0,0)v = 0, then v = 0 and so we can apply the implicitfunction theorem to obtain x̂= x̂(x̃,λ) as the solution to Qf (x,λ) = 0 for x̃,λsmall where here x̃ is in ker(L). Since everything in sight is Ck, one can use Tay-lor series for functions of many variables to approximate the solution in these twoequations. See the Taylor formula 7.19. This is the general idea in the above twoproblems.