7.7. EXERCISES 177

Thus you can’t say f (x,y,λ ) = 0 defines (x,y) as a function of λ near (0,0,0).However, let

Q

β

)≡

(1/2 1/21/2 1/2

)(α

β

)=

(α+β

2α+β

2

)

(I−Q)

β

)=

β

)−

(α+β

2α+β

2

)=

(12 α− 1

2 β

12 β − 1

2 α

)

The equation f (x,y,λ ) = 0 can be written in the form

Qf (x,y,λ ) =

(− 1

2 x2 + 12 xy+ x+ y2 + 1

2 λ + 12 sinλ

− 12 x2 + 1

2 xy+ x+ y2 + 12 λ + 1

2 sinλ

)= 0 (7.24)

(I−Q)f (x,y,λ ) =( 1

2 x2 + 12 yx− 1

2 λ + 12 sinλ

− 12 x2− 1

2 yx+ 12 λ − 1

2 sinλ

)= 0

DxQf (0,0,0) =(

11

)which is one to one on R. Indeed, if

(11

)u =

(00

),

then u = 0. By Theorem 7.2.1, the first equation in 7.24 defines x = x(y,λ ) forsmall y,λ . Also, you know it is a Ck function for every k so you can use Taylorapproximation for functions of many variables to approximate x(y,λ ). In the topequation, xy = 0. Also xλ = −1 so x(y,λ ) ≈ −λ other than higher order terms forsmall y,λ . Now plug in to the bottom equation

12

x2 (y,λ )+12

yx(y,λ )− 12

λ +12

sinλ

=12(−λ )2 +

12

y(−λ )− 12

λ +12

sinλ = 0

Solve this for y to find y(λ ) = −1+ sin(λ )λ

+λ at least approximately. This kind ofprocedure is called the Lyapunov Schmidt procedure. It deals with the case wherethe partial derivative used in the statement of the implicit function theorem is notinvertible. Note how it was possible to solve for a solution f (x,y,λ ) = 0 in thisexample.

5. Let f ((x,y) ,λ ) =(

x+ xy+ y2 + xsin(λ )x+ y2− x2 + xλ

). One solution to f ((x,y) ,λ ) = 0 is

x(λ ) = y(λ ) = 0. Use the above procedure to show there is a nonzero solution tothis non-linear system of equations for small λ .

6. Let X ,Y be finite dimensional vector spaces and let L ∈L (X ,Y ). Let {Lx1, ...,Lxm}be a basis for L(X). Show that if {z1, ...,zr} is a basis for ker(L) , then a basisfor X is {x1, ...,xm,z1, ...,zr} is a basis for X . Show that L is one to one on X1 ≡span(x1, ...,xm) .

7. Go through the details of the following argument. Let f : U ⊆Rn×Rm→Rn whereU is open in Rn×Rm,(0,0) ∈U . Let f be Ck for k ≥ 1. Also suppose f (0,0) =0. If L = D1f (0,0) and if L−1 exists, then by the implicit function theorem, the

7.7. EXERCISES 177Thus you can’t say f (x,y,A) = 0 defines (x,y) as a function of A near (0,0,0).However, let(5) = (BIG)08) = (ee) (i)2lIaNERwl wo}tm =eNeThe equation f (x,y, A) = 0 can be written in the formOf (x,y,4) =1,21 2,1 le5x + 5xytxtye+5A45sina( zh ee ) =0 (7.24)5x4 sxytxty?+ 5A +5sind1,2 1 1 lg_ _ ax + 5yx—5A+5sind _(1—Q) f(y.) ( —5x° —Zyx+5A—4sind 0D,Qf (0,0,0) = ( ) which is one to one on R. Indeed, if ( e = ( ° ;then u = 0. By Theorem 7.2.1, the first equation in 7.24 defines x = x(y,A) forsmall y,A. Also, you know it is a C* function for every k so you can use Taylorapproximation for functions of many variables to approximate x(y,4). In the topequation, x, = 0. Also x, = —1 so x(y,A) & —A other than higher order terms forsmall y, 2. Now plug in to the bottom equationI 4 1 1 1.3% (A) + 5x14) — 54+ 5 sind1 ay I a= 5 Xd) + 5y( i) x4 +5 sind =0sin(A)Solve this for y to find y(A) = —1+ > +A at least approximately. This kind ofprocedure is called the Lyapunov Schmidt procedure. It deals with the case wherethe partial derivative used in the statement of the implicit function theorem is notinvertible. Note how it was possible to solve for a solution f (x,y,A) = 0 in thisexample.xtxy+y? +xsin (A)x+y? —2x7+4+xdx(A) = y(A) =0. Use the above procedure to show there is a nonzero solution tothis non-linear system of equations for small A.5. Let f ((x,y),A) = ( ) . One solution to f ((x,y),A) = 0 is6. Let X,Y be finite dimensional vector spaces and let L€ Y (X,Y). Let {Lx,...,Lxm}be a basis for L(X). Show that if {z,,...,z-} is a basis for ker(L), then a basisfor X is {x1,...,%m,Z1,---,Zr} is a basis for X. Show that L is one to one on X; =span (x1, ...,%m)-7. Go through the details of the following argument. Let f : U C R” x R” — R” whereU is open in R” x R”, (0,0) € U. Let f be C* for k > 1. Also suppose f (0,0) =0. If L= D)f (0,0) and if L~! exists, then by the implicit function theorem, the