220 CHAPTER 8. IMPLICIT FUNCTION THEOREM

=n

∑j=1

u2jλ j ≥ δ

2n

∑j=1

u2j = δ

2 |u|2 .

From 8.23 and the continuity of H, if v is small enough,

f (x+v)≥ f (x)+12

δ2 |v|2− 1

4δ

2 |v|2 = f (x)+δ

2

4|v|2 .

This shows the first claim of the theorem. The second claim follows from similar reason-ing. Suppose H (x) has a positive eigenvalue λ

2. Then let v be an eigenvector for thiseigenvalue. Then from 8.23,

f (x+tv) = f (x)+12

t2vT H (x)v+

12

t2 (vT (H (x+tv)−H (x))v)

which implies

f (x+tv) = f (x)+12

t2λ

2 |v|2 + 12

t2 (vT (H (x+tv)−H (x))v)

≥ f (x)+14

t2λ

2 |v|2

whenever t is small enough. Thus in the direction v the function has a local minimum atx. The assertion about the local maximum in some direction follows similarly. This provesthe theorem. ■

This theorem is an analogue of the second derivative test for higher dimensions. As inone dimension, when there is a zero eigenvalue, it may be impossible to determine from theHessian matrix what the local qualitative behavior of the function is. For example, consider

f1 (x,y) = x4 + y2, f2 (x,y) =−x4 + y2.

Then D fi (0,0) = 0 and for both functions, the Hessian matrix evaluated at (0,0) equals(0 00 2

)but the behavior of the two functions is very different near the origin. The second has asaddle point while the first has a minimum there.

8.8 The Rank TheoremThis is a very interesting result. The proof follows Marsden and Hoffman. First here issome linear algebra.

Theorem 8.8.1 Let L : Rn→ RN have rank m. Then there exists a basis

{u1, · · · ,um,um+1, · · · ,un}

such that a basis for ker(L) is {um+1, · · · ,un} .