6.7 Second Derivative Test
Now consider the case where U ⊆ ℝn and f : U → ℝ is C2
. Then from Taylor’s theorem,
is small enough, there exists t ∈
are the usual basis vectors. Letting
the second derivative term in 6.22 reduces to
Definition 6.7.1 The matrix whose ijth entry is
is called the Hessian
matrix, denoted as H
From Theorem 5.10.1, this is a symmetric real matrix, thus self adjoint. By the continuity
of the second partial derivative,
where the last two terms involve ordinary matrix multiplication and
for vi the components of v relative to the standard basis.
Definition 6.7.2 Let f : D → ℝ where D is a subset of some normed vector
space. Then f has a local minimum at x ∈ D if there exists δ > 0 such that for all
y ∈ B
f has a local maximum at x ∈ D if there exists δ > 0 such that for all y ∈ B
Theorem 6.7.3 If f : U → ℝ where U is an open subset of ℝn and f is C2,
. Then if H
has all positive eigenvalues, x is a local minimum.
If the Hessian matrix H
has all negative eigenvalues, then x is a local maximum.
has a positive eigenvalue, then there exists a direction in which f has a local
minimum at x, while if H
has a negative eigenvalue, there exists a direction in which
has a local maximum at x.
Proof: Since Df
= 0, formula
holds and by continuity of the second derivative,
is a symmetric matrix. Thus
has all real eigenvalues. Suppose first that
all positive eigenvalues and that all are larger than
0. Then by Theorem A.4.1
an orthonormal basis of eigenvectors,
and if u
is an arbitrary vector, such that
= u ⋅ vj
From 6.23 and the continuity of H, if v is small enough,
This shows the first claim of the theorem. The second claim follows from similar reasoning.
has a positive eigenvalue
. Then let v
be an eigenvector for this eigenvalue.
Then from 6.23
is small enough. Thus in the direction v
the function has a local minimum at x.
The assertion about the local maximum in some direction follows similarly. This proves the
This theorem is an analogue of the second derivative test for higher dimensions. As in one
dimension, when there is a zero eigenvalue, it may be impossible to determine from the
Hessian matrix what the local qualitative behavior of the function is. For example,
and for both functions, the Hessian matrix evaluated at
but the behavior of the two functions is very different near the origin. The second has a
saddle point while the first has a minimum there.