Advanced Calculus of Several Variables (1973)

Part II. Multivariable Differential Calculus

Chapter 8. THE CLASSIFICATION OF CRITICAL POINTS

Let f be a real-valued function of class on an open set , and let be a critical point of f. In order to apply Theorem 7.5 to determine the nature of the critical point a, we need to decide the character of the quadratic form defined by

where the n × n matrix A = (a_ij) is defined by , and

Note that the matrix A is symmetric, that is a_ij = a_ji, since D_i D_j = D_j D_i.

We have seen that, in order to determine whether q is positive-definite, negative-definite, or nondefinite, it suffices to determine the maximum and minimum values of q on the unit sphere Sⁿ⁻¹. The following result rephrases this problem in terms of the linear mapping defined by

We shall refer to the quadratic form q, the symmetric matrix A, and the linear mapping L as being “associated” with one another.

Theorem 8.1 If the quadratic form q attains its maximum or minimum value on Sⁿ⁻¹ at the point , then there exists a real number λ such that

where L is the linear mapping associated with q.

Such a nonzero vector v, which upon application of the linear mapping is simply multiplied by some real number λ, L(v) = λv, is called an eigenvector of L, with associated eigenvalue λ.

PROOF We apply the Lagrange multiplier method. According to Theorem 5.5, there exists a number such that

where defines the sphere Sⁿ⁻¹.

But D_k g(x) = 2x_k, and

where the row vector A_k is the kth row of A, that is, A_k = (a_k1 · · · a_kn). Consequently Eq. (1) implies that

for each i = 1, . . . , n. But A_k v is the ith row of Av, while λv_k is the kth element of the column vector λv, so we have

as desired.

Thus the problem of maximizing and minimizing q on Sⁿ⁻¹ involves the eigenvectors and eigenvalues of the associated linear mapping. Actually, it will turn out that if we can determine the eigenvalues without locating the eigenvectors first, then we need not be concerned with the latter.

Lemma 8.2 If is an eigenvector of the linear mapping associated with the quadratic form q, then

where λ is the eigenvalue associated with v.

PROOF

since .

The fact that we need only determine the eigenvalues of L (and not the eigenvectors) is a significant advantage, because the following theorem asserts that we therefore need only solve the so-called characteristic equation

Here A − λI denotes the determinant of the matrix A − λI (where I is the n × n identity matrix).

Theorem 8.3 The real number λ is an eigenvalue of the linear mapping , defined by L(x) = Ax, if and only if λ satisfies the equation

PROOF Suppose first that λ is an eigenvalue of L, that is,

for some v ≠ 0. Then (A − λI)v = 0.

Enumerating the rows of the matrix A − λI, we obtain

where e₁, . . . , e_n are the standard unit vectors in ⁿ. Hence the n vectors A₁ − λe₁, . . . , A_n − λe_n all lie in the (n − 1)-dimensional hyperplane through 0 perpendicular to v, and are therefore linearly dependent, so it follows from Theorem I.6.1 that

as desired.

Conversely, if λ is a solution of A − λI = 0, then the row vectors A₁ − λe₁, . . . , A_n − λe_n of A − λI are linearly dependent, and therefore all lie in some (n − 1)-dimensional hyperplane through the origin. If v is a unit vector perpendicular to this hyperplane, then

Hence (A − λI)v = 0, so

as desired.

Corollary 8.4 The maximum (minimum) value attained by the quadratic form q(x) = x^tAx on Sⁿ⁻¹ is the largest (smallest) real root of the equation

PROOF If q(v) is the maximum (minimum) value of q on Sⁿ⁻¹, then v is an eigenvector of L(x) = Ax by Theorem 8.1, and q(v) = λ, the associated eigenvalue, by Lemma 8.2. Theorem 8.3 therefore implies that q(v) is a root of A − λI = 0. On the other hand, it follows from 8.2 and 8.3 that every root of A − λI = 0 is a value of q on Sⁿ⁻¹. Consequently q(v) must be the largest (smallest) root of A − λI = 0.

Example 1 Suppose a is a critical point of the function f: and that the quadratic form of f at a is

so the matrix of q is

The characteristic equation of A is then

with roots λ = −1, 1, 3. It follows from Corollary 8.4 that the maximum and minimum values of q in S² are + 3 and − 1 respectively. Since q attains both positive and negative values, it is nondefinite. Hence Theorem 7.5 implies that f has neither a maximum nor a minimum at a.

The above example indicates how Corollary 8.4 reduces the study of the quadratic form q(x) = x^tAx to the problem of finding the largest and smallest real roots of the characteristic equation A − λI = 0. However this problem itself can be difficult. It is therefore desirable to carry our analysis of quadratic forms somewhat further.

Recall that the matrix A of a quadratic form is symmetric, that is, a_ij = a_ji. We say that the linear mapping is symmetric if and only if L(x) · y = x · L(y) for all x and y in ⁿ. The motivation for this terminology is the following result.

Lemma 8.5 If A is the matrix of the linear mapping , then L is symmetric if and only if A is symmetric.

PROOF If , then

Hence L is symmetric if and only if

for all i, j = 1, . . . , n. But

and

The following theorem gives the precise character of a quadratic form q in terms of the eigenvectors and eigenvalues of its associated linear mapping L.

Theorem 8.6 If q is a quadratic form on ⁿ, then there exists an orthogonal set of unit eigenvectors v₁, . . . , v_n of the associated linear mapping L. If y₁, . . . , y_n are the coordinates of with respect to the basis v₁, . . . , v_n, that is,

then

where λ₁, . . . , λ_n are the eigenvalues of v₁, . . . , v_n.

PROOF This is trivially true if n = 1, for then q is of the form q(x) = λx², with L(x) = λx. We therefore assume by induction that it is true in dimension n − 1.

If q attains its maximum value on the unit sphere Sⁿ⁻¹ at , then see Theorem 8.1 says that v_n is an eigenvector of L. Fig. 2.40.

Figure 2.40

Let be the subspace of ⁿ perpendicular to v_n. Then L maps into itself, because, given , we have

using the symmetry of L (Lemma 8.5).

Let and be the “restrictions” of L and q to , that is,

If (u₁, . . . , u_n−1) are the coordinates of x with respect to an orthonormal basis for ⁿ⁻¹, then by Lemma 8.5,

where B is a symmetric (n − 1) × (n − 1) matrix. Then

so q^* is a quadratic form in u₁, . . . , u_n−1.

The inductive hypothesis therefore gives an orthogonal set of unit eigenvectors v₁, . . . , v_n−1 for L^*. Then v₁, . . . , v_n is the desired orthogonal set of unit eigenvectors for L.

To prove the last statement of the theorem, let . Then

since v_i · v_j = 0 unless i = j, and v_i · v_i = 1.

Equation (2) may be written in the form

The process of finding an orthogonal set of unit eigenvectors, with respect to which the matrix of q is diagonal, is sometimes called diagonalization of the quadratic form q.

Note that, independently of Corollary 8.4, it follows immediately from Eq. (2) that

(a)q is positive-definite if all the eigenvalues are positive,

(b)q is negative-definite if all the eigenvalues are negative, and

(c)q is nondefinite if some are positive and some are negative.

If one or more eigenvalues are zero, while the others are either all positive or all negative, then q is neither positive-definite nor negative-definite nor nondefinite. According to Lemma 8.7 below, this can occur only if the determinant of the matrix of q vanishes.

Example 2 If q(x, y, z) = xy + yz, the matrix of q is

The characteristic equation is then

with solutions . By substituting each of these eigenvalues in turn into the matrix equation

(from the proof of Theorem 8.3), and solving for a unit vector v, we obtain the eigenvectors

If u, v, w are coordinates with respect to coordinate axes determined by v₁, v₂, v₃, that is, x = uv₁ + vv₂ + wv₃, then Eq. (2) gives

Lemma 8.7 If λ₁, . . . , λ_n are eigenvalues of the symmetric n × n matrix A, corresponding to orthogonal unit eigenvectors v₁, . . . , v_n, then

PROOF Let

Then, since Av_j = λ_iv_j, we have

It follows that

Since V ≠ 0 because the vectors v₁, . . . , v_n are orthogonal and therefore linearly independent, this gives Eq. (3).

We are finally ready to give a complete analysis of a quadratic form q(x) = x^t Ax on ⁿ for which A ≠ 0. Writing A = (a_ij) as usual, we denote by Δ_k the determinant of the upper left-hand k × k submatrix of A, that is,

Thus

Theorem 8.8 Let q(x) = x^t Ax be a quadratic form on ⁿ with A ≠ 0. Then q is

(a)positive-definite if and only if Δ_k >0 for each k = 1, . . . , n,

(b)negative-definite if and only if (−1)^k Δ_k >0 for each k = 1, . . . , n,

(c)nondefinite if neither of the previous two conditions is satisfied.

PROOF OF (a) This is obvious if n = 1; assume by induction its truth in dimension n − 1.

Since 0 < Δ_n = A = λ₁λ₂ · · · λ_n by Lemma 8.7, all the eigenvalues λ₁, . . . , λ_n of A are nonzero. Either they are all positive, in which case we are finished, or an even number of them are negative. In the latter case, let λ_i and λ_j be two negative eigenvalues, corresponding to eigenvectors v_i and v_j.

Let be the 2-plane in ⁿ spanned by v_i and v_j. Then, given , we have

so q is negative on minus the origin.

Now let q_* : ^{n − 1} be the restriction of q to ⁿ⁻¹, defined by

where

Since Δ₁ >0, . . . , Δ_{n − 1} >0, the inductive assumption implies that q_* is positive-definite on ⁿ⁻¹. But this is a contradiction, since the (n − 1)-plane ⁿ⁻¹ and the 2-plane , on which q is negative, must have at least a line in common. We therefore conclude that q has no negative eigenvalues, and hence is positive-definite.

Conversely, if q is positive-definite, we want to prove that each Δ_k is positive. If m is the minimum value of q on Sⁿ⁻¹, then m >0.

Let q_k : ^k be the restriction of q to ^k, that is,

where A_k is the upper left-hand k × k submatrix of A, so Δ_k = A_k, and define L_k : ^{k k} by L_k(x) = A_k x. Then L_k is symmetric by Lemma 8.5.

Let ¼₁, . . . , ¼_k be the eigenvalues of L_k given by Theorem 8.6. Since by Lemma 8.2 each ¼_i is the value of q_k at some point of S^k−1, the unit sphere in ^k, and q_k(x) = q(x, 0), it follows that each . Finally Lemma 8.7implies that

PROOF OF (b) Define the quadratic form q^* : ⁿ by q^*(x) = −q(x). Then clearly q is negative-definite if and only if q^* is positive-definite. By part (a), q^* is positive-definite if and only if each Δ_k^* is positive, where Δ_k^* = −A_k, A_k being the upper left-hand k × k submatrix of A. But, since each of the k rows of − A_k is −1 times the corresponding row of A_k, it follows that

PROOF OF (c) Since Δ_n = λ₁ λ₂ · · · λ_n ≠ 0, all of the eigenvalues λ₁, . . . , λ_n are nonzero. If all of them were positive, then part (a) would imply that Δ_k >0 for all k, while if all of them were negative, then part (b) would imply that (−1)^k Δ_k >0 for all k. Since neither is the case, we conclude that q has both positive eigenvalues and negative eigenvalues, and is therefore nondefinite.

Recalling that, if a is a critical point of the class function f : , the quadratic form of f at a is , Theorems 7.5 and 8.8 enable us to determine the behavior of f in a neighborhood of a,provided that the determinant

often called the Hessian determinant of f at a, is nonzero. But if D_i D_jf(a) = 0, then these theorems do not apply, and ad hoc methods must be applied.

Note that the classical second derivative test for a function of two variables (Theorem 4.4) is an immediate corollary to Theorems 7.5 and 8.8.

It will clarify the picture at this point to generalize the discussion at the end of Section 6. Suppose for the sake of simplicity that the origin is a critical point of the class function f: , and that f(0) = 0. Then

with

Let v₁, . . . , v_n be an orthogonal set of unit eigenvectors (provided by Theorem 8.6), with associated eigenvalues λ₁, . . . , λ_n. Assume that D_i D_jf(0) ≠ 0, so Lemma 8.7 implies that each λ_i is nonzero. Let λ₁, . . . , λ_k be the positive eigenvalues, and λ_k+1, . . . , λ_n the negative ones (writing k = 0 if all the eigenvalues are negative). Since

if x = y₁v₁ + · · · + y_n v_n, it is clear that q has a minimum at 0 if k = n and a maximum if k = 0.

If , so that q has both positive and negative eigenvalues, denote by the subspace of spanned by v₁, . . . , v_k, and by the subspace of ⁿ spanned by v_k+1, . . . , v_n. Then the graph of z = q(x) in and looks like Fig. 2.41; q has a minimum on and a maximum on .

Given < min_i λ_i, choose δ >0 such that

or

since . This inequality and Eq. (4) yield

Using (5), we obtain

Figure 2.41

where ¼_i = −λ_i >0, i = k + 1, . . . , n. Since λ_i − >0 and ¼_i − >0 because < min λ_i, inequalities (6) show that, for x < δ, the graph of z = f(x) lies between two very close quadratic surfaces (hyperboloids) of the same type as z = q(x). In particular, if , then f has a local minimum on the subspace and a local maximum on the subspace . This makes it clear that the origin is a “saddle point” for f if q has both positive and negative eigenvalues. Thus the general case is just like the 2-dimensional case except that, when there is a saddle point, the subspace on which the critical point is a maximum point may have any dimension from 1 to n − 1.

It should be emphasized that, if the Hessian determinant D_iD_jf(a) is zero, then the above results yield no information about the critical point a. This will be the situation if the quadratic form q of f at a is either positive-semidefinite but not positive-definite, or negative-semidefinite but not negative-definite. The quadratic form q is called positive-semidefinite if for all x, and negative-semidefinite if for all x. Notice that the terminology “q is nondefinite,” which we have been using, actually means that q is neither positive-semidefinite nor negative-semidefinite (so we might more descriptively have said “ nonsemidefinite”).

Example 3 Let f(x, y) = x² − 2xy + y² + x⁴ + y⁴, and g(x, y) = x² − 2xy + y² − x⁴ − y⁴. Then the quadratic form for both at the critical point (0, 0) is

which is positive-semidefinite but not positive-definite. Since

f has a relative minimum at (0, 0). However, since

it is clear that g has a saddle point at (0, 0).

Thus, if the Hessian determinant vanishes, then anything can happen.

We conclude this section with discussion of a “second derivative test” for constrained (Lagrange multiplier) maximum–minimum problems. We first recall the setting for the general Lagrange multiplier problem: We have a differentiable function f: and a continuously differentiable mapping G : ^m (m < n), and denote by M the set of all those points at which the gradient vectors ∇G₁(x), . . . , ∇G_m(x), of the component functions of G, are linearly independent. By Theorem 5.7, M is an (n − m)-dimensional manifold in ⁿ, and by Theorem 5.6 has an (n − m)-dimensional tangent space T_x at each point . Recall that, by definition, T_x is the subspace of ⁿ consisting of all velocity vectors (at x) to differentiable curves on M which pass through x. This notation will be maintained throughout the present discussion.

We are interested in locating and classifying those points of M at which the function f attains its local maxima and minima on M. According to see Theorem 5.1, in order for f to have a local maximum or minimum on M at , it is necessary that a be a critical point for f on M in the following sense. The point is called a critical point for f on M if and only if the gradient vector ∇f(a) is orthogonal to the tangent space T_a of M at a. Since the linearly independent gradient vectors ∇G₁(a), . . . , ∇G_m(a) generate the orthogonal complement to T_a (Fig. 2.42), it follows as in Theorem 5.8 that there exist unique numbers λ₁, . . . , λ_m such that

Figure 2.42

We will obtain sufficient conditions for f to have a local extremum at a by considering the auxiliary function H : for f at a defined by

Notice that (7) simply asserts that ∇H(a) = 0, so a is an (ordinary) critical point for H. We are, in particular, interested in the quadratic form

of H at a.

Theorem 8.9 Let a be a critical point for f on M, and denote by q : the quadratic form at a of the auxiliary function , as above. If f and G are both of class in a neighborhood of a, then f has

(a)a local minimum on M at a if q is positive-definite on the tangent space T_a to M at a,

(b)a local maximum on M at a if q is negative-definite on T_a,

(c)neither if q is nondefinite on T_a.

The statement, that “q is positive-definite on T_a,” naturally means that q(h) >0 for all nonzero vectors ; similarly for the other two cases.

PROOF The proof of each part of this theorem is an extension of the proof of the corresponding part of Theorem 7.5. We give the details only for part (a).

We start with the Taylor expansion

for the auxiliary function H. Since , it clearly suffices to find a δ >0 such that

if

Let m be the minimum value attained by q on the unit sphere in the tangent plane T_a. Then m >0 because q is positive-definite on T_a. Noting that

we choose δ >0 so small that

and also so small that 0 < h < δ and together imply that h/h is sufficiently near to S^* that

Then (9) follows as desired. The latter condition on δ can be satisfied because S^* is (by definition) the set of all tangent vectors at a to unit-speed curves on M which pass through a. Therefore, if and h is sufficiently small, it follows that h/h is close to a vector in S^*.

The following two examples provide illustrative applications of Theorem 8.9.

Example 4 In Example 4 of Section 5 we wanted to find the box with volume 1000 having least total surface area, and this involved minimizing the function

on the 2-manifold defined by the constraint equation

We found the single critical point a = (10, 10, 10) for f on M, with . The auxiliary function is then defined by

We find by routine computation that the matrix of second partial derivatives of h at a is

so the quadratic form of h at a is

It is clear that q is nondefinite on ³. However it is the behavior of q on the tangent plane T_a that interests us.

Since the gradient vector ∇g(a) = (100, 100, 100) is orthogonal to M at a = (10, 10, 10), the tangent plane T_a is generated by the vectors

Given , we may therefore write

and then find by substitution into (10) that

Since the quadratic form s² + st + t² is positive-definite (by the ac − b² test), it follows that q is positive-definite on T_a. Therefore Theorem 8.9 assures us that f does indeed have a local minimum on M at the critical point a = (10, 10, 10).

Example 5 In Example 9 of Section 5 we sought the minimum distance between the circle x² + y² = 1 and the straight line x + y = 4. This involved minimizing the function

on the 2-manifold defined by the constraint equations

We found the geometrically obvious critical points (see Fig. 2.43), with It is obvious that f has a minimum at a, and neither a maximum nor a minimum at b, but we want to verify this using Theorem 8.9.

The auxiliary function H is defined by

Figure 2.43

By routine computation, we find that the matrix of second partial derivatives of H is

Substituting , we find that the quadratic form of H at the critical point is

Computing the subdeterminants Δ₁, Δ₂, Δ₃, Δ₄ of Theorem 8.8, we find that all are positive, so this quadratic form q is positive-definite on ⁴, and hence on the tangent plane T_a (this could also be verified by the method of the previous example). Hence Theorem 8.9 implies that f does indeed attain a local minimum at a.

Substituting , we find that the quadratic form of H at the other critical point is

Now the vectors v₁ = (1, −1, 0, 0) and v₂ = (0, 0, 1, −1) obviously lie in the tangent plane T_b. Since

we see that q is nondefinite on T_b, so Theorem 8.9 implies that f does not have a local extremum at b.

Exercises

8.1This exercise gives an alternative proof of Theorem 8.8 in the 3-dimensional case. If

where A = (a_ij) is a symmetric 3 × 3 matrix with Δ₁ ≠ 0, Δ₂ ≠ 0, show that

Conclude that q is positive-definite if Δ₁, Δ₂, Δ₃ are all positive, while q is negative-definite if Δ₁ < 0, Δ₂ >0, Δ₃ < 0.

8.2Show that q(x, y, z) = 2x² + 5y² + 2z² + 2xz is positive-definite by

(a)applying the previous exercise,

(b)solving the characteristic equation.

8.3Use the method of Example 2 to diagonalize the quadratic form of Exercise 8.2, and find its eigenvectors. Then sketch the graph of the equation 2x² + 5y² + 2z² + 2xz = 1.

8.4This problem deals with the function defined by

Note that 0 = (0, 0, 0) is a critical point of f.

(a)Show that q(x, y, z) = 2x² +5y² + 2z² + 2xz is the quadratic form of f at 0 by substituting the expansion , collecting all second degree terms, and verifying the appropriate condition on the remainder (x, y, z). State the uniqueness theorem which you apply.

(b)Write down the symmetric 3 × 3 matrix A such that

By calculating determinants of appropriate submatrices of A, determine the behavior of f at 0.

(c)Find the eigenvalues λ₁, λ₂, λ₃ of q. Let v₁, v₂, v₃ be the eigenvectors corresponding to λ₁, λ₂, λ₃ (do not solve for them). Express q in the uvw-coordinate system determined by v₁, v₂, v₃. That is, write q(uv₁ + uv₂ + wv₃) in terms of u, v, w. Then give a geometric description (or sketch) of the surface

8.5Suppose you want to design an ice-cube tray of minimal cost, which will make one dozen ice “cubes” (not necessarily actually cubes) from 12 in.³ of water. Assume that the tray is divided into 12 square compartments in a 2 × 6 pattern as shown in Fig. 2.44, and that

Figure 2.44

the material used costs 1 cent per square inch. Use the Lagrange multiplier method to minimize the cost function f(x, y, z) = xy + 3xz + 7yz subject to the constraints x = 3y and xyz = 12. Apply Theorem 8.9 to verify that you do have a local minimum.

8.6Apply Theorem 8.8 to show that the quadratic form

is nondefinite. Solve the characteristic equation for its eigenvalues, and then find an ortho-normal set of eigenvectors.

8.7Let f(x) = x^t Ax be a quadratic form on ⁿ (with A symmetric), and let v₁, . . . , v_n be the orthonormal eigenvectors given by Theorem 8.6. If P = (v₁, . . . , v_n) is the n × n matrix having these eigenvectors as its column vectors, show that

Hint: Let x = Py = y₁v₁ + · · · + y_n v_n. Then

by the definition of f, while

by Theorem 8.6. Writing B = P^tAP, we have

for all y = (y₁, . . . , y_n). Using the obvious symmetry of B prove that b_ii = λ_i, while b_ij = 0 if i ≠ j. Finally note that P⁻¹ = P^t by Exercise 6.11 of Chapter I.