Math. 311 (Fulling) 6 May 2002

Final Examination – Solutions

Name:

Calculators may be used for simple arithmetic operations only!

1. (28 pts.) The eigenvalues of N = ( \begin{matrix}−1 & −3 & 0 \\ −3 & −1 & 0 \\ 0 & 0 & 2 \end{matrix} ) are 2 and −4 .

(a) If ∇f(0) = 0 and N is the Hessian (second-derivative) matrix of f at 0 , is that

point a maximum, a minimum, or a saddle point of f ?

A saddle point, because the eigenvalues have different signs.

(b) Find an orthonormal basis of eigenvectors of N .

For λ = 2 :

( \begin{matrix}−3 & −3 & 0 \\ −3 & −3 & 0 \\ 0 & 0 & 0 \end{matrix} ) ⟶ ( \begin{matrix}1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) ⇒ z is arbitrary, x + y = 0.

Two orthonormal eigenvectors are ( − \frac{1}{\sqrt{2}} ; \frac{1}{\sqrt{2}} ; 0 ) , ( \begin{matrix}0 \\ 0 \\ 1 \end{matrix} ). For λ = −4 :

( \begin{matrix}3 & −3 & 0 \\ −3 & 3 & 0 \\ 0 & 0 & 6 \end{matrix} ) ⟶ ( \begin{matrix}1 & −1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} ) ⇒ z = 0, x − y = 0.

A normalized eigenvector is ( \frac{1}{\sqrt{2}} ; \frac{1}{\sqrt{2}} ; 0 ). It is automatically orthogonal to the other two. These three eigenvectors make up an orthonormal basis.

(c) Find an orthogonal matrix U and a diagonal matrix D so that N = UDU−1 or

N = U−1DU . State WHICH of these two formulas applies!

D = ( \begin{matrix}−4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix} ) , U = ( \begin{matrix}\frac{1}{\sqrt{2}} & − \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{matrix} ) .

The question does not demand that U^{−1} be written down, but it is trivial to do so, because it is the transpose of U :

U^{−1} = ( \begin{matrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ − \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{matrix} ) .

Since U maps coordinates with respect to the eigenbasis into coordinates with respect to the natural basis, the correct formula is N = UDU^{−1} , and with an extra sheet of paper one can check that this is right.

311F-S02 Page 2 There are many other correct answers. The sign of the second eigenvector could be changed (yielding a U that is its own transpose and inverse). The order of the eigenvalues on the diagonal could be changed. One must be careful to keep each complete solution internally consistent, however.

2. (24 pts.) Determine whether each set is linearly independent. If not, find a linearly independent set with the same span.

(a) {(0, 1, 0), (1, 1, 1), (1, 2, 3)}

Independent, because | \begin{matrix}0 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{matrix} | = − | \begin{matrix}1 & 1 \\ 1 & 3 \end{matrix} | = −2 ≠ 0 (or because the matrix row-reduces to the identity).

(b) {(1, 1, −2, 3), (1, 4, 2, 1), (1, 7, 6, −1)}

Dependent:

( \begin{matrix}1 & 1 & −2 & 3 \\ 1 & 4 & 2 & 1 \\ 1 & 7 & 6 & −1 \end{matrix} ) ⟶ ( \begin{matrix}1 & 1 & −2 & 3 \\ 0 & 3 & 4 & −2 \\ 0 & 6 & 8 & −4 \end{matrix} ) ⟶ ( \begin{matrix}1 & 1 & −2 & 3 \\ 0 & 3 & 4 & −2 \\ 0 & 0 & 0 & 0 \end{matrix} )

The two nonzero rows are a basis for the span:

{(1, 1, −2, 3), (0, 3, 4, −2)}.

(In fact, any two of the original rows would also do.)

(c) {1, ex, e2x}

Independent (by common knowledge).

(d) {1, ex, 2e−x, cosh x, sinh x − 4 cosh x}

Dependent, because the last two elements can be expressed (linearly) in terms of e^{x} and e^{−x} . Suitable bases for the span include

{1, e^{x}, 2e^{−x}},  {1, e^{x}, e^{−x}},  or {1, cosh x, sinh x}.

3. (18 pts.) Determine whether each formula defines an inner product on R3 . If not, explain why not. ( r⃗1 = (x1, y1, z1) , etc.)

(a) ⟨r⃗1, r⃗2⟩ = x1x2 + 2y1y2 + 3z1z2

This is an inner product (bilinear, symmetric, positive definite).

(b) ⟨r⃗1, r⃗2⟩ = x1x2 + y1z2 − z1y2

No — not symmetric. (It’s also not positive definite: if x = 0 , then ⟨r⃗, r⃗⟩ = yz − zy = 0 .)

(c) ⟨r⃗1, r⃗2⟩ = x1x2 + y1z2 + z1y2

No — not positive. If x = 0 and z = −y , then ⟨r⃗, r⃗⟩ = −2y^{2} < 0 .

311F-S02 Page 3

4. (25 pts.) Suppose that x2 + yz − z2 = 0 and x = s cosh t, y = s sinh t.

(a) Find \frac{∂z}{∂x} and \frac{∂z}{∂y} at all points where x = 2 and y = 0 .

Differentiate x^{2} + yz − z^{2} = 0 with respect to x and y :

2x + y \frac{∂z}{∂x} − 2z \frac{∂z}{∂x} = 0, z + y \frac{∂z}{∂y} − 2z \frac{∂z}{∂y} = 0.

Solve:

\frac{∂z}{∂x} = \frac{2x}{2z − y} , \frac{∂z}{∂y} = \frac{z}{2z − y} .

When x = 2 and y = 0 , we have 4 = z^{2} , or z = ±2 . Then \frac{∂z}{∂x} = ±1 , \frac{∂z}{∂y} = \frac{1}{2} .

(b) Find the best affine approximation to x and y (as functions of s and t) in the neighborhood of the point where s = 2 and t = 0 .

( \begin{matrix}\frac{∂x}{∂s} & \frac{∂x}{∂t} \\ \frac{∂y}{∂s} & \frac{∂y}{∂t} \end{matrix} ) = ( \begin{matrix}cosh t & s sinh t \\ sinh t & s cosh t \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ) at the point.

At the point, x = 2 and y = 0 . So

( \begin{matrix}x \\ y \end{matrix} ) ≈ ( \begin{matrix}2 \\ 0 \end{matrix} ) + ( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ) ( \begin{matrix}s − 2 \\ t − 0 \end{matrix} ) ,

or x ≈ 2 + (s − 2) = s, y ≈ 2t.

(c) Find \frac{∂z}{∂s} and \frac{∂z}{∂t} when s = 2 and t = 0 . (Use the results from (a) and (b).)

Chain rule:

( \frac{∂z}{∂s} \frac{∂z}{∂t} ) = ( ±1 \frac{1}{2} ) ( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ) = ( ±1 1 ) ,

where the sign ± is the sign of z . Or, in more classical notation,

\frac{∂z}{∂s} = \frac{∂z}{∂x} \frac{∂x}{∂s} + \frac{∂z}{∂y} \frac{∂y}{∂s} = ±1 + 0 = ±1 , \frac{∂z}{∂t} = \frac{∂z}{∂x} \frac{∂x}{∂t} + \frac{∂z}{∂y} \frac{∂y}{∂t} = 0 + \frac{1}{2} · 2 = 1.

5. (25 pts.) (a) Explain what a subspace is.

A subspace of a vector space is a (nonempty) subset that is closed under addition and scalar multiplication. That is, {0} ≠ S ⊆ V , and, for all v⃗_{1} and v⃗_{2} in S and all numbers r , one has rv⃗_{1} + v⃗_{2} ∈ S .

(b) Do the solutions of x2 + 2y − z = 3 form a subspace of R3 ?

No — this is a nonlinear equation, and the solution set is not closed under addition or scalar multiplication.

(c) Do the solutions of x + 5y + 2z = 0 form a subspace of R3 ?

Yes — this is a homogeneous linear equation. (The solutions are the kernel of a linear function.)

(d) Do the solutions of \frac{d2y}{dt2} + t \frac{dy}{dt} + 4y = 0 form a subspace of C2(−∞, ∞) ?

Yes — same remarks as for (c). (Solutions of a second-order ordinary differential equation with smooth coefficients will be smooth, certainly in C^{2} .)

6. (20 pts.) (a) Show, for vector fields A⃗(x⃗) and B⃗(x⃗) , that

∇ · (A⃗ × B⃗) = B⃗ · ∇ × A⃗ − A⃗ · ∇ × B⃗.

Method 1: Clearly there will be two sorts of terms, one with derivatives of A and one with derivatives of B . The algebraic structure of each set of terms can be deduced from the antisymmetry and cyclic symmetry of the scalar triple product:

C⃗ · (A⃗ × B⃗) = B⃗ · (C⃗ × A⃗) and C⃗ · (A⃗ × B⃗) = A⃗ · (B⃗ × C⃗) = −A⃗ · (C⃗ × B⃗).

Replacing C⃗ by ∇, we get the two terms claimed.

Method 2 (brute force):

∇ · (A⃗ × B⃗) = \frac{∂}{∂x} (A_{y}B_{z} − B_{y}A_{z}) + \frac{∂}{∂y} (A_{z}B_{x} − B_{z}A_{x}) + \frac{∂}{∂z} (A_{x}B_{y} − B_{x}A_{y})

= B_{x} (\frac{∂A_{z}}{∂y} − \frac{∂A_{y}}{∂z} ) + B_{y} (\frac{∂A_{x}}{∂z} − \frac{∂A_{z}}{∂x} ) + B_{z} (\frac{∂A_{y}}{∂x} − \frac{∂A_{x}}{∂y} )

− (similar terms with A and B interchanged)

= B⃗ · (∇ × A⃗) − A⃗ · (∇ × B⃗).

(b) Show, for scalar functions f(x⃗) and g(x⃗) and a closed surface S , that

∫∫S(∇f × ∇g) · dS⃗ = 0.

By Gauss’s theorem, the integral is

∫ ∫ ∫_{V} ∇ · (∇f × ∇g) dx dy dz,

where V is the interior of S (i.e., S is the boundary of V ). The integrand can be evaluated by (a), and since the curl of a gradient is always zero, the integrand vanishes. Therefore, the original integral is 0 .

7. (10 pts.) The normalized Legendre polynomials are defined by applying the Gram–Schmidt algorithm with respect to the inner product ⟨f, g⟩ = ∫{1 \atop −1} f(t)g(t) dt. The first two of these polynomials are p0(t) = \sqrt{\frac{1}{2}} , p1(t) = \sqrt{\frac{3}{2}} t. What is the third one?

⟨p_{0}, t^{2}⟩ = ∫_{−1}^{1} \sqrt{\frac{1}{2}} t^{2} dt = \frac{\sqrt{2}}{3} ,

⟨p_{1}, t^{2}⟩ = ∫_{−1}^{1} \sqrt{\frac{3}{2}} t^{3} dt = 0 .

Therefore, the part of t^{2} perpendicular to span(p_{0}, p_{1}) is t^{2} − ^{\frac{\sqrt{2}}{3}} p_{0} = t^{2} − \frac{1}{3} . The normalization integral is

∫_{−1}^{1} (t^{4} − \frac{2}{3} t^{2} + \frac{1}{9}) dt = \frac{2}{5} − \frac{4}{9} + \frac{2}{9} = \frac{8}{45} .

Thus

p_{2}(t) = \sqrt{\frac{45}{8}} (t^{2} − \frac{1}{3}) .

(This is also worked out on p. 288 of the textbook.)

8. (25 pts.) (a) Find all the eigenvalues and eigenvectors of M = ( \begin{matrix}3 & 4 \\ 1 & 6 \end{matrix} ).

det(M − λ) = | \begin{matrix}3 − λ & 4 \\ 1 & 6 − λ \end{matrix} | = λ^{2} − 9λ + 14 = (λ − 2)(λ − 7) .

The eigenvalues are 2 and 7 . Find the eigenvector for λ = 2 :

( \begin{matrix}1 & 4 \\ 1 & 4 \end{matrix} ) ⟶ ( \begin{matrix}1 & 4 \\ 0 & 0 \end{matrix} ) ⇒ x = −4y .

A suitable eigenvector is ( \begin{matrix}−4 \\ 1 \end{matrix} ). (There is no need to normalize it, since we don’t expect this eigenbasis to be orthogonal anyway.) Find the eigenvector for λ = 7 :

( \begin{matrix}−4 & 4 \\ 1 & −1 \end{matrix} ) ⟶ ( \begin{matrix}1 & −1 \\ 0 & 0 \end{matrix} ) ⇒ x = y .

A suitable eigenvector is ( \begin{matrix}1 \\ 1 \end{matrix} ).

311F-S02 Page 6

(b) Find the matrix-valued function etM that solves the initial-value problem for the

differential-equation system \frac{dx}{dt} = 3x + 4y, \frac{dy}{dt} = x + 6y.

Method 1: In the coordinates relative to the eigenbasis, the solution operator is

e^{tD} = exp ( \begin{matrix}2 & 0 \\ 0 & 7 \end{matrix} ) t = ( \begin{matrix}e^{2t} & 0 \\ 0 & e^{7t} \end{matrix} ) .

The operators of the coordinate transformation M = UDU^{−1} are

U = ( \begin{matrix}−4 & 1 \\ 1 & 1 \end{matrix} ) , U^{−1} = \frac{1}{−5} ( \begin{matrix}1 & −1 \\ −1 & −4 \end{matrix} ) = \frac{1}{5} ( \begin{matrix}−1 & 1 \\ 1 & 4 \end{matrix} ) .

Therefore, in the natural coordinate system the solution operator is

\begin{aligned}e^{tM} & = Ue^{tD}U^{−1} = \frac{1}{5} ( \begin{matrix}−4 & 1 \\ 1 & 1 \end{matrix} ) ( \begin{matrix}e^{2t} & 0 \\ 0 & e^{7t} \end{matrix} ) ( \begin{matrix}−1 & 1 \\ 1 & 4 \end{matrix} ) \\ & = \frac{1}{5} ( \begin{matrix}4e^{2t} + e^{7t} & −4e^{2t} + 4e^{7t} \\ −e^{2t} + e^{7t} & e^{2t} + 4e^{7t} \end{matrix} ) . \end{aligned}

Method 2: In vectorial notation, the most general solution is

( \begin{matrix}x(t) \\ y(t) \end{matrix} ) = c_{1} ( \begin{matrix}−4 \\ 1 \end{matrix} ) e^{2t} + c_{2} ( \begin{matrix}1 \\ 1 \end{matrix} ) e^{7t} .

It is necessary to solve for c_{1} and c_{2} in terms of the initial data, x(0) and y(0) :

( \begin{matrix}x(0) \\ y(0) \end{matrix} ) = c_{1} ( \begin{matrix}−4 \\ 1 \end{matrix} ) + c_{2} ( \begin{matrix}1 \\ 1 \end{matrix} ) .

The solution of this is

c_{1} = − \frac{1}{5} x(0) + \frac{1}{5} y(0) , c_{2} = \frac{1}{5} x(0) + \frac{4}{5} y(0) .

(This is equivalent to finding U^{−1} in the other method.) Substitute into the solution and combine terms:

( \begin{matrix}x(t) \\ y(t) \end{matrix} ) = \frac{1}{5} ( \begin{matrix}4e^{2t} + e^{7t} \\ −e^{2t} + e^{7t} \end{matrix} ) x(0) + \frac{1}{5} ( \begin{matrix}−4e^{2t} + 4e^{7t} \\ e^{2t} + 4e^{7t} \end{matrix} ) y(0) = e^{tM} ( \begin{matrix}x(0) \\ y(0) \end{matrix} )

with the same matrix e^{tM} as above.

311F-S02 Page 7

9. (25 pts.) Derive the numerical integration formula (quadrature rule) of the form

∫{e \atop 0} f(x) ln x dx = c1f(0) + c2f(1) + c3f(e)

that gives the exact answer whenever f is a quadratic polynomial ( f ∈ P2 ). (Note that the integral, although improper, is convergent for any continuous f .) Hint:

∫ xn ln x dx = xn+1 [\frac{ln x}{n + 1} − \frac{1}{(n + 1)2} ] for n = 0 , 1 , 2 , . . ..

c_{1} + c_{2} + c_{3} = ∫_{0}^{e} ln x dx = 0,
c_{2} + ec_{3} = ∫_{0}^{e} x ln x dx = \frac{e^{2}}{4} ,
c_{2} + e^{2}c_{3} = ∫_{0}^{e} x^{2} ln x dx = \frac{2e^{3}}{9} .

The easiest way to solve the system is Cramer’s rule. The determinant is

| \begin{matrix}1 & 1 & 1 \\ 0 & 1 & e \\ 0 & 1 & e^{2} \end{matrix} | = e^{2} − e.

\begin{aligned}c_{1} & = \frac{1}{e^{2} − e} | \begin{matrix}0 & 1 & 1 \\ \frac{e^{2}}{4} & 1 & e \\ \frac{2e^{3}}{9} & 1 & e^{2} \end{matrix} | = \frac{1}{e^{2} − e} [− | \begin{matrix}\frac{e^{2}}{4} & e \\ \frac{2e^{3}}{9} & e^{2} \end{matrix} | + | \begin{matrix}\frac{e^{2}}{4} & 1 \\ \frac{2e^{3}}{9} & 1 \end{matrix} |] \\ & = \frac{1}{e^{2} − e} (− \frac{e^{4}}{4} + \frac{2e^{4}}{9} + \frac{e^{2}}{4} − \frac{2e^{3}}{9} ) = \frac{− \frac{e^{3}}{36} − \frac{2e^{2}}{9} + \frac{e}{4}}{e − 1} . \end{aligned}

c_{2} = \frac{1}{e^{2} − e} | \begin{matrix}1 & 0 & 1 \\ 0 & \frac{e^{2}}{4} & e \\ 0 & \frac{2e^{3}}{9} & e^{2} \end{matrix} | = \frac{1}{e^{2} − e} (\frac{e^{4}}{4} − \frac{2e^{4}}{9} ) = \frac{\frac{e^{3}}{36}}{e − 1} .

c_{3} = \frac{1}{e^{2} − e} | \begin{matrix}1 & 1 & 0 \\ 0 & 1 & \frac{e^{2}}{4} \\ 0 & 1 & \frac{2e^{3}}{9} \end{matrix} | = \frac{1}{e^{2} − e} (\frac{2e^{3}}{9} − \frac{e^{2}}{4} ) = \frac{\frac{2e^{2}}{9} − \frac{e}{4}}{e − 1} .

Addendum: One student showed that the solution by row reduction not only is feasible, it reveals that the numerator in c_{1} can be factored to cancel the denominator: c_{1} = − \frac{e^{2}}{36} − \frac{e}{4} .