Advanced-Calculus-Sol--James-Callahan

The microscope values are given by the tangent line to the graph of x=ϕs =√ s at s= 100.. Because the graph of x=ϕs = tan s is concave up at π/4 i.e.,ϕ′′π/4 > 0, the graph lies above its

Advanced Calculus

A Geometric View

James Callahan

Preface

This book consists of detailed solutions to the exercises in my text Advanced Calculus: A Geometric

View (Springer, New York, 2010, ISBN 978-1-4419-7331-3) My Web site,

Given the nature of the material, it is likely that the solutions themselves still have some errors I

invite readers of this book to contact me at

callahan@math.smith.edu

to let me know about any errors or questions concerning the solutions found here I will post a

solutions errata file on my Web site

This solutions manual is available only directly from Springer, via its Web site

http://www.springer.com/instructors?SGWID=0-115-12-333200-0

J CallahanApril 2011

iv

vi CONTENTS

Solutions: Chapter 1

Starting Points

1 1 The integrals are “improper,”, because there is an

endpoint at±∞ However, we know from the text that

a

;

basic facts about improper integrals (cf Chap 9 of the

text) then allow us to write

a

=π/4 − (−π/2) = 3π/4

1 2 We use the push-forward substitution u = 1 + x2, so

that du = 2x dx and then

1 3 We use the pullback substitution x = R sinθ, so that

dx = R cosθdθ, R2− x2= R2cos2θandθ= ±π/2 when

(1 − p)(lnw) p−1, otherwise

1 5 b For the value of I to be finite, the expression just

given in part (a) must be finite as w→∞ This rules out

the first case (wherep = 1) and requires p − 1 > 0 in the second case Thus, I is finite precisely when p> 1, andthen

1 6 a Hereϕ′= −1/s2, soϕis invertible on each of the

rays s > 0 and s < 0 The inverse is s = 1/x on each ray.

1 6 b Hereϕ′= 1 + 3s2> 0 on the entire s-axis The inverse is the unique real root of s3+ s − x = 0 that is

provided by Cardano’s formula:

s= 3

vu

2 SOLUTIONS: CHAPTER 1 STARTING POINTS

1 6 c Hereϕ′= (1 − s2)/(1 + s2)2, soϕ has an inverse

on each of the three domains s ≤ −1, −1 ≤ s ≤ 1, and

1≤ s The graph ofϕ, shown below, makes it clear thatϕ

has an inverse on each of the three sets

- 20 - 10 10 20

- 0.4

- 0.2

0.2 0.4

To find a formula for the inverse, we need to solve the

equation x = s/(1 + s2) for s We have

x + xs2= s or xs2− s + x = 0.

The roots of this quadratic equation are

s=1±√1− 4x2

2x , |x| ≤ 1/2.

This formula defines all three inverse functions first of

all, with the minus sign, the graph is:

- 0.4 - 0.2 0.2 0.4

- 1.0

- 0.5

0.5 1.0

Clearly, this is the inverse for−1 ≤ s ≤ 1 Note that there

is an indeterminate form when x→ 0, but l’Hopital’s rule

shows that s→ 0 With the plus sign, the formula defines

the inverse on each of the other two domains:

−1/2 ≤ x < 0 ⇒ s ≤ −1; 0 < x ≤ 1/2 ⇒ 1 ≤ s.

1 6 d Hereϕ′= cosh s > 0 for all s, so ϕ is invertible

for all s To get the formula for the inverse, we first write

2x= es− e−s or 2xe s= e2s− 1

This is the quadratic equation(es)2− 2xe s− 1 = 0, and

has the roots

es=2x±√4x2+ 4

2 = x ±p

x2+ 1

The minus sign is untenable, because es> 0, so we obtain

the formula for the inverse as

s= ln

x+p

x2+ 1

1 6 e First of all,ϕis defined only when−1 < s < 1 In

that case,

(1 − s2)3 /2 > 0

soϕ is invertible for−1 < s < 1 To get the formula for

the inverse, note that

1 6 f Hereϕ(s) = ms + b is invertible everywhere

pro-videdϕ′= m 6= 0 In that case, the inverse is

s=x − b

m

1 6 g Hereϕ′= sinh s; becauseϕ′is nonzero on each of

the rays s < 0 and s > 0,ϕ will have an inverse on each

of these domains To find the formulas, we just adapt the

procedure for sinh s, above We have 2x= es+ e−s or (es)2− 2xe s+ 1 = 0.This time the roots give us the two inverse functions

In each case the inverse is a root of the cubic equation

s3− s + x = 0 as provided by Cardano’s formulas.

1 6 i Hereϕ′= sech2s≥ 1, soϕis invertible for all s.

To find the inverse, we have

1 6 j Hereϕ′= −2/(1 + s)2< 0 for all s 6= −1 Thus

ϕ has an inverse on each ray s < −1, −1 < s To get the

gives the same result

Applying the chain rule to tan(arcsin s), we get

gives the same result

1 8 From the previous exercise we know that if we take

1 9 d The microscope values are given by the tangent

line to the graph of x=ϕ(s) =√

s at s= 100 The graphitself is concave down there (becauseϕ′′(100) < 0), so itlies below its tangent line at that point

1 10 a Hereϕ′(s) = −1/s2, soϕ′(2) = −1/4 and themicroscope equation is∆x≈ −∆s/4 Therefore the ap-proximation is given by

−0.0001 and −0.00005, respectively

1 10 c This time the tangent line (which gives the

micro-scope estimates) lies below the graph, because the graph

is concave up:ϕ′′(2) > 0

1 11 Letϕ(s) =√

s, and let s0= 1 The approximation

we use (see the solution to Exercise 1.9) is

ϕ(s0+∆s) ≈ϕ(s0) +ϕ′(s0)∆s, ∆s≈ 0.Here∆s = 2h,ϕ(s0) = 1, andϕ′(s0) = 1/2, so our equa-tion becomes

√

1+ 2h ≈ 1 + h, h ≈ 0.

4 SOLUTIONS: CHAPTER 1 STARTING POINTS

1 12 a When x=ϕ(s) = tan s, thenϕ′(s) = sec2s, so

ϕ′(π/4) = (√

2)2= 2 and the microscope equation is

∆x≈ 2∆s

1 12 b Because tan(π/4) = 1, the approximating

equa-tion (Ex 1.9) takes the form

tan(π/4 +∆s) ≈ 1 + 2∆s, ∆s≈ 0;

with∆s = h, this can be rewritten as

tan(h +π/4) ≈ 1 + 2h, h ≈ 0.

Because the graph of x=ϕ(s) = tan s is concave up at

π/4 (i.e.,ϕ′′(π/4) > 0), the graph lies above its tangent

line, so actual values of the tangent function are greater

than their approximations using the microscope equation

1 13 The local length multiplier at a point is the value of

the derivative at that point, in this case the value of cos s.

Thus

length multiplier at s 1 1/√

2 0 −1/2 −1

1 14 At a point where the local length multiplier of the

function x=ϕ(s) is negative, the mapϕ: s → x reverses

′

=e

s+ e−s

2 = cosh s, (cosh s)′=

es+ e−s2

2

−



es− e−s2

1 17 In each case, work done is the dot product of force

and displacement Thus

(a) W = (2, −3)·(1,2) = −4, (b) W = 8, (c) W = −2.

1 18 Again, work done is the dot product of force and

displacement:

(a) W= −1 + 2 = 1, (b) W= 9, (c) W = 2

1 19 Write the unknown force as F= (P, Q) The given

information of work done on two particular displacementsprovides two equations:

F· (2,−1) = 2P − Q = 7,

F· (4,1) = 4P + Q = −3.

Solving these equations, we find P = 2/3, Q = −17/3, so

F= (2/3, −17/3) This allows us to determine

W = F · (1,0) = 2/3, W = F · (0,1) = −17/3.

The work done will be zero if∆x is perpendicular to F;

thus∆x can be any nonzero multiple of(17/3, 2/3)

1 20 Consider first W= F ·∆x as function of its first

ar-gument, F We have

W(F1+ F2) = (F1+ F2) ·∆x

= F1·∆x + F2·∆x= W (F1) + W (F2),

W (rF) = (rF) ·∆x= rF ·∆x= rW (F).

This shows W is a linear function of its first argument,

i.e., F Symmetry of the dot product then shows that W is

likewise a linear function of its second argument,∆x.

1 21 Ifθis the angle between F and∆u, then a formula

for the dot product allows us to express the work W in

ment that maximizes W is the one in the same direction as the force, and the one that minimizes W is in the opposite

direction

1 22 Work as dot product gives us two linear equations

in the unknowns P and Q:



=



A B

.The equations define the matrix equation shown above on

the right; to find P and Q, invert the matrix:



P Q

where∆= ad −bc To find P and Q, these equations must

be solvable, that is, the determinant∆must be nonzero

1 23 a.

x y

The five parameter points are marked; the curve is the

en-tire unit circle minus the point(x, y) = (0, −1).

1 23 b We can determine the limits by rewriting the

ex-pressions for x and y; then

By construction,α(x(t), y(t)) is the square of the distance

from the point(x(t), y(t)) to the origin;α≡ 1 means that

all points on the curve lie on the unit circle

1 24 In every case, the work done by F= (P, Q) is the

1 24 c This problem requires us to determine the value

of W no matter what path is chosen between the

end-points Therefore we cannot simply chose our own path,but must instead use an arbitrary parametrization

a

= x(b)y(b) − x(a)y(a) = 77 − 10 = 67.

The key is to recognize that the original integrand is the

derivative of the product x (t) · y(t).

1 24 d The path is now in(x, y, z)-space, where the work

done by the force F= (P, Q, R) is

(sin2θ+ cos2θ+ 3) dθ= 4A.

1 25 a We need to parametrize ~C; one possibility is

x(t) = (x(t), y(t)) = (3t − 2,4t + 3), 0 ≤ t ≤ 1 Then x + 2y = 11t + 4, x − y = −t − 5, and

6 SOLUTIONS: CHAPTER 1 STARTING POINTS

1 25 c Because x2+ y2= R2here, the integral becomes

1 26 a We show (cf Exercise 1.23) that r(u) lies on the

circle of radius 2 centered at the origin:

2±√3 = ∓√3

Thus r(u) begins at (1, −√3) and ends at (1,√

3) The

graphs of x (u) and y(u), (below right and left), show that

x increases from 1 to 2 and then decreases back to 1 while

y increases monotonically from−√3 to+√

This confirms that r(u) traverses the circular arc once

32u2(u2+ 1)3du=√

3+4π

3 ;

we have used a computer algebra system to evaluate the

integral This value of the integral agrees with those found

in the text using different parametrizations

1 27 a We have∂Φ/∂x=∂Φ/∂y= 0 while∂Φ/∂z=

−gm, showing that gradΦ= (0, 0, −mg) = F.

1 27 b Because the gravitational force F has the

poten-tialΦ(x, y, z), the work done is just the potential difference

W=Φ(x, y, z)

(α,β,γ)

(a,b,c)

= − gmz

(α,β,γ)

(a,b,c)

= gm(c −γ)

This is negative if c−γ< 0, i.e., if c <γ Negative work

by gravity can be interpreted as positive work that has to

be done by another force acting against gravity to lift theobject from the lower point to the higher

1 27 c Because the points are at the same vertical height,

c=γ, so no work is done by gravity The gravitationalforce near the surface of the earth is conservative in thesense that the work done by gravity in moving an objectover any closed path is zero; more generally, the path needmerely begin and end at the same vertical height

1 28 a The explicit form of F(x, y, z) is



−µx (x2+ y2+ z2)3 /2, −µy

(x2+ y2+ z2)3 /2, −µz

(x2+ y2+ z2)3 /2



1 28 e Because the gravitational force is conservative,

the work done in one complete orbit (a closed path) iszero (The net energy expenditure is zero: energy is con-served.)

1

=85

3 /2− 133 /2

27 = 27.2885

1 29 c Here

x′= (et cost− et sin t, et sin t+ et cost)

= et (cost − sint,sint + cost),

kx′k2= e2t cos2t − 2cost sint + sin2t

+ sin2t + 2 sint cost + cos2t

= 2e2t,so

−1

=π.(This can also be seen by noting that the curve is a semi-

circle of radius 1; see Exercise 1.23.)

1 29 f One possible parametrization is

The value is determined by a computer algebra system

1 30 a We have kx′(t)k = R; therefore, if we take t = 0

and the initial point, then

s (t) =

Z t0

R dt = Rt.

1 30 b The inverse of s = s(t) is simply t =σ(s) = s/R.

The arc-length parametrization is

y(s) = R cos(s/R), R sin(s/R)

1 31 Using kx′(t)k from the solution to Exercise 1.29.e

and taking the initial value as t = 0 (so s(0) = 0), we have

1 32 We use Definition 1.6 and Theorem 1.5 (text

page 18) to express total mass as the path integral of mass

density over the wire If we use x(t) = (R cost, R sint) as

the parametrization of the curve, thenkx′(t)k = R, mass

density isρ(x(t)) = 1 + R2cos2t gm/cm, and

total mass=

Z 2 π

0 (1 + R2cos2t ) · R dt = 2πR+πR3gm

1 33 This exercise is similar to the example worked in

the text on pp 18–19 In particular, ds=√

1 34 Suppose C is a curve in the plane; then in an

ex-pression of the form

Z

C

f (s) ds, the value f (s) is independent of the position of C Thus

if C has arc length L, we can define

where the integral on the right is just an ordinary one

Therefore, if C is a circle of radius 5, then L= 10πand

I

C

cos s ds=

Z 10π0

cos2s ds= 5π

8 SOLUTIONS: CHAPTER 1 STARTING POINTS

1 35 The change from Cartesian to polar coordinates,

Z √10 1

sin(r2) r dr dθ

= −cos(r2)2

√ 10

1

Z π/2 0

At a critical point, g′µ , σ(x) = 0; but because gµ , σ> 0 and

σ2> 0, g′µ,σ(x) = 0 only when x −µ= 0, i.e., x =µ

In the graph above (withµ= 5,σ= 2), the horizontal and

vertical axes have the same scale

For the general function, shown below, we use differentscales on the two axes, and mark the inflection points at

1 38 a The narrowest and tallest graph below is the one

withσ = 1/2; the widest and shortest has σ = 3 Forclarity the vertical and horizontal scales are different

- 4 - 2 0 2 4

0.2 0.4 0.6 0.8 1.0

1 38 b The graphs are just horizontal translates of one

another; from left to right, they haveµ= −2,1,10 Thevertical and horizontal scales are different

0.1 0.2 0.3 0.4 0.5

1 39 The procedure here is similar to the one used to

solve Exercise 1.37.b If we use the suggested substitution

z = (x −µ)/σ, then dz = dx/σ, z = 0 when x =µ, and

e−z2/2dz

= Prob(0 ≤ Z0 ,1≤ (b −µ)/σ)

1 40 When a<µ, we need to add Prob(a ≤ Xµ , σ≤µ)

to the probabilities that must be determined Now the

nor-mal density function Xµ , σis symmetric around x=µ, and

so are the pair of points x = a and x = 2µ− a (because

their average isµ); therefore we can writeProb(a ≤ Xµ , σ≤µ) = Prob(µ≤ Xµ, σ≤ 2µ− a).

Solutions: Chapter 2

Geometry of Linear Maps

2 1 a By definition of M1, the image of a line of slope

m=∆v/∆u in the (u, v)-plane is a line in the (x, y)-plane

2 2 According to the text, page 33, and the later

discus-sion, pages 35–36, the coordinate change matrix G that

we need for M5= G−1M5G has for its columns the

eigen-vectors of M5 From the work on page 34 of the text, we

A straightforward calculation shows that G−1M5G = M5

2 3 By hypothesis, there are n × n matrices H and G for

which

C = H−1BH and B = G−1AG

But then C = H−1G−1AGH = K−1AK with K = GH (and

K−1= H−1G−1), showing that C is equivalent to A.

2 4 a The area multiplier is det M= 18 − 4 = 14

2 4 b.

x y

2 4 c We must show the image of(2, 1) lies in the tion of(2, 1) We have



=

147



= 7

21



2 4 d To show the image of(−1,2) lies in the direction



=



−24



= 2



−12



2 4 e.

x y

2 4 f The image of a single square of the new grid is a

large black rectangle in the figure above In terms of thegrid of gray unit squares, the black rectangle’s dimensionsare 7×2 Because the image of a gray unit square is made

up of 14 such squares, the area multiplier is again 14

2 5 a We have tr M = −1, detM = −6, so the

character-istic polynomial isλ2+λ− 6 = (λ+ 3)(λ− 2) Thus theeigenvalues areλ = −3,+2 Forλ= −3, an eigenvector



=

00

.Thus for an eigenvector we can take



−12



10 SOLUTIONS: CHAPTER 2 GEOMETRY OF LINEAR MAPS

The eigenvector equation forλ= +2 is



=

00



;

For an eigenvector we can take

21



2 5 b Now tr M = 2and detM = −3, so the

characteris-tic polynomial isλ2− 2λ− 3 = (λ− 3)(λ+ 1) For the

eigenvalueλ= +3 we have the eigenvector equation



=

00



;

For an eigenvector we can take

11

.For the eigenvalueλ= −1, the eigenvector equation is



=

00



2 5 c Here tr M = −1 and detM = −6, so the

charac-teristic polynomial isλ2+λ− 6 = (λ+ 3)(λ− 2) The

eigenvalues areλ = −3,+2 (as they were for the matrix

in part a) Forλ= −3,



=

00



;For an eigenvector we can take

.For the eigenvalueλ= +2, the eigenvector equation is



=

00

√

62



=

 3

√6



2 6 The special grid for M is one built on the

eigenvec-tors of M.

2 6 a The eigenvectors and eigenvalues of M were

de-termined in the solution to Exercise 2.5a; they are

forλ= −3

Hence the grid we choose is the same as the grid of graysquares used in Exercise 2.4

x y

The image of the grid of gray squares is the grid of blackrectangles The gray square with its lower left corner atthe origin (shown slightly darker) has its boundary ori-ented counterclockwise Its image is the black rectanglewith its boundary oriented clockwise Each gray square isstretched by the factor+2 in one direction; in the other,the stretch, by a factor of 3, is combined with a flip Thelinear multipliers are therefore 2 and−3 The image ofeach oriented gray unit square is a 2× 3 rectangle of theopposite orientation, so the area multiplier is−6

2 6 b We first need to find the eigenvalues and

eigenvec-tors of M We have tr M = 1, det M = −2, so the

charac-teristic polynomial isλ2−λ− 2 = (λ− 2)(λ+ 1) For

λ= 2 the eigenvector equation is



=

00

,



x y



=

11

.Forλ= −1 the equation is



=

00

,



x y



=



−12



In the figure below (on the next page), the gray gridconsists of parallelograms that have their sides parallel tothe vectors(1, 1) and (−1,2) The black parallelogramsare their images Each one is made up of two gray paral-lelograms The gray parallelogram with its lower left cor-ner at the origin has its sides oriented counterclockwise

Its image has its sides oriented clockwise, so M reverses

orientation The area multiplier is therefore−2 In onedirection, a gray parallelogram is stretched double; in the

other, it is merely flipped Thus the linear multipliers are

2 and−1

x y

2 7 a The image of M8− 2I consists of all vectors X of



=



2u − 2v 2u − 2v



=



x y



Points in the image satisfy the condition y = x; that is,

the image is the line y = x in the (x, y)-plane and thus is

1-dimensional The kernel of M8−2I is, by definition, the

set of eigenvectors of M8with eigenvalue 2 According to

the rank–nullity theorem (text pp 46–47, applied to the

2 7 b Straightforward calculation shows M8(u) = 2x

and M8(v) = 2x + 2y Thus u is an eigenvector for M8

with eigenvalue 2

2 7 c Suppose the vector W has coordinates (u, v) based

on the standard basis vectors e1and e2and has coordinates

(u, v) based on u and v That is



= W = u

11



+ v

10

.From these equations we can read off the following:

u = u + v and v = u.

The same relation holds for the x- and y-variables.

2 7 d Using the various relations and the definition of

M8we can write

x = y = 2u = 2u + 2v

and then

y = x−y = (4u−2v)−2u = 2u−2v = 2(u+v)−2u = 2v.

This confirms that

2 8 a We have tr Rθ= 2 cosθand det Rθ= 1 Therefore

the characteristic equation isλ2− (2cosθ)λ+ 1 = 0 andits roots are

−sin2θ= cosθ± isinθ

Thus the eigenvalues of Rθare cosθ± isinθ= e±iθ

2 8 b The addition formulas for the sine and cosine

functions establish this equality We have

cosθ −sinθ

r(cosθcosα− sinθsinα)

r(sinθcosα+ cosθsinα)

The computation says Rθmaps the line that makes theangleαwith the positive u-axis to the line that makes the

angleα+θ with that axis If θ 6= nπ, these two lines

are different; in particular, Rθ can map no line to a real

multiple of itself This means Rθhas no real eigenvalues

2 8 c For the eigenvalueλ = cosθ+ i sinθ, the vector equation is



= sinθ−i −11 −ix y=

00

.For an eigenvector we can take

i1

.For the eigenvalueλ = cosθ− isinθ, the eigenvectorequation is



= sinθ1i −1i x y=

00

.For an eigenvector we can take



−i1



12 SOLUTIONS: CHAPTER 2 GEOMETRY OF LINEAR MAPS

Note that ifθ= nπ, then sinθ= 0 so these eigenvector

Π

2

In particular, note that arctan(q) has the same sign as q

and arctan(q) → ±π/2 as q → ±∞

For clarity, write arctan(y/x) asθ(x, y) With q = y/x,

we can consider the definition ofθto be

arctan(q), first and fourth quadrants,

arctan(q) +π, second quadrant,

arctan(q) −π, third quadrant,

±π/2, positive, negative y-axis, resp

Thus, to check continuity ofθ on the positive y-axis, we

must therefore show

θ(x, y) →θ(0, y) = +π

2 as x→ 0

If x→ 0+(i.e., x approaches 0 through positive values),

then(x, y) is in the first quadrant (q > 0) and

2 10 b The spiral-ramp graph of z=θ(x, y) is shown in

the text on page 430

2 11 a Because u1 and u2 are eigenvectors, they arenonzero Therefore, if they were not linearly independent,

we could write u2= ku1, k6= 0 But then

λ2u2= M(u2) = M(ku1) = kM(u1) = kλ1u1=λ1u2.Becauseλ26=λ1, we have u2= 0, a contradiction.

2 11 b It is a basic result of linear algebra that any

square matrix whose columns are linearly independent is

invertible Suppose we write G as a a row of column tors and G−1as a column of row vectors:

2 12 a According to the discussion after Theorem 2.2



b

d−λ

,

= b

a + d2

Linearity of M and M−λI establishes that every nonzero

2 12 b By definition, the eigenvectors of M for

eigen-valueλ are the elements of the kernel of M−λI By

hypothesis, there is just a single [linearly independent]

eigenvector, so the kernel of M−λI is 1-dimensional

Ac-cording to the rank–nullity theorem (text pp 46–47), the

image of M−λI is also 1-dimensional.

We established in part (a) that the image of M−λI

con-sists of eigenvectors of M Given that u is an eigenvector,

every nonzero multiple, in particularλu, is also an

eigen-vector (recall thatλ6= 0) Thus, by part (a) again, there is

some vector v for which

(M −λI)v =λu, or Mv=λu+λv

If we assume u and v are linearly dependent, then v= ku

for some k, and

λu= (M −λI )v = k(M −λI )u = k · 0.

This is a contradiction, so u and v are linearly

indepen-dent

2 12 c Any square matrix with linearly independent

columns, such as G, is invertible If we write

G= u v

and G−1=



r s

,

by analogy with the previous solution, then G−1G = I

im-plies the following row–column products

2 13 a Equating the real and the imaginary parts of

Mu + iMv = M(u + iv) = (a − ib)(u + iv)

14 SOLUTIONS: CHAPTER 2 GEOMETRY OF LINEAR MAPS

To compute GC a ,b, we write out the components of the

2 13 c Suppose u= 0; then the equation Mu = au + bv

implies that bv = 0 Now b 6= 0, so we must have v = 0

and thus u + iv = 0 But u + iv 6= 0 because it is an

eigen-vector of M; this contradiction implies u6= 0.

An entirely similar argument establishes that v 6= 0.

Following the suggestion, we now assume ru + sv = 0 and

deduce

0= M(ru + sv) = r(au + bv) + s(−bu + av)

= a(ru + sv) + b(−su + rv) = b(−su + rv).

But then−su + rv = 0 because b 6= 0 Therefore we also

have

0= r(ru + sv) − s(−su + rv) = (r2+ s2)u,

but since u6= 0, we must have r2+ s2= 0, implying that

r = s = 0 It follows that u and v are linearly independent.

2 13 d As in the previous exercise, the matrix G whose

columns are the linearly independent vectors u and v must

be invertible Therefore the equation MG = GC a ,bleads

to

G−1MG = G−1GC a ,b = C a ,b

2 14 a In the discussion following Theorem 2.2 in the

text (p 36), the equation for the eigenvalues implies the

eigenvalues will be real if the discriminant of the

char-acteristic polynomial, tr2(M) − 4det(M), is nonnegative.

For the given matrix M, we have

tr2(M) − 4det(M) = (a + c)2− 4ac + 4b2

= a2− 2ac + c2+ 4b2

= (a − c)2+ 4b2≥ 0,

so the eigenvalues are always real

2 14 b Suppose xi is an eigenvector for M with

eigen-valueλi (i= 1, 2), andλ16=λ2 Let x†i denote the

trans-pose of xi(as in the text itself); then, using the symmetry

M†= M, we have

λ1(x1· x2) =λ1x†1x2= (Mx1)†x2= x†1M†x2

= x†1Mx2=λ2x†1x2=λ2(x1· x2)

Becauseλ16=λ2, this equality will hold only if x · x2= 0,

that is, only if x1and x2are orthogonal

2 14 c If a symmetric matrix has equal eigenvalues, then

the discriminant of the characteristic polynomial must bezero; that is,(a −c)2+ 4b2= 0 This holds only if a −c =

2 15 d When v ∧w is the positively oriented unit square,

that is, when

detV = det MV = detM × detV.

2 16 b By definition, M(v ∧ w) is the image of the allelogram v∧ w under the linear map M; by part (a), this

par-is the parallelogram M (v) ∧ M(w) Thus

area M (v ∧ w) = areaM(v) ∧ M(w).

According to Exercise 2.15, the last area equals

detV = detMV = det M × detV.

Finally, detV= areav ∧ w, so we have

area M (v ∧ w) = detM × areav ∧ w.

2 17 The aim here is to assume only the three “bulleted”

properties of D and then deduce that D must be the

for arbitrary x and y By property 2, the left hand side

and the first and last terms on the right-hand side are zero

This leaves

0= D(y, x) + D(x, y),

from which it follows that D (y, x) = −D(x,y).

2 17 b By property 1, we know D(e1, e2) = +1; by

part (a), D(e2, e1) = −D(e1, e2) = −1

2 17 c This time, bilinearity gives

−1 5

2 3