ALL 10 CHAPTERS COVERED
Answer key
,2 Solutions to Exercises
Problem Set 1.1, page 8
1 The combinations give (a) a line in R3 (b) a plane in R3 (c) all of R3.
2 v + w = (2, 3) and v − w = (6, −1) will be the diagonals of the parallelogram with
v and w as two sides going out from (0, 0).
3 This problem gives the diagonals v + w and v − w of the parallelogram and asks for the sides:
The opposite of Problem 2. In this example v = (3, 3) and w = (2, −2).
4 3v + w = (7, 5) and cv + dw = (2c + d, c + 2d).
5 u+v = (−2, 3, 1) and u+v+w = (0, 0, 0) and 2u+2v+w = ( add first answers) = (−2, 3, 1). The
vectors u, v, w are in the same plane because a combination gives (0, 0, 0). Stated another
way: u = −v − w is in the plane of v and w.
6 The components of every cv + dw add to zero because the components of v and of w add to
zero. c = 3 and d = 9 give (3, 3, −6). There is no solution to cv+dw = (3, 3, 6) because 3 + 3 + 6 is
not zero.
7 The nine combinations c(2, 1) + d(0, 1) with c = 0, 1, 2 and d = (0, 1, 2) will lie on a lattice. If we
took all whole numbers c and d, the lattice would lie over the whole plane.
8 The other diagonal is v − w (or else w − v). Adding diagonals gives 2v (or 2w).
9 The fourth corner can be (4, 4) or (4, 0) or (−2, 2). Three possible parallelograms!
10 i − j = (1, 1, 0) is in the base (x-y plane). i + j + k = (1, 1, 1) is the opposite corner from (0, 0, 0).
Points in the cube have 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.
11 Four more corners (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1). The center point is (1 , 1 , 1 ).
2 2 2
Centers of faces are ( 1 , 1 , 0), ( 1 , 1 , 1) and (0, 1 , 1 ), (1, 1 , 1 ) and ( 1 , 0, 1 ), (1 , 1, 1 ).
2 2 2 2 2 2 2 2 2 2 2 2
12 The combinations of i = (1, 0, 0) and i + j = (1, 1, 0) fill the xy plane in xyz space.
13 Sum = zero vector. Sum = −2:00 vector = 8:00 vector. 2:00 is 30◦ from horizontal
√
= (cos π , sin π ) = ( 3/2, 1/2).
6 6
14 Moving the origin to 6:00 adds j = (0, 1) to every vector. So the sum of twelve vectors changes
from 0 to 12j = (0, 12).
,Solutions to Exercises 3
3 1
15 The point v + w is three-fourths of the way to v starting from w. The vector
4 4
1 1 1 1
v + w is halfway to u = 4 v+ w. The vector v + w is 2u (the far corner of the
4 2 2
parallelogram).
16 All combinations with c + d = 1 are on the line that passes through v and w. The point
V = −v + 2w is on that line but it is beyond w.
17 All vectors cv + cw are on the line passing through (0, 0) and u = 1 v + 1 w. That
2 2
line continues out beyond v + w and back beyond (0, 0). With c ≥ 0, half of this line is removed,
leaving a ray that starts at (0, 0).
18 The combinations cv + dw with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 fill the parallelogram with sides v and w.
For example, if v = (1, 0) and w = (0, 1) then cv + dw fills the unit square. But when v = (a, 0) and
w = (b, 0) these combinations only fill a segment of a line.
19 With c ≥ 0 and d ≥ 0 we get the infinite “cone” or “wedge” between v and w. For example, if v
= (1, 0) and w = (0, 1), then the cone is the whole quadrant x ≥ 0, y ≥
0. Question: What if w = −v? The cone opens to a half-space. But the combinations of v = (1,
0) and w = (−1, 0) only fill a line.
20 (a) 1 u + 1 v + 1 w is the center of the triangle between u, v and w; 1 u + 1 w lies
3 3 3 2 2
between u and w (b) To fill the triangle keep c ≥ 0, d ≥ 0, e ≥ 0, and c +d + e = 1.
21 The sum is (v − u)+ (w − v)+ (u− w) = zero vector. Those three sides of a triangle are in the same
plane!
22 The vector 1 (u + v + w) is outside the pyramid because c + d + e = 1
+1+ 1
> 1.
2 2 2 2
23 All vectors are combinations of u, v, w as drawn (not in the same plane). Start by seeing
that cu + dv fills a plane, then adding ew fills all of R3.
24 The combinations of u and v fill one plane. The combinations of v and w fill another plane.
Those planes meet in a line: only the vectors cv are in both planes.
25 (a) For a line, choose u = v = w = any nonzero vector (b) For a plane, choose
u and v in different directions. A combination like w = u + v is in the same plane.
,4 Solutions to Exercises
26 Two equations come from the two components: c + 3d = 14 and 2c + d = 8. The solution is c
= 2 and d = 4. Then 2(1, 2) + 4(3, 1) = (14, 8).
27 A four-dimensional cube has 24 = 16 corners and 2 · 4 = 8 three-dimensional faces and 24 two-
dimensional faces and 32 edges in Worked Example 2.4 A.
28 There are 6 unknown numbers v1, v2, v3, w1, w2, w3. The six equations come from the
components of v + w = (4, 5, 6) and v − w = (2, 5, 8). Add to find 2v = (6, 10, 14) so v = (3, 5, 7) and
w = (1, 0, −1).
29 Fact : For any three vectors u, v, w in the plane, some combination cu + dv + ew is the zero
vector (beyond the obvious c = d = e = 0). So if there is one combination Cu+ Dv + Ew that
produces b, there will be many more—just add c, d, e or 2c, 2d, 2e to the particular solution C, D,
E.
The example has 3u − 2v + w = 3(1, 3) − 2(2, 7) + 1(1, 5) = (0, 0). It also has
−2u + 1v + 0w = b = (0, 1). Adding gives u − v + w = (0, 1). In this case c, d, e
equal 3, −2, 1 and C, D, E = −2, 1, 0.
Could another example have u, v, w that could NOT combine to produce b ? Yes. The vectors (1,
1), (2, 2), (3, 3) are on a line and no combination produces b. We can easily solve cu + dv + ew =
0 but not Cu + Dv + Ew = b.
30 The combinations of v and w fill the plane unless v and w lie on the same line through (0, 0). Four
vectors whose combinations fill 4-dimensional space: one example is the “standard basis” (1, 0,
0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1).
31 The equations cu + dv + ew = b are
2c −d =1
−c +2d −e = 0 So d = 2e c = 3/4 d
−d +2e = 0 then c = 3e = 2/4 e =
then 4e = 1 1/4
,Solutions to Exercises 5
Problem Set 1.2, page 18
1 u · v = −2.4 + 2.4 = 0, u · w = −.6 + 1.6 = 1, u · (v + w) = u · v + u · w =
0 + 1, w · v = 4 + 6 = 10 = v · w.
√ . Then |u · v| = 0 < (1)(5) and |v · w| = 10 <
2 kuk = 1 and kvk = 5 and kwk = 5
√5, confirming the Schwarz inequality.
5
3 Unit vectors v/kvk = ( 4 , 3 ) = (0.8, 0.6). The vectors w, (2, −1), and −w make
5 5
√ √
0◦ , 90 ◦, 180 ◦ angles with w and w/kwk = (1/ 5, 2/ 5 ). The cosine of θ is v k·vk
√
w
kwk = 10/5 5.
4 (a) v · (−v) = −1 (b) (v + w) · (v − w) = v · v + w · v − v · w − w · w = 1+( )−( )−1 = 0 so θ = 90◦
(notice v·w = w·v) (c) (v−2w)·(v+2w) = v · v − 4w · w = 1 − 4 = −3.
√ √ 0 is
5 u1 = v/kvk = (1, 3)/ 1 0 and u2 = w/kwk = (2, 1, 2)/3. U1 = (3, −1)/ 1
√ √ : There is a
perpendicular to u1 (and so is (−3, 1)/ 10). U2 could be (1, −2, 0)/ 5
whole plane of vectors perpendicular to u2, and a whole circle of unit vectors in that plane.
6 All vectors w = (c, 2c) are perpendicular to v. They lie on a line. All vectors (x, y, z) with x + y + z
= 0 lie on a plane. All vectors perpendicular to (1, 1, 1) and (1, 2, 3) lie on a line in 3-
dimensional space.
7 (a) cos θ = v · w/kvkkwk = 1/(2)(1) so θ = 60◦ or π/3 radians (b) cos θ = 0 so θ = 90◦ or
π/2 radians (c) cos θ = 2/(2)(2) = 1/2 so θ = 60◦ or π/3
√ ◦
(d) cos θ = −1/ 2 so θ = 135 or 3π/4.
8 (a) False: v and w are any vectors in the plane perpendicular to u (b) True: u · (v + 2w) = u ·
v + 2u · w = 0 (c) True, ku − vk2 = (u − v) · (u − v) splits into u · u + v · v = 2 when u · v = v · u = 0.
9 If v2w2/v1w1 = −1 then v2w2 = −v1w1 or v1w1 +v2w2 = v· w = 0: perpendicular! The vectors (1, 4) and
(1, − 4 ) are perpendicular.
1
,6 Solutions to Exercises
10 Slopes 2/1 and −1/2 multiply to give −1: then v · w = 0 and the vectors (the direc- tions) are
perpendicular.
11 v · w < 0 means angle > 90◦; these w’s fill half of 3-dimensional space.
12 (1, 1) perpendicular to (1, 5) − c(1, 1) if (1, 1) · (1, 5) − c(1, 1) · (1, 1) = 6 − 2c = 0 or c = 3; v · (w − cv) =
0 if c = v · w/v · v. Subtracting cv is the key to constructing a perpendicular vector.
13 The plane perpendicular to (1, 0, 1) contains all vectors (c, d, −c). In that plane, v = (1, 0, −1) and
w = (0, 1, 0) are perpendicular.
14 One possibility among many: u = (1, −1, 0, 0), v = (0, 0, 1, −1), w = (1, 1, −1, −1)
and (1, 1, 1, 1) are perpendicular to each other. “We can rotate those u, v, w in their
3D hyperplane and they will stay perpendicular.” √ √
15 1 (x + y) = (2 + 8)/2 = 5 and 5 > 4; cos θ = 2 6/ 0 √ 10 = 8/10.
2 2 1 1
16 kvk = 1 + 1 + · · · + 1 = 9 so kvk = 3; u = v/3 = (1 , . . . , 1 ) is a unit vector in 9D;
3 3
√
2 vector in the
w = (1, −1,√0, . . . , 0)/ is a unit √ 8D hyperplane perpendicular to v.
. For any vector v = (v1, v2, v3) the
17 cos α = 1/ , cos β = 0, cos γ = −1/
2 2
cosines with (1, 0, 0) and (0, 0, 1) are cos2 α+cos2 β+cos2 γ = (v2+v 2+v2)/kvk2 = 1.
1 2 3
18 kvk2 = 42 + 22 = 20 and kwk2 = (−1)2 + 22 = 5. Pythagoras is k(3, 4)k2 = 25 = 20 + 5 for the length
of the hypotenuse v + w = (3, 4).
19 Start from the rules (1), (2), (3) for v · w = w · v and u · (v + w) and (cv) · w. Use rule (2) for (v + w)
· (v + w) = (v + w) · v + (v + w) · w. By rule (1) this is v · (v + w) + w · (v + w). Rule (2) again
gives v · v + v · w + w · v + w · w = v · v + 2v · w + w · w. Notice v · w = w · v! The main point is to
feel free to open up parentheses.
20 We know that (v − w) · (v − w) = v · v − 2v · w + w · w. The Law of Cosines writes kvkkwk cos θ for v ·
w. Here θ is the angle between v and w. When θ < 90◦ this v · w is positive, so in this case v ·
v + w · w is larger than kv − wk2.
Pythagoras changes from equality a2 +b2 = c2 to inequality when θ < 90 ◦ or θ > 90 ◦.
,Solutions to Exercises 7
21 2v· w ≤ 2kvkkwk leads to kv+ wk2 = v· v + 2v· w +w· w ≤ kvk2 + 2kvkkwk +
2 2
kwk . This is (kvk + kwk) . Taking square roots gives kv + wk ≤ kvk + kwk.
22 v2w2 + 2v1w1v2w2 + v2w2 ≤ v2w2 + v2w2 + v2w2 + v2w2 is true (cancel 4 terms)
1 1 2 2 1 1 1 2 2 1 2 2
because the difference is v2w2 + v2w2 − 2v1w1v2w2 which is (v1w2 − v2w1)2 ≥ 0.
1 2 2 1
23 cos β = w1/kwk and sin β = w2/kwk. Then cos(β −a) = cos β cos α+sin β sin α =
v1w1/kvkkwk + v2w2/kvkkwk = v · w/kvkkwk. This is cos θ because β − α = θ.
24 Example 6 gives |u1||U1| ≤ 1 (u2 + U 2) and |u2||U2| ≤ 1 (u2 + U 2). The whole line
2 1 1 2 2 2
1 2 2 1 2 2
becomes .96 ≤ (.6)(.8) + (.8)(.6) ≤ 2 (.6 + .8 ) + 2 (.8 + .6 ) = 1. True: .96 < 1.
p
25 The cosine of θ is x/ x2 + y 2, near side over hypotenuse. Then | cos θ|2 is not greater than 1:
x2/(x2 + y 2 ) ≤ 1.
26–27 (with apologies for that typo !) These two lines add to 2||v||2 + 2||w||2 :
2
||v+w|| = (v + w) · (v + w) = v · v + v · w + w · v + w · w
2
= (v − w) · (v − w) = v · v − v · w − w · v + w · w
||v − w||
28 The vectors w = (x, y) with (1, 2) · w = x + 2y = 5 lie on a line in the xy plane. The
√
shortest w on that line is (1, 2). (The Schwarz inequality kwk ≥ v · w/kvk = is 5
√
an equality when cos θ = 0 and w = (1, 2) and kwk = 5.)
29 The length kv − wk is between 2 and 8 (triangle inequality when kvk = 5 and kwk = 3). The dot
product v · w is between −15 and 15 by the Schwarz inequality.
30 Three vectors in the plane could make angles greater than 90◦ with each other: for
example (1, 0), (−1, 4), (−1, −4). Four vectors could not do this (360◦ total angle). How many
can do this in R3 or Rn? Ben Harris and Greg Marks showed me that the answer is n + 1. The
vectors from the center of a regular simplex in Rn to its n + 1 vertices all have negative dot
products. If n+2 vectors in Rn had negative dot products, project them onto the plane orthogonal
to the last one. Now you have n + 1 vectors in Rn−1 with negative dot products. Keep going to 4
vectors in R2 : no way!
31 For a specific example, pick v = (1, 2, −3) and then w = (−3, 1, 2). In this example
√
√ 1 14 = −1/2 and θ = 120◦ . This always
cos θ = v · w/kvkkwk = −7/ 4
happens when x + y + z = 0:
,8 Solutions to Exercises
1 1
v · w = xz + xy + yz = (x + y + z)2 − (x2 + y 2 + z 2 )
2 2
1 1
This is the same as v · w = 0 − kvkkwk. Then cos θ = .
2 2
32 Wikipedia gives this proof of geometric mean G = √
3 xyz ≤ arithmetic mean
A = (x + y + z)/3. First there is equality in case x = y = z. Otherwise A is somewhere
between the three positive numbers, say for example z < A < y.
Use the known inequality g ≤ a for the two positive numbers x and y + z − A. Their mean a =
1
(x + y + z − 2A) is 1 (3A − A) = same as
2 A! So a ≥ g says that
A ≥ g A = x(y + z − A)A. But (y + z − A)A = (y − A)(A − z) + yz > yz. Substitute to find A3
3 2
> xyz = G3 as we wanted to prove. Not easy!
There are many proofs of G = (x1x2 · · · xn)1/n ≤ A = (x1 + x2 + · · · + xn)/n. In
calculus you are maximizing G on the plane x1 + x2 + · · · + xn = n. The maximum occurs when
all x’s are equal.
1
33 The columns of the 4 by 4 “Hadamard matrix” (times ) are perpendicular
2
unit
vectors:
1 1 1 1
1 1 1 −1 1 −1
H= .
2 2 1 1 −1 −1
1 −1 −1 1
34 The commands V = randn (3, 30); D = sqrt (diag (V ′ ∗ V )); U = V \D; will give 30 random unit
vectors in the columns of U . Then u ′ ∗ U is a row matrix of 30 dot products whose average
absolute value should be close to 2/π.
Problem Set 1.3, page 29
1 3s1 + 4s2 + 5s3 = (3, 7, 12). The same vector b comes from S times x = (3, 4, 5):
,Solutions to Exercises 9
(row 1) · x
1 10 0 34 = (row 2) · x = 3 7.
1 1 1 5 (row 2) · x 12
2 The solutions are y1 = 1, y2 = 0, y3 = 0 (right side = column 1) and y1 = 1, y2 = 3,
y3 = 5. That second example illustrates that the first n odd numbers add to n2.
yy1 + y == B
B12 y1 = B
3 1 2 gives y2 = −B1 1+B2 1 0 0 B1
−
= 1 1 0 B2
y1 + y2 + y3 = B 3 y3 = −B2 +B3 0 −1 1
B3
1 0 0 1 0 0
The inverse of S = 1 1 0is A = −1 1 0 : independent columns in A and S!
1 1 1 0 −1 1
4 The combination 0w1 + 0w2 + 0w3 always gives the zero vector, but this problem looks for
other zero combinations (then the vectors are dependent, they lie in a plane): w2 = (w1 + w3)/2
so one combination that gives zero is w1 − 2w2 + w3 = 0.
5 The rows of the 3 by 3 matrix in Problem 4 must also be dependent: r2 = 1 (r1 + r3). 2
The column and row combinations that produce 0 are the same: this is unusual. Two
solutions to y1r1 + y2r2 + y3r3 = 0 are (Y1, Y2, Y3) = (1, −2, 1) and (2, −4, 2).
1 1 0
6 c=3 3 2 1 has column 3 = column 1 − column 2
7 4 3
1 0 −1
c = −1 1 1 0 has column 3 = − column 1 + column 2
0 1 1
0 0 0
c=0 2 1 5 has column 3 = 3 (column 1) − column 2
3 3 6
, THOSE WERE PREVIEW PAGES
1
0
Solutions to Exercises
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
Answer key
,2 Solutions to Exercises
Problem Set 1.1, page 8
1 The combinations give (a) a line in R3 (b) a plane in R3 (c) all of R3.
2 v + w = (2, 3) and v − w = (6, −1) will be the diagonals of the parallelogram with
v and w as two sides going out from (0, 0).
3 This problem gives the diagonals v + w and v − w of the parallelogram and asks for the sides:
The opposite of Problem 2. In this example v = (3, 3) and w = (2, −2).
4 3v + w = (7, 5) and cv + dw = (2c + d, c + 2d).
5 u+v = (−2, 3, 1) and u+v+w = (0, 0, 0) and 2u+2v+w = ( add first answers) = (−2, 3, 1). The
vectors u, v, w are in the same plane because a combination gives (0, 0, 0). Stated another
way: u = −v − w is in the plane of v and w.
6 The components of every cv + dw add to zero because the components of v and of w add to
zero. c = 3 and d = 9 give (3, 3, −6). There is no solution to cv+dw = (3, 3, 6) because 3 + 3 + 6 is
not zero.
7 The nine combinations c(2, 1) + d(0, 1) with c = 0, 1, 2 and d = (0, 1, 2) will lie on a lattice. If we
took all whole numbers c and d, the lattice would lie over the whole plane.
8 The other diagonal is v − w (or else w − v). Adding diagonals gives 2v (or 2w).
9 The fourth corner can be (4, 4) or (4, 0) or (−2, 2). Three possible parallelograms!
10 i − j = (1, 1, 0) is in the base (x-y plane). i + j + k = (1, 1, 1) is the opposite corner from (0, 0, 0).
Points in the cube have 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.
11 Four more corners (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1). The center point is (1 , 1 , 1 ).
2 2 2
Centers of faces are ( 1 , 1 , 0), ( 1 , 1 , 1) and (0, 1 , 1 ), (1, 1 , 1 ) and ( 1 , 0, 1 ), (1 , 1, 1 ).
2 2 2 2 2 2 2 2 2 2 2 2
12 The combinations of i = (1, 0, 0) and i + j = (1, 1, 0) fill the xy plane in xyz space.
13 Sum = zero vector. Sum = −2:00 vector = 8:00 vector. 2:00 is 30◦ from horizontal
√
= (cos π , sin π ) = ( 3/2, 1/2).
6 6
14 Moving the origin to 6:00 adds j = (0, 1) to every vector. So the sum of twelve vectors changes
from 0 to 12j = (0, 12).
,Solutions to Exercises 3
3 1
15 The point v + w is three-fourths of the way to v starting from w. The vector
4 4
1 1 1 1
v + w is halfway to u = 4 v+ w. The vector v + w is 2u (the far corner of the
4 2 2
parallelogram).
16 All combinations with c + d = 1 are on the line that passes through v and w. The point
V = −v + 2w is on that line but it is beyond w.
17 All vectors cv + cw are on the line passing through (0, 0) and u = 1 v + 1 w. That
2 2
line continues out beyond v + w and back beyond (0, 0). With c ≥ 0, half of this line is removed,
leaving a ray that starts at (0, 0).
18 The combinations cv + dw with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 fill the parallelogram with sides v and w.
For example, if v = (1, 0) and w = (0, 1) then cv + dw fills the unit square. But when v = (a, 0) and
w = (b, 0) these combinations only fill a segment of a line.
19 With c ≥ 0 and d ≥ 0 we get the infinite “cone” or “wedge” between v and w. For example, if v
= (1, 0) and w = (0, 1), then the cone is the whole quadrant x ≥ 0, y ≥
0. Question: What if w = −v? The cone opens to a half-space. But the combinations of v = (1,
0) and w = (−1, 0) only fill a line.
20 (a) 1 u + 1 v + 1 w is the center of the triangle between u, v and w; 1 u + 1 w lies
3 3 3 2 2
between u and w (b) To fill the triangle keep c ≥ 0, d ≥ 0, e ≥ 0, and c +d + e = 1.
21 The sum is (v − u)+ (w − v)+ (u− w) = zero vector. Those three sides of a triangle are in the same
plane!
22 The vector 1 (u + v + w) is outside the pyramid because c + d + e = 1
+1+ 1
> 1.
2 2 2 2
23 All vectors are combinations of u, v, w as drawn (not in the same plane). Start by seeing
that cu + dv fills a plane, then adding ew fills all of R3.
24 The combinations of u and v fill one plane. The combinations of v and w fill another plane.
Those planes meet in a line: only the vectors cv are in both planes.
25 (a) For a line, choose u = v = w = any nonzero vector (b) For a plane, choose
u and v in different directions. A combination like w = u + v is in the same plane.
,4 Solutions to Exercises
26 Two equations come from the two components: c + 3d = 14 and 2c + d = 8. The solution is c
= 2 and d = 4. Then 2(1, 2) + 4(3, 1) = (14, 8).
27 A four-dimensional cube has 24 = 16 corners and 2 · 4 = 8 three-dimensional faces and 24 two-
dimensional faces and 32 edges in Worked Example 2.4 A.
28 There are 6 unknown numbers v1, v2, v3, w1, w2, w3. The six equations come from the
components of v + w = (4, 5, 6) and v − w = (2, 5, 8). Add to find 2v = (6, 10, 14) so v = (3, 5, 7) and
w = (1, 0, −1).
29 Fact : For any three vectors u, v, w in the plane, some combination cu + dv + ew is the zero
vector (beyond the obvious c = d = e = 0). So if there is one combination Cu+ Dv + Ew that
produces b, there will be many more—just add c, d, e or 2c, 2d, 2e to the particular solution C, D,
E.
The example has 3u − 2v + w = 3(1, 3) − 2(2, 7) + 1(1, 5) = (0, 0). It also has
−2u + 1v + 0w = b = (0, 1). Adding gives u − v + w = (0, 1). In this case c, d, e
equal 3, −2, 1 and C, D, E = −2, 1, 0.
Could another example have u, v, w that could NOT combine to produce b ? Yes. The vectors (1,
1), (2, 2), (3, 3) are on a line and no combination produces b. We can easily solve cu + dv + ew =
0 but not Cu + Dv + Ew = b.
30 The combinations of v and w fill the plane unless v and w lie on the same line through (0, 0). Four
vectors whose combinations fill 4-dimensional space: one example is the “standard basis” (1, 0,
0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1).
31 The equations cu + dv + ew = b are
2c −d =1
−c +2d −e = 0 So d = 2e c = 3/4 d
−d +2e = 0 then c = 3e = 2/4 e =
then 4e = 1 1/4
,Solutions to Exercises 5
Problem Set 1.2, page 18
1 u · v = −2.4 + 2.4 = 0, u · w = −.6 + 1.6 = 1, u · (v + w) = u · v + u · w =
0 + 1, w · v = 4 + 6 = 10 = v · w.
√ . Then |u · v| = 0 < (1)(5) and |v · w| = 10 <
2 kuk = 1 and kvk = 5 and kwk = 5
√5, confirming the Schwarz inequality.
5
3 Unit vectors v/kvk = ( 4 , 3 ) = (0.8, 0.6). The vectors w, (2, −1), and −w make
5 5
√ √
0◦ , 90 ◦, 180 ◦ angles with w and w/kwk = (1/ 5, 2/ 5 ). The cosine of θ is v k·vk
√
w
kwk = 10/5 5.
4 (a) v · (−v) = −1 (b) (v + w) · (v − w) = v · v + w · v − v · w − w · w = 1+( )−( )−1 = 0 so θ = 90◦
(notice v·w = w·v) (c) (v−2w)·(v+2w) = v · v − 4w · w = 1 − 4 = −3.
√ √ 0 is
5 u1 = v/kvk = (1, 3)/ 1 0 and u2 = w/kwk = (2, 1, 2)/3. U1 = (3, −1)/ 1
√ √ : There is a
perpendicular to u1 (and so is (−3, 1)/ 10). U2 could be (1, −2, 0)/ 5
whole plane of vectors perpendicular to u2, and a whole circle of unit vectors in that plane.
6 All vectors w = (c, 2c) are perpendicular to v. They lie on a line. All vectors (x, y, z) with x + y + z
= 0 lie on a plane. All vectors perpendicular to (1, 1, 1) and (1, 2, 3) lie on a line in 3-
dimensional space.
7 (a) cos θ = v · w/kvkkwk = 1/(2)(1) so θ = 60◦ or π/3 radians (b) cos θ = 0 so θ = 90◦ or
π/2 radians (c) cos θ = 2/(2)(2) = 1/2 so θ = 60◦ or π/3
√ ◦
(d) cos θ = −1/ 2 so θ = 135 or 3π/4.
8 (a) False: v and w are any vectors in the plane perpendicular to u (b) True: u · (v + 2w) = u ·
v + 2u · w = 0 (c) True, ku − vk2 = (u − v) · (u − v) splits into u · u + v · v = 2 when u · v = v · u = 0.
9 If v2w2/v1w1 = −1 then v2w2 = −v1w1 or v1w1 +v2w2 = v· w = 0: perpendicular! The vectors (1, 4) and
(1, − 4 ) are perpendicular.
1
,6 Solutions to Exercises
10 Slopes 2/1 and −1/2 multiply to give −1: then v · w = 0 and the vectors (the direc- tions) are
perpendicular.
11 v · w < 0 means angle > 90◦; these w’s fill half of 3-dimensional space.
12 (1, 1) perpendicular to (1, 5) − c(1, 1) if (1, 1) · (1, 5) − c(1, 1) · (1, 1) = 6 − 2c = 0 or c = 3; v · (w − cv) =
0 if c = v · w/v · v. Subtracting cv is the key to constructing a perpendicular vector.
13 The plane perpendicular to (1, 0, 1) contains all vectors (c, d, −c). In that plane, v = (1, 0, −1) and
w = (0, 1, 0) are perpendicular.
14 One possibility among many: u = (1, −1, 0, 0), v = (0, 0, 1, −1), w = (1, 1, −1, −1)
and (1, 1, 1, 1) are perpendicular to each other. “We can rotate those u, v, w in their
3D hyperplane and they will stay perpendicular.” √ √
15 1 (x + y) = (2 + 8)/2 = 5 and 5 > 4; cos θ = 2 6/ 0 √ 10 = 8/10.
2 2 1 1
16 kvk = 1 + 1 + · · · + 1 = 9 so kvk = 3; u = v/3 = (1 , . . . , 1 ) is a unit vector in 9D;
3 3
√
2 vector in the
w = (1, −1,√0, . . . , 0)/ is a unit √ 8D hyperplane perpendicular to v.
. For any vector v = (v1, v2, v3) the
17 cos α = 1/ , cos β = 0, cos γ = −1/
2 2
cosines with (1, 0, 0) and (0, 0, 1) are cos2 α+cos2 β+cos2 γ = (v2+v 2+v2)/kvk2 = 1.
1 2 3
18 kvk2 = 42 + 22 = 20 and kwk2 = (−1)2 + 22 = 5. Pythagoras is k(3, 4)k2 = 25 = 20 + 5 for the length
of the hypotenuse v + w = (3, 4).
19 Start from the rules (1), (2), (3) for v · w = w · v and u · (v + w) and (cv) · w. Use rule (2) for (v + w)
· (v + w) = (v + w) · v + (v + w) · w. By rule (1) this is v · (v + w) + w · (v + w). Rule (2) again
gives v · v + v · w + w · v + w · w = v · v + 2v · w + w · w. Notice v · w = w · v! The main point is to
feel free to open up parentheses.
20 We know that (v − w) · (v − w) = v · v − 2v · w + w · w. The Law of Cosines writes kvkkwk cos θ for v ·
w. Here θ is the angle between v and w. When θ < 90◦ this v · w is positive, so in this case v ·
v + w · w is larger than kv − wk2.
Pythagoras changes from equality a2 +b2 = c2 to inequality when θ < 90 ◦ or θ > 90 ◦.
,Solutions to Exercises 7
21 2v· w ≤ 2kvkkwk leads to kv+ wk2 = v· v + 2v· w +w· w ≤ kvk2 + 2kvkkwk +
2 2
kwk . This is (kvk + kwk) . Taking square roots gives kv + wk ≤ kvk + kwk.
22 v2w2 + 2v1w1v2w2 + v2w2 ≤ v2w2 + v2w2 + v2w2 + v2w2 is true (cancel 4 terms)
1 1 2 2 1 1 1 2 2 1 2 2
because the difference is v2w2 + v2w2 − 2v1w1v2w2 which is (v1w2 − v2w1)2 ≥ 0.
1 2 2 1
23 cos β = w1/kwk and sin β = w2/kwk. Then cos(β −a) = cos β cos α+sin β sin α =
v1w1/kvkkwk + v2w2/kvkkwk = v · w/kvkkwk. This is cos θ because β − α = θ.
24 Example 6 gives |u1||U1| ≤ 1 (u2 + U 2) and |u2||U2| ≤ 1 (u2 + U 2). The whole line
2 1 1 2 2 2
1 2 2 1 2 2
becomes .96 ≤ (.6)(.8) + (.8)(.6) ≤ 2 (.6 + .8 ) + 2 (.8 + .6 ) = 1. True: .96 < 1.
p
25 The cosine of θ is x/ x2 + y 2, near side over hypotenuse. Then | cos θ|2 is not greater than 1:
x2/(x2 + y 2 ) ≤ 1.
26–27 (with apologies for that typo !) These two lines add to 2||v||2 + 2||w||2 :
2
||v+w|| = (v + w) · (v + w) = v · v + v · w + w · v + w · w
2
= (v − w) · (v − w) = v · v − v · w − w · v + w · w
||v − w||
28 The vectors w = (x, y) with (1, 2) · w = x + 2y = 5 lie on a line in the xy plane. The
√
shortest w on that line is (1, 2). (The Schwarz inequality kwk ≥ v · w/kvk = is 5
√
an equality when cos θ = 0 and w = (1, 2) and kwk = 5.)
29 The length kv − wk is between 2 and 8 (triangle inequality when kvk = 5 and kwk = 3). The dot
product v · w is between −15 and 15 by the Schwarz inequality.
30 Three vectors in the plane could make angles greater than 90◦ with each other: for
example (1, 0), (−1, 4), (−1, −4). Four vectors could not do this (360◦ total angle). How many
can do this in R3 or Rn? Ben Harris and Greg Marks showed me that the answer is n + 1. The
vectors from the center of a regular simplex in Rn to its n + 1 vertices all have negative dot
products. If n+2 vectors in Rn had negative dot products, project them onto the plane orthogonal
to the last one. Now you have n + 1 vectors in Rn−1 with negative dot products. Keep going to 4
vectors in R2 : no way!
31 For a specific example, pick v = (1, 2, −3) and then w = (−3, 1, 2). In this example
√
√ 1 14 = −1/2 and θ = 120◦ . This always
cos θ = v · w/kvkkwk = −7/ 4
happens when x + y + z = 0:
,8 Solutions to Exercises
1 1
v · w = xz + xy + yz = (x + y + z)2 − (x2 + y 2 + z 2 )
2 2
1 1
This is the same as v · w = 0 − kvkkwk. Then cos θ = .
2 2
32 Wikipedia gives this proof of geometric mean G = √
3 xyz ≤ arithmetic mean
A = (x + y + z)/3. First there is equality in case x = y = z. Otherwise A is somewhere
between the three positive numbers, say for example z < A < y.
Use the known inequality g ≤ a for the two positive numbers x and y + z − A. Their mean a =
1
(x + y + z − 2A) is 1 (3A − A) = same as
2 A! So a ≥ g says that
A ≥ g A = x(y + z − A)A. But (y + z − A)A = (y − A)(A − z) + yz > yz. Substitute to find A3
3 2
> xyz = G3 as we wanted to prove. Not easy!
There are many proofs of G = (x1x2 · · · xn)1/n ≤ A = (x1 + x2 + · · · + xn)/n. In
calculus you are maximizing G on the plane x1 + x2 + · · · + xn = n. The maximum occurs when
all x’s are equal.
1
33 The columns of the 4 by 4 “Hadamard matrix” (times ) are perpendicular
2
unit
vectors:
1 1 1 1
1 1 1 −1 1 −1
H= .
2 2 1 1 −1 −1
1 −1 −1 1
34 The commands V = randn (3, 30); D = sqrt (diag (V ′ ∗ V )); U = V \D; will give 30 random unit
vectors in the columns of U . Then u ′ ∗ U is a row matrix of 30 dot products whose average
absolute value should be close to 2/π.
Problem Set 1.3, page 29
1 3s1 + 4s2 + 5s3 = (3, 7, 12). The same vector b comes from S times x = (3, 4, 5):
,Solutions to Exercises 9
(row 1) · x
1 10 0 34 = (row 2) · x = 3 7.
1 1 1 5 (row 2) · x 12
2 The solutions are y1 = 1, y2 = 0, y3 = 0 (right side = column 1) and y1 = 1, y2 = 3,
y3 = 5. That second example illustrates that the first n odd numbers add to n2.
yy1 + y == B
B12 y1 = B
3 1 2 gives y2 = −B1 1+B2 1 0 0 B1
−
= 1 1 0 B2
y1 + y2 + y3 = B 3 y3 = −B2 +B3 0 −1 1
B3
1 0 0 1 0 0
The inverse of S = 1 1 0is A = −1 1 0 : independent columns in A and S!
1 1 1 0 −1 1
4 The combination 0w1 + 0w2 + 0w3 always gives the zero vector, but this problem looks for
other zero combinations (then the vectors are dependent, they lie in a plane): w2 = (w1 + w3)/2
so one combination that gives zero is w1 − 2w2 + w3 = 0.
5 The rows of the 3 by 3 matrix in Problem 4 must also be dependent: r2 = 1 (r1 + r3). 2
The column and row combinations that produce 0 are the same: this is unusual. Two
solutions to y1r1 + y2r2 + y3r3 = 0 are (Y1, Y2, Y3) = (1, −2, 1) and (2, −4, 2).
1 1 0
6 c=3 3 2 1 has column 3 = column 1 − column 2
7 4 3
1 0 −1
c = −1 1 1 0 has column 3 = − column 1 + column 2
0 1 1
0 0 0
c=0 2 1 5 has column 3 = 3 (column 1) − column 2
3 3 6
, THOSE WERE PREVIEW PAGES
1
0
Solutions to Exercises
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
