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Solutions Manual for Introduction to Linear Algebra 5th 习题解答.pdf
Chapter 1
Problem Set 1.1, page 8
Problem Set 1.2, page 18
Problem Set 1.3, page 29
Chapter 2
Problem Set 2.1, page 41
Problem Set 2.2, page 53
Problem Set 2.3, page 66
Problem Set 2.4, page 77
Problem Set 2.5, page 92
Problem Set 2.6, page 104
Problem Set 2.7, page 117
Chapter 3
Problem Set 3.1, page 131
Problem Set 3.2, page 142
Problem Set 3.3, page 158
Problem Set 3.4, page 175
Problem Set 3.5, page 190
Chapter 4
Problem Set 4.1, page 202
Problem Set 4.2, page 214
Problem Set 4.3, page 229
Problem Set 4.4, page 242
Chapter 5
Problem Set 5.1, page 25
Problem Set 5.2, page 266
Problem Set 5.3, page 283
Chapter 6
Problem Set 6.1, page 298
Problem Set 6.2, page 314
Problem Set 6.3, page 332
Problem Set 6.4, page 345
Problem Set 6.5, page 358
Problem Set 6.6, page 360
Problem Set 6.7, page 37
Chapter 7
Problem Set 7.1, page 370
Problem Set 7.2, page 379
Problem Set 7.3, page 391
Problem Set 7.4, page 398
Chapter 8
Problem Set 8.1, page 407
Problem Set 8.2, page 418
Problem Set 8.3, page 429
Chapter 9
Problem Set 9.1, page 436
Problem Set 9.2, page 443
Problem Set 9.3, page 450
Chapter 10
Problem Set 10.1, page 459
Problem Set 10.2, page 472
Problem Set 10.3, page 480
Problem Set 10.4, page 489
Problem Set 10.5, page 494
Problem Set 10.6, page 50
Problem Set 10.7, page 507
Chapter 11
Problem Set 11.1, page 516
Problem Set 11.2, page 522
Problem Set 11.3, page 531
Chapter 12
Problem Set 12.1, page 544
Problem Set 12.2, page 554
Problem Set 12.3, page 560
INTRODUCTION
TO
LINEAR
ALGEBRA
Fifth Edition
MANUAL FOR INSTRUCTORS
Gilbert Strang
Massachusetts Institute of Technology
math.mit.edu/linearalgebra
web.mit.edu/18.06
video lectures: ocw.mit.edu
math.mit.edu/∼gs
www.wellesleycambridge.com
email: linearalgebrabook@gmail.com
Wellesley - Cambridge Press
Box 812060
Wellesley, Massachusetts 02482
2
SolutionstoExercises
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
2 ), ( 1
Centers of faces are ( 1
2 , 1) and (0, 1
2 , 1
2 , 0), ( 1
2 ) and ( 1
2 , 1
2 ), (1, 1
2 , 1
2 , 0, 1
2 , 1
2 , 1
2 , 1
2 ).
2 , 1, 1
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
6 , sin π
6 ) = (√3/2, 1/2).
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).
SolutionstoExercises
3
15 The point
v +
w is three-fourths of the way to v starting from w. The vector
3
4
1
4
w is halfway to u =
1
4
v +
1
4
parallelogram).
1
2
v +
1
2
w. The vector v + w is 2u (the far corner of the
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
2 w. That
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).
2 v + 1
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
3 u + 1
3 v + 1
between u and w
3 w is the center of the triangle between u, v and w; 1
2 w lies
(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
2 u + 1
are in the same plane!
22 The vector 1
2 (u + v + w) is outside the pyramid because c + d + e = 1
2 + 1
2 + 1
2 > 1.
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
SolutionstoExercises
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
= 1
2c −d
−c +2d −e = 0
−d +2e = 0
So d = 2e
then c = 3e
then 4e = 1
c = 3/4
d = 2/4
e = 1/4
SolutionstoExercises
Problem Set 1.2, page 18
5
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.
2 kuk = 1 and kvk = 5 and kwk = √5. Then |u · v| = 0 < (1)(5) and |v · w| = 10 <
5√5, confirming the Schwarz inequality.
5 , 3
3 Unit vectors v/kvk = ( 4
5 ) = (0.8, 0.6). The vectors w, (2,−1), and −w make
0 ◦, 90 ◦, 180 ◦ angles with w and w/kwk = (1/√5, 2/√5). The cosine of θ is v
kvk ·
w
kwk
= 10/5√5.
)−(
4 (a) v · (−v) = −1
(b) (v + w) · (v − w) = v · v + w · v − v · w − w · w =
(c) (v−2w)·(v+2w) =
)−1 = 0 so θ = 90◦ (notice v·w = w·v)
1+(
v · v − 4w · w = 1 − 4 = −3.
5 u1 = v/kvk = (1, 3)/√10 and u2 = w/kwk = (2, 1, 2)/3. U 1 = (3,−1)/√10 is
perpendicular to u1 (and so is (−3, 1)/√10). U 2 could be (1,−2, 0)/√5: There is a
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.
0 so θ = 90◦ or π/2 radians
(c) cos θ = 2/(2)(2) = 1/2 so θ = 60◦ or π/3
7 (a) cos θ = v · w/kvkkwk = 1/(2)(1) so θ = 60◦ or π/3 radians
(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 ·
(c) True, ku − vk2 = (u − v) · (u − v) splits into
(b) cos θ =
(v + 2w) = u · v + 2u · w = 0
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,− 1
4 ) are perpendicular.
6
SolutionstoExercises
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.”
2 (x + y) = (2 + 8)/2 = 5 and 5 > 4; cos θ = 2√16/√10√10 = 8/10.
15 1
16 kvk2 = 1 + 1 +··· + 1 = 9 so kvk = 3; u = v/3 = ( 1
3 ) is a unit vector in 9D;
w = (1,−1, 0, . . . , 0)/√2 is a unit vector in the 8D hyperplane perpendicular to v.
17 cos α = 1/√2, cos β = 0, cos γ = −1/√2. For any vector v = (v1, v2, v3) the
3)/kvk2 = 1.
18 kvk2 = 42 + 22 = 20 and kwk2 = (−1)2 + 22 = 5. Pythagoras is k(3, 4)k2 = 25 =
cosines with (1, 0, 0) and (0, 0, 1) are cos2 α+cos2 β+cos2 γ = (v2
3 , . . . , 1
1+v2
2+v2
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 ◦.
SolutionstoExercises
7
21 2v· w ≤ 2kvkkwk leads to kv + wk2 = v · v + 2v· w + w· w ≤ kvk2 + 2kvkkwk +
kwk2. This is (kvk + kwk)2. Taking square roots gives kv + wk ≤ kvk + kwk.
1w2
2w2
because the difference is v2
1 + 2v1w1v2w2 + v2
2 + v2
1 + v2
1w2
2w2
2w2
22 v2
2 is true (cancel 4 terms)
2 ≤ v2
1w2
1 + v2
1 − 2v1w1v2w2 which is (v1w2 − v2w1)2 ≥ 0.
2w2
2 + v2
1w2
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.
2 ). The whole line
2 (.82 + .62) = 1. True: .96 < 1.
24 Example 6 gives |u1||U1| ≤ 1
1 + U 2
becomes .96 ≤ (.6)(.8) + (.8)(.6) ≤ 1
25 The cosine of θ is x/px2 + y2, near side over hypotenuse. Then | cos θ|2 is not greater
than 1: x2/(x2 + y2) ≤ 1.
26–27 (with apologies for that typo !) These two lines add to 2||v||2 + 2||w||2 :
||v + w||2 = (v + w) · (v + w) = v · v + v · w + w · v + w · w
||v − w||2 = (v − w) · (v − w) = v · v − v · w − w · v + w · w
1 ) and |u2||U2| ≤ 1
2 (.62 + .82) + 1
2 + U 2
2 (u2
2 (u2
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 = √5 is
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
cos θ = v · w/kvkkwk = −7/√14√14 = −1/2 and θ = 120◦ . This always
happens when x + y + z = 0:
8
SolutionstoExercises
v · w = xz + xy + yz =
This is the same as v · w = 0 −
(x + y + z)2 −
(x2 + y2 + z2)
1
2 kvkkwk. Then cos θ =
1
2
1
2
1
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
2 (3A − A) = same as A! So a ≥ g says that
A3 ≥ g2A = x(y + z − A)A. But (y + z − A)A = (y − A)(A − z) + yz > yz.
Substitute to find A3 > xyz = G3 as we wanted to prove. Not easy!
2 (x + y + z − A) is 1
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.
33 The columns of the 4 by 4 “Hadamard matrix” (times 1
2 ) are perpendicular unit
vectors:
1
2
H =
1
2
1
1
1
1
1 −1
1
1 −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):
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