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线性代数导论第五版答案--Gilbert Strang

线性代数

Gilbert

Stra

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Gilbert Strang 线性代数导论第五版答案 MIT 公开课：Gilbert Strang《线性代数》

INTRODUCTION

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

Solutions to Exercises

Problem Set 1.1, page 8

1 T

he combinations give (a) a line in R

(b) a plane in R

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 (

Centers of faces are (

, 0), (

, 1) and (0,

), (1,

) and (

, 0,

), (

, 1,

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

, sin

) = (

√

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 12 j = (0, 12).

Solutions to Exercises

15 The point

v +

w is three-fourths of the way to v starting from w. The vector

v +

w is halfway to u =

v +

w. The vector v + w is 2u (the far corner of the

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 =

v +

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).

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)

u +

v +

w is the center of the triangle between u, v and w;

u +

w lies

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

(u + v + w) is outside the pyramid because c + d + e =

> 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 R

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.

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 2

= 1 6 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 v

, v

, w

. 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 Two combinations out of inﬁnitely many that produce b = (0, 1) are −2u + v and

w −

v. No, three vectors u, v, w in the x-y plane could fail to produce b if all

three lie on a line that does not contain b. Yes, if one combination produces b then two

(and inﬁnitely many) combinations will produce b. This is true even if u = 0; the

combinations can have different cu.

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

−d +2e = 0

So d = 2e

then c = 3e

then 4e = 1

c = 3/4

d = 2/ 4

e = 1/4

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 = −2 = v · w.

2 kuk = 1 and kvk = 5 and kwk =

√

5. Then |u ·v| = 0 < (1)(5) and |v · w| = 10 <

√

5, conﬁrming the Schwarz inequality.

Solutions to Exercises

3 Unit vectors v/kvk = (

) = (0.8, 0.6). The vectors w, (2, −1), and −w make

◦

, 90

◦

, 180

◦

angles with w and w/kwk = (1/

√

5, 2/

√

5). The cosine of θ is

kvk

kwk

= 10/5

√

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.

5 u

= v/kvk = (1, 3)/

√

10 and u

= w/kwk = (2 , 1, 2)/3. U

= (3, −1)/

√

10 is

perpendicular to u

(and so is (−3, 1)/

√

10). U

could be (1, −2, 0)/

√

5: There is a

whole plane of vectors perpendicular to u

, 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) co s 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 −vk

= (u −v) ·(u −v) splits into

u · u + v · v = 2 when u · v = v ·u = 0.

9 If v

= −1 then v

= −v

or v

= v·w = 0: perpendicular!

The vectors (1, 4) and (1, −

) are perpendicular.

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 −2 c = 0 or

c = 3; v · (w − cv) = 0 if c = v · w/v · v. Subtracting cv is the key to constructing

a perpendicular vector.

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