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MIT线代第五版习题答案.pdf
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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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