Chapter 5
ISM: Linear Algebra
Chapter 5 5.1
√ √ √ 72 + 112 = 49 + 121 = 170 ≈ 13.04 √ √ √ 2. v = 22 + 32 + 42 = 4 + 9 + 16 = 29 ≈ 5.39 √ √ √ 3. v = 22 + 32 + 42 + 52 = 4 + 9 + 16 + 25 = 54 ≈ 7.35 1. v = 4. θ = arccos 5. θ = arccos 6. θ = arccos
u·v u v u·v u v u·v u v √ = arccos √7+11 = arccos √18 ≈ 0.219 (radians) 2 170 340 2+6+12 = arccos √14√29 ≈ 0.122 (radians) √ √ = arccos 2−3+8−10 ≈ 1.700 (radians) 10 54
7. Use the fact that u · v = u v cos θ, so that the angle is acute if u · v > 0, and obtuse if u · v < 0. Since u · v = 10 − 12 = −2, the angle is obtuse. 8. Since u · v = 4 − 24 + 20 = 0, the two vectors enclose a right angle. 9. Since u · v = 3 − 4 + 5 − 3 = 1, the angle is acute (see Exercise 7). 10. u · v = 2 + 3k + 4 = 6 + 3k. The two vectors enclose a right angle if u · v = 6 + 3k = 0, that is, if k = −2.
u·v u v 1 = arccos √n π 4 (=
11. a. θn = arccos
1 θ2 = arccos √2 =
45◦ )
1 θ3 = arccos √3 ≈ 0.955 (radians)
θ4 = arccos 1 = 2
π 3 (=
60◦ )
b. Since y = arccos(x) is a continuous function,
n→∞
lim θn = arccos
n→∞
lim
1 √ n
= arccos(0) =
π 2 (=
90◦ )
12. v + w = v ≤ v
2 2
2
= (v + w) · (v + w) (by hint)
2 2
+ w + w
+ 2(v · w) (by definition of length) +2 v w (by Cauchy-Schwarz) 234
ISM: Linear Algebra = ( v + w )2 , so that v+w
2
Section 5.1
≤ ( v + w )2
Taking square roots of both sides, we find that v + w ≤ v + w , as claimed. 13. Figure 5.1 shows that F2 + F3 = 2 cos
θ 2
F2 = 20 cos
θ 2
θ 2
.
It is required that F2 + F3 = 16, so that 20 cos
= 16, or θ = 2 arccos(0.8) ≈ 74◦ .
Figure 5.1: for Problem 5.1.13. 14. The horizontal components of F1 and F2 are − F1 sin β and F2 sin α, respectively (the horizontal component of F3 is zero). Since the system is at rest, the horizontal components must add up to 0, so that − F1 sin β+ F2 sin α = 0 or F1 sin β = F2 sin α or
EA , note that EA = ED tan α EB F1 cos β cos β cos α = F2 cos α . Since α and β F1 F2 F1 F2
=
sin α sin β .
To find
sin α sin β EA EB
and EB = ED tan β so that
EA EB
=
tan α tan β
=
· =
are two distinct acute angles, it follows that
, so that Leonardo was mistaken.
15. The subspace consists of all vectors x in R4 such that x1 1 x2 2 x · v = · = x1 + 2x2 + 3x3 + 4x4 = 0. x3 3 x4 4 −2r r These are vectors of the form −3s −4t s −2 −3 −4 1 0 0 =r +s +t . 0 1 0 t 0 0 1
The three vectors to the right form a basis. 235
Chapter 5
ISM: Linear Algebra
16. You may be able to find the solutions by educated guessing. Here is the systematic approach: we first find all vectors x that are orthogonal to v1 , v2 , and v3 , then we identify the unit vectors among them. Finding the vectors x with x · v1 = x · v2 = x · v3 = 0 amounts to solving the system x1 + x 2 + x 3 + x 4 = 0 x1 + x 2 − x 3 − x 4 = 0 x1 − x 2 + x 3 − x 4 = 0 we can omit all the coefficients
1 2
.
17. The orthogonal complement W ⊥ of W consists of the vectors x in R4 such that x1 1 x1 5 x2 2 x2 6 · = 0 and · = 0. x3 3 x3 7 x4 4 x4 8 Finding these vectors amounts to solving the system The solutions are of the form 2 1 s + 2t x1 −3 −2 x2 −2s − 3t . +t = s = 0 1 s x3 1 0 t x4 18. a. x
2 1 1 1 = 1+ 4 + 16 + 64 +· · · = 2 √ 3 1 1− 1 4
Since x = 2|t|, we have a unit vector if t = choices for v4 : 1 1 −2 2 1 1 −2 2 − 1 and 1 . 2 2 1 1 −2 2
t x1 x2 −t The solutions are of the form x = = . −t x3 t x4
1 2
1 or t = − 2 . Thus there are two possible
x1 + 2x2 + 3x3 + 4x4 = 0 . 5x1 + 6x2 + 7x3 + 8x4 = 0
The two vectors to the right form a basis of W ⊥ . =
4 3
use the formula for a geometric series, with a =
1 4
,
so that x =
≈ 1.155.
236
ISM: Linear Algebra b. If we let u = (1, 0, 0, . . .) and v = 1, 1 , 1 , · · · , then 2 4 θ = arccos
u·v u v
Section 5.1
= arccos
1
2 √ 3
= arccos
√ 3 2
=
π 6 (=
30◦ ).
1 1 1 c. x = 1, √2 , √3 , · · · , √n , · · · does the job, since the harmonic series 1+ 1 + 1 +· · · diverges 2 3
(a fact discussed in introductory calculus classes).
d. If we let v = (1, 0, 0, . . .), x = 1, 1 , 1 , · · · and u = 2 4 projL v = (u · v)u = 19. See Figure 5.2.
3 4 1 1, 2 , 1 , · · · . 4
x x
=
√ 3 2
1, 1 , 1 , · · · then 2 4
Figure 5.2: for Problem 5.1.19. 20. On the line L spanned by x we want to find the vector mx closest to y (that is, we want mx − y to be minimal). We want mx − y to be perpendicular to L (that is, to x), which x·y 4182.9 means that x · (mx − y) = 0 or m(x · x) − x · y = 0 or m = x·x ≈ 198.532 ≈ 0.106. Recall that the correlation coefficient r is r =
x·y x y
, so that m =
y x
r. See Figure 5.3.
21. Call the three given vectors v1 , v2 , and v3 . Since v2 is required to be a unit vector, we must have b = g = 0. Now v1 · v2 = d must be zero, so that d = 0. Likewise, v2 · v3 = e must be zero, so that e = 0. Since v3 must be a unit vector, we have v3
2 1 4 √ 3 2 .
= c2 +
= 1, so that c = ±
√ 3 2 .
Since we are asked to find just one solution, let us pick c = 237
Chapter 5
ISM: Linear Algebra
Figure 5.3: for Problem 5.1.20. The condition v1 · v3 = 0 now implies that Finally, it is required that v1
2 √ 3 2 . √ 3 2 a
√ + 1 f = 0, or f = − 3a. 2
1 = a2 + f 2 = a2 + 3a2 = 4a2 = 1, so that a = ± 2 .
1 Let us pick a = 2 , so that f = −
√ 3 0 2 0 v1 = √ , v2 = 1 , v3 = 0 1 0 − 3
1 2 2 2
Summary:
There are other solutions; some components will have different signs.
22. Let W = {x in Rn : x · vi = 0 for all i = 1, . . . , m}. We are asked to show that V ⊥ = W , that is, any x in V ⊥ is in W , and vice versa. If x is in V ⊥ , then x · v = 0 for all v in V ; in particular, x · vi = 0 for all i (since the vi are in V ), so that x is in W . Conversely, consider a vector x in W . To show that x is in V ⊥ , we have to verify that x · v = 0 for all v in V . Pick a particular v in V . Since the vi span V , we can write 238
ISM: Linear Algebra
Section 5.1
v = c1 v1 + · · · + cm vm , for some scalars ci . Then x · v = c1 (x · v1 ) + · · · + cm (x · vm ) = 0, as claimed. 23. We will follow the hint. Let v be a vector in V . Then v · x = 0 for all x in V ⊥ . Since (V ⊥ )⊥ contains all vectors y such that y · x = 0, v is in (V ⊥ )⊥ . So V is a subspace of (V ⊥ )⊥ . Then, by Fact 5.1.8c, dim (V ) + dim(V ⊥ ) = n and dim(V ⊥ ) + dim((V ⊥ )⊥ ) = n, so dim (V ) + dim(V ⊥ ) = dim(V ⊥ ) + dim((V ⊥ )⊥ ) and dim (V ) = dim((V ⊥ )⊥ ). Since V is a subspace of (V ⊥ )⊥ , it follows that V = (V ⊥ )⊥ , by Exercise 3.3.41. 24. Write T (x) = projV (x) for simplicity. To prove the linearity of T we will use the definition of a projection: T (x) is in V , and x − T (x) is in V ⊥ . To show that T (x+y) = T (x)+T (y), note that T (x)+T (y) is in V (since V is a subspace), and x + y − (T (x) + T (y)) = (x − T (x)) + (y − T (y)) is in V ⊥ (since V ⊥ is a subspace, by Fact 5.1.8a). To show that T (kx) = kT (x), note that kT (x) is in V (since V is a subspace), and kx − kT (x) = k(x − T (x)) is in V ⊥ (since V ⊥ is a subspace).
25. a. kv
2
√ Now take square roots of both sides; note that k 2 = |k|, the absolute value of k (think about the case when k is negative). kv = |k| v , as claimed. b. u =
1 v
= (kv) · (kv) = k 2 (v · v) = k 2 v
2
v = ↑
1 v
v = 1, as claimed.
by part a
26. The two given vectors spanning the subspace are orthogonal, but they are not unit vectors: both have length 7. To obtain an orthonormal basis u1 , u2 of the subspace, we divide by 7: 2 3 1 u1 = 7 3 , u2 = 1 −6 . 7 6 2 239
Chapter 5 49 Now we can use Fact 5.1.5, with x = 49 : 49
ISM: Linear Algebra
27. Since the two given vectors in the subspace are orthogonal, we have the orthonormal basis 2 1 2 u1 = 3 , u2 = 1 0 −2 1 2 3 0. 1
2 3 19 projV x = (u1 · x)u1 + (u2 · x)u2 = 11 3 − −6 = 39 . 6 2 64
Now we can use Fact 5.1.5, with x = 9e1 : projV x = (u1 · x)u1 + (u2 · x)u2 8 −2 2 2 0 2 = 2 − 2 = . 2 0 1 −2 1 0 28. Since the three given vectors in the subspace are orthogonal, we have the orthonormal basis 1 1 1 1 1 1 −1 u1 = 1 , u2 = 1 , u3 = 2 . 2 2 1 −1 −1 1 −1 1 Now we can use Fact 5.1.5, with x = e1 : projV x = (u1 · x)u1 + (u2 · x)u2 + (u3 · x)u3 = 3 1 1 4 −1 . 1 29. By the Pythagorean theorem (Fact 5.1.9), x
2
= 7u1 − 3u2 + 2u3 + u4 − u5 2 = 7u1 2 + 3u2 2 + 2u3 2 + u4 2 + u5 = 49 + 9 + 4 + 1 + 1 = 64, so that x = 8.
2
.
30. Since y = projV x, the vector x − y is orthogonal to y, by definition of a projection (see Fact 5.1.4): (x − y) · y = 0 or x · y − y 2 = 0 or x · y = y 2 . See Figure 5.4. 240
ISM: Linear Algebra
Section 5.1
Figure 5.4: for Problem 5.1.30.
31. If V = span(u1 , . . . , um ), then projV x = (u1 · x)u1 + · · · + (um · x)um , by Fact 5.1.5, and projV x 2 = (u1 · x)2 + · · · + (um · x)2 = p, by the Pythagorean theorem (Fact 5.1.9). Therefore p ≤ x 2 , by Fact 5.1.10. The two quantities are equal if (and only if) x is in V. 32. By Fact 2.3.6a, the matrix G is invertible if (and only if) (v1 · v1 )(v2 · v2 ) − (v1 · v2 )2 = v1 2 v2 2 − (v1 · v2 )2 = 0. The Cauchy-Schwarz inequality (Fact 5.1.11) tells us that v1 2 v2 2 − (v1 · v2 )2 ≥ 0; equality holds if (and only if) v1 and v2 are parallel (that is, linearly dependent). x1 33. Let x = · · · be a vector in Rn whose components add up to 1, that is, x1 +· · ·+xn = 1. xn 1 Let y = · · · (all n components are 1). The Cauchy-Schwarz inequality (Fact 5.1.11) 1 √ 1 tells us that |x·y| ≤ x y , or, |x1 +· · ·+xn | ≤ x n, or x ≥ √n . By Fact 5.1.11, the 1 equation x = √n holds if (and only if) the vectors x and y are parallel, that is, x1 = x2 = 1 · · · = xn =
1 n.
Thus the vector of minimal length is x = · · · all components are
1 n
n
1 n
.
Figure 5.5 illustrates the case n = 2. 1 34. Let x be a unit vector in Rn , that is, x = 1. Let y = . . . (all n components 1 are 1). The Cauchy-Schwarz inequality (Fact 5.1.11) tells us that |x · y|√ x y , or, ≤ √ √ |x1 +. . .+xn | ≤ x n = n. By Fact 5.1.11, the equation x1 +. . n = n holds if x = .+x k ky for positive k. Thus x must be a unit vector of the form x = . . . for some positive k 241
Chapter 5
x2
ISM: Linear Algebra
1 → = x X
1 2 1 2
1
x1
x1 + x2 = 1
Figure 5.5: for Problem 5.1.33.
1 √ n 1 √ n
k. It is required that nk 2 = 1, or, k =
1 √ . n
Thus x = . . . all components are
1 √ n
.
Figure 5.6 illustrates the case n = 2.
x2 1 √2 1 √2 x1 x1 + x2 =√ 2
→= x X
1
Figure 5.6: for Problem 5.1.34.
x 1 35. Applying the Cauchy-Schwarz inequality to u = y and v = 2 gives |u·v| ≤ u v , z 3 √ √ attained when u = kv or |x + 2y + 3z| ≤ 14. The minimal value x + 2y + 3z = − 14 is k for negative k. Thus u must be a unit vector of the form u = 2k , for negative k. It is 3k 242
ISM: Linear Algebra − √1 14 − √3 14
Section 5.1
required that 14k 2 = 1, or, k = − √1 . Thus u = − √2 . 14 14 a 0.2 36. Let x = b and y = 0.3 . It is required that x · y = 0.2a + 0.3b + 0.5c = 76. c 0.5 Our goal is to minimize quantity x · x = a2 + b2 + c2 . The Cauchy-Schwarz inequality (squared) tells us that (x · y)2 ≤ x 2 y 2 , or 762 ≤ (a2 + b2 + c2 )(0.22 + 0.32 + 0.52 ) or 762 762 a2 + b2 + c2 ≥ 0.38 . Quantity 2 + b2 + c2 is minimal when a2 + b2 + c2 = 0.38 . This a a 0.2k is the case when x = b = 0.3k for some positive constant k. It is required that c 0.5k 0.2a + 0.3b + 0.5c = (0.2)2 k + (0.3)2 k + (0.5)2 k = 0.38k = 76, so that k = 200. Thus a = 40, b = 60, c = 100: The student must study 40 hours for the first exam, 60 hours for the second, and 100 hours for the third. 37. Using Definition 2.2.2 as a guide, we find that ref V x = 2(projV x) − x = 2(u1 · x)u1 + 2(u2 · x)u2 − x.
1 38. Since v1 and v2 are unit vectors, the condition v1 · v2 = v1 v2 cos(α) = cos(α) = 2 π ◦ implies that v1 and v2 enclose an angle of 60 = 3 . The vectors v1 and v3 enclose an angle of 60◦ as well.
In the case n = 2 there are two possible scenarios: either v2 = v3 , or v2 and v3 enclose 1 an angle of 120◦ . Therefore, either v2 · v3 = 1 or v2 · v3 = cos(120◦ ) = − 2 . In the case n = 3, the vectors v2 and v3 could enclose any angle between 0◦ (if v2 = v3 ) 1 and 120◦ , as illustrated in Figure 5.7. We have − 2 ≤ v2 · v3 ≤ 1.
Figure 5.7: for Problem 5.1.38.
243
Chapter 5 0 0 √ For example, consider v1 = 0 , v2 = 23 , v3 = 1 1
2 1 4 √ 3 2 √ 3 2 1 2
ISM: Linear Algebra cos θ
Note that v2 · v3 = 3 sin θ + 4 (when sin θ = 1), as claimed.
1 could be anything between − 2 (when sin θ = −1) and 1
sin θ
If n exceeds three, we can consider the orthogonal projection w of v3 onto the plane E spanned by v1 and v2 .
1 Since projv1 w = (v1 · w)v1 = 2 v1 , and since w ≤ v3 = 1, (by Fact 5.1.10), the tip of w will be on the line segment in Figure 5.8. Note that θ is between 0◦ and 120◦ , so that 1 cos θ is between − 2 and 1.
Therefore, v2 · v3 = v2 · w = w cos θ is between − 1 and 1. 2 This implies that ∠(v2 , v3 ) is between 0◦ and 120◦ as well. To see that all these values are attained, add (n − 3) zeros to the three vectors v1 , v2 , v3 in R3 given above.
Figure 5.8: for Problem 5.1.38. 39. No! By definition of a projection, the vector x − projL x is perpendicular to projL x, so that (x − projL x) · (projL x) = x · projL x − projL x 2 = 0 and x · projL x = projL x 2 ≥ 0. (See Figure 5.9.)
Figure 5.9: for Problem 5.1.39. 244
ISM: Linear Algebra 40. ||v2 || = √ √ v2 · v2 = a22 = 3. √ √ a11 + 2a12 + a22 = 22.
Section 5.2
a√ 20 41. θ =arccos( ||vv2 ·v3 3 || ) =arccos( √a2223 a33 ) =arccos( 21 ) ≈ 0.31 radians. 2 ||||v
42. ||v1 + v2 || = 43. Let u =
v2 ||v2 ||
(v1 + v2 ) · (v1 + v2 ) = =
v2 3 .
Then, u is an orthonormal basis for span(v2 ). Using Fact 5.1.5,
20 49 v3 .
1 2 2 2 projv2 (v1 ) = (u · v1 )u = ( v3 · v1 ) v2 = 1 (v2 · v1 ) v3 = 3 (a12 ) v3 = 5 v2 . 3 3 9
44. One method to solve this is to take v = v2 − projv3 v2 = v2 −
45. Write the projection as a linear combination of v2 and v3 , c2 v2 + c3 v3 . Now you want v1 − c2 v2 − c3 v3 to be perpendicular to V , that is, perpendicular to both v2 and v3 . Using dot products, this boils down to two linear equation in two unknowns, 9c2 + 20c3 = 5, 1 and 20c2 + 49c3 = 11, with the solution c2 = 25 and c3 = − 41 . Thus the answer is 41 1 25 41 v2 − 41 v3 . 46. Write the projection as a linear combination of v1 and v2 : c1 v1 + c2 v2 . Now we want v3 − c1 v1 + c2 v2 to be perpendicular to V , that is, perpendicular to both v1 and v2 . Using dot products, this boils down to two linear equations in two unknowns, 11 = 3c1 + 5c2 1 5 and 20 = 5c1 + 9c2 , with the solution c1 = − 1 , c2 = 5 . Thus, the answer is − 2 v1 + 2 v2 . 2 2
5.2
In Exercises 1–14, we will refer to the given vectors as v1 , . . . , vm , where m = 1, 2, or 3.
1. u1 =
1 v1
2. u1 =
1 v1
2 1 v1 = 3 1 −2 6 1 v1 = 7 3 2 =
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
u2 =
⊥ v2 ⊥ v2
Note that u1 · v2 = 0. 4 1 1 3. u1 = v1 v1 = 5 0 3
2 1 = 7 −6 3
245
Chapter 5 3 1 = 5 0 −4
ISM: Linear Algebra
u2 =
⊥ v2 ⊥ v2
=
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
4 4. u1 = 1 0 and u2 = 5 3
3 1 0 as in Exercise 3. 5 −4
1 v3
Since v3 is orthogonal to u1 and u2 , u3 = 2 v1 = 1 2 3 1 =
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
0 v3 = −1 . 0
5. u1 =
1 v1
u2 =
⊥ v2 ⊥ v2
6. u1 =
1 v1
1 v1 = 0 = e 1 0 =
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
−1 = √1 −1 = 18 4
−1 1 √ −1 3 2 4
u2 =
⊥ v2 ⊥ v2
0 = 1 = e2 0
u3 =
⊥ v3 ⊥ v3
=
v3 −(u1 ·v3 )u1 −(u2 ·v3 )u2 v3 −(u1 ·v3 )u1 −(u2 ·v3 )u2
2 1 7. Note that v1 and v2 are orthogonal, so that u1 = v1 v1 = 1 2 and u2 = 3 1 −2 2 1 ⊥ v3 v3 −(u1 ·v3 )u1 −(u2 ·v3 )u2 1 1 . Then u3 = v⊥ = v3 −(u1 ·v3 )u1 −(u2 ·v3 )u2 = √1 −4 = 1 −2 . 3 3 36 3 2 4 2 8. u1 =
1 v1
0 = 0 = e3 1
1 v2
v2 =
5 4 v1 = 1 7 2 2
246
ISM: Linear Algebra −2 1 2 v2 = 7 5 −4 1 1 1 v1 = 2 1 1 =
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
Section 5.2
u2 =
1 v2
9. u1 =
1 v1
u2 =
⊥ v2 ⊥ v2
−1 1 7 = 10 −7 1
10. u1 =
1 v1
1 1 1 v1 = 2 1 1 =
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
u2 =
⊥ v2 ⊥ v2
1 1 −1 = 2 1 −1
11. u1 =
1 v1
4 1 0 v1 = 5 0 3 =
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
u2 =
⊥ v2 ⊥ v2
−3 1 2 = √225 = 14 4
−3 1 2 15 14 4
2 1 3 12. u1 = 7 0 6 u2 =
⊥ v2 ⊥ v2
=
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
0 1 −2 = 3 2 1 247
Chapter 5 1 1 v1 = 1 2 1 1 =
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
ISM: Linear Algebra
13. u1 =
1 v1
u2 =
⊥ v2 ⊥ v2
=
1 2 −1 2 1 −2 1 2
=
u3 =
⊥ v3 ⊥ v3
=
v3 −(u1 ·v3 )u1 −(u2 ·v3 )u2 v3 −(u1 ·v3 )u1 −(u2 ·v3 )u2
14. u1 =
1 v1
1 7 1 v1 = 10 1 7 =
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
1 2 1 2 −1 2 −1 2
u2 =
⊥ v2 ⊥ v2
−1 1 0 = √2 1 0
u3 =
⊥ v3 ⊥ v3
=
v3 −(u1 ·v3 )u1 −(u2 ·v3 )u2 v3 −(u1 ·v3 )u1 −(u2 ·v3 )u2
0 1 1 = √2 0 −1
In Exercises 15–28, we will use the results of Exercises 1–14 (note that Exercise k, where k = 1, . . . , 14, gives the QR factorization of Exercise (k + 14)). We can set Q = [u 1 . . . um ]; the entries of R are r11 r22 r33 rij 2 15. Q = 1 1 , R = [3] 3 −2 248 = v1 ⊥ = v2 ⊥ = v3
= v2 − (u1 · v2 )u1 = v3 − (u1 · v3 )u1 − (u2 · v3 )u2 = ui · vj , where i < j.
ISM: Linear Algebra 6 2 7 0 16. Q = 1 3 −6 , R = 7 0 7 2 3 4 3 5 5 0,R = 17. Q = 1 0 5 0 35 3 −4 4 3 0 5 5 0 18. Q = 1 0 0 −5 , R = 0 35 0 5 3 −4 0 0 0 2 19. Q =
1 3
Section 5.2
1 1 2 − √ , R = 3 1 √ 2 2 0 4 √ 1 2 2 v2 v3 ] = 0 0 1 3 −2 , R = 0 2 0 3 5 4 6 0 7
1 2 − √2
20. Q = I3 , R = [ v1 2 −2 1 1 21. Q = 3 2 1 2
5 −2 2 7 7 4 22. Q = 1 , R = 7 2 5 0 7 2 −4
0 12 3 −12 0 6
0.5 −0.1 2 4 0.7 0.5 23. Q = , R = 0 10 0.5 −0.7 0.5 0.1 1 1 2 10 1 −1 24. Q = 1 , R = 2 1 1 0 2 1 −1 12 −3 2 1 0 25. Q = 15 , R = 0 14 9 4 5 10 0 15 249
Chapter 5 2
7 3 7 6 7
ISM: Linear Algebra 0
26. Q = 0
−2 7 14 3 2 , R = 0 3 3
1 3
1 1 1 2 1 1 1 −1 1 27. Q = 1 , R = 0 1 −2 2 1 −1 −1 0 0 1 1 1 −1 28. Q = 29. u1 = u2 =
1 10 7 10 1 10 7 10 1 − √2 0 1 √ 2
0
1 √ 2 1 − √2
0 v1 = =
1 5
10 √ 10 , R = 0 2 0 0 0
10 √0 2
1 v1
−3 4 =
1 5
⊥ v2 ⊥ v2
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
4 . (See Figure 5.10.) 3
Figure 5.10: for Problem 5.2.29. 30. See Figure 5.11.
1 v1
31. u1 =
1 v1 = 0 = e 1 0 250
ISM: Linear Algebra
Section 5.2
Figure 5.11: for Problem 5.2.30. b b 0 ⊥ v2 = v2 − projV1 v2 = c − 0 = c , so that u2 = 0 0 0 0 = 1 = e2 0 0 = 0 = e3 . 1
⊥ v2 ⊥ v2
Here V1 = span(e1 ) = x axis. d d 0 ⊥ v3 = v3 − projV2 v3 = e − e = 0 , so that u3 = f 0 f Here V2 = span(e1 , e2 ) = x-y plane. (See Figure 5.12.)
⊥ v3 ⊥ v3
Figure 5.12: for Problem 5.2.31. −1 −1 32. A basis of the plane is v1 = 1 , v2 = 0 . 0 1 251
Chapter 5 Now apply the Gram-Schmidt process. u1 =
1 v1
ISM: Linear Algebra
v1
u2
=
⊥ v2 ⊥ v2
=
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
−1 1 = √2 1 0 −1 1 = √6 −1 2
Your solution may be different if you start with a different basis v1 , v2 of the plane. 33. rref(A) = 1 0 0 1 0 1 1 0
−1 0 0 −1 A basis of ker(A) is v1 = , v2 = . 0 1 1 0 −1 1 0 Since v1 and v2 are orthogonal already, we obtain u1 = √2 , u 2 = 0 1 34. rref(A) = 1 0 −1 −2 0 1 2 3 0 1 −1 √ . 2 1 0
1 2 −2 −3 A basis of ker(A) is v1 = , v2 = . 1 0 0 1 We apply the Gram-Schmidt process and obtain 1 1 −2 = √6 1 0 2 −1 = √1 30 −4 3
u1
=
1 v1
v1
u2
=
⊥ v2 ⊥ v2
=
v2 −(u1 ·v2 )u1 v2 −(u1 ·v2 )u1
252
ISM: Linear Algebra 1 0 0 1 35. rref(A) = 0 0
1 3 1 3
Section 5.2
0
The non-redundant columns of A give us a basis of im(A): 2 1 v1 = 2 , v2 = 1 −2 2
1 1 1 1 −1 −1 36. Write M = 1 2 1 −1 1 1 1 −1 ↑ Q0
1 Since v1 and v2 are orthogonal already, we obtain u1 = 1 2 , u2 = 3 2 2 3 5 0 −4 6 0 0 7 ↑ R0
2 1 1 . 3 −2
This is almost the QR factorization of M : the matrix Q0 has orthonormal columns and R0 is upper triangular; the only problem is the entry −4 on the diagonal of R0 . Keeping in mind how matrices are multiplied, we can change all the signs in the second column of Q0 and in the second row of R0 to fix this problem: 1 −1 1 2 3 5 1 −1 1 0 4 −6 M= 1 2 1 1 −1 0 0 7 1 −1 1 ↑ Q ↑ R 3 0 0 0 4 5 0 0 ↑ R0
1 1 1 1 1 1 −1 −1 37. Write M = 1 2 1 −1 1 −1 1 1 −1 −1 ↑ Q0
Note that the last two columns of Q0 and the last two rows of R0 have no effect on the product Q0 R0 ; if we drop them, we have the QR factorization of M : 253
Chapter 5 1 1 1 1 −1 M= 2 1 −1 1 1 ↑ Q
ISM: Linear Algebra
3 4 0 5 ↑ R
38. Since v1 = 2e3 , v2 = −3e1 and v3 = 4e4 are orthogonal, we have 0 −1 0 v1 0 0 0 0 v2 v3 v = v2 Q = v1 and R = 0 v2 v3 1 1 0 0 0 0 0 0 1 1 1 −5 1 39. u1 = √1 2 , u2 = √3 1 , u3 = u1 × u2 = √1 4 14 42 3 −1 −1 40. If v1 , . . . , vn are the columns of A, then Q =
v1 v1
0 2 0 0 0 = 0 3 0. v3 0 0 4
···
vn vn
(See Exercise 38 as an example.)
and R =
v1 .. 0 .
0 vn
.
41. If all diagonal entries of A are positive, then we have Q = In and R = A. A small modification is necessary if A has negative entries on the diagonal: if aii < 0 we let rij = −aij for all j, and we let qii = −1; if aii > 0 we let rij = aij and qii = 1. Furthermore, qij = 0 if i = j (that is, Q is diagonal). −1 2 3 −1 0 0 1 −2 −3 For example, 0 4 5 = 0 1 0 0 4 5 0 0 −6 0 0 −1 0 0 6 ↑ A ↑ Q ↑ R
⊥ 42. We have r11 = v1 and r22 = v2 = v2 − projL v2 , so that r11 r22 is the area of the parallelogram defined by v1 and v2 . See Figure 5.13.
43. Partition the matrices Q and R in the QR factorization of A as follows: [ A1 A2 ] = A = QR = [ Q1 Q2 ] R1 0 R2 R3 = [ Q 1 R1 Q1 R2 + Q 2 R3 ] ,
where Q1 is n × m1 , Q2 is n × m2 , R1 is m1 × m1 , and R3 is m2 × m2 . 254
ISM: Linear Algebra
Section 5.3
Figure 5.13: for Problem 5.2.42. Then, A1 = Q1 R1 is the QR factorization of A1 : note that the columns of A1 are orthonormal, and R1 is upper triangular with positive diagonal entries. 44. No! If m exceeds n, then there is no n × m matrix Q with orthonormal columns (if the columns of a matrix are orthonormal, then they are linearly independent). 45. Yes. Let A = [ v1 · · · vm ]. The idea is to perform the Gram-Schmidt process in reversed order, starting with um = v1 vm . m Then we can express vj as a linear combination of uj , . . . , um , so that [ v1 [ u1 · · · uj · · · um ] L for some lower triangular matrix L, with l1j ··· · · · um ] ljj = ljj uj + · · · + lmj um . ··· lmj · · · vj · · · vm ] =
vj = [ u1
· · · uj
5.3
1. Not orthogonal, the column vectors fail to be perpendicular to each other. 2. This matrix is orthogonal. Check that the column vectors are unit vectors, and that they are perpendicular to each other. 3. This matrix is orthogonal. Check that the column vectors are unit vectors, and that they are perpendicular to each other. 4. Not orthogonal, the first and third column vectors fail to be perpendicular to each other. 5. 3A will not be orthogonal, because the length of the column vectors will be 3 instead of 1, and they will fail to be unit vectors. 6. −B will certainly be orthogonal, since the columns will be perpendicular unit vectors. 7. AB is orthogonal by Fact 5.3.4a. 255
Chapter 5
ISM: Linear Algebra
8. A + B will not necessarily be orthogonal, because the columns may not be unit vectors. For example, if A = B = In , then A + B = 2In , which is not orthogonal. 9. B −1 is orthogonal by Fact 5.3.4b. 10. This matrix will be orthogonal, by Fact 5.3.4. 11. AT is orthogonal. AT = A−1 , by Fact 5.3.7, and A−1 is orthogonal by Fact 5.3.4b. 13. 3A is symmetric, since (3A)T = 3AT = 3A. 14. −B is symmetric, since (−B)T = −B T = −B. 15. AB is not necessarily symmetric, since (AB)T = B T AT = BA, which is not necessarily the same as AB. (Here we used Fact 5.3.9a.) 16. A + B is symmetric, since (A + B)T = AT + B T = A + B. 17. B −1 is symmetric, because (B −1 )T = (B T )−1 = B −1 . In the first step we have used 5.3.9b. 18. A10 is symmetric, since (A10 )T = (AT )10 = A10 . 19. This matrix is symmetric. First note that (A2 )T = (AT )2 = A2 for a symmetric matrix A. T Now we can use the linearity of the transpose, (2In +3A−4A2 )T = 2In +3AT −(4A2 )T = T 2 2 2In + 3A − 4(A ) = 2In + 3A − 4A . 20. AB 2 A is symmetric, since (AB 2 A)T = (ABBA)T = (BA)T (AB)T = AT B T B T AT = AB 2 A. 21. Symmetric. (AT A)T = AT (AT )T = AT A. 22. BB T is symmetric: (BB T )T = (B T )T B T = BB T . 23. Not necessarily symmetric. (A − AT )T = AT − A = −(A − AT ). 24. Not necessarily symmetric. (AT BA)T = AT (AT B)T = AT B T A. 25. Symmetric, because (AT B T BA)T = AT B T (B T )T (AT )T = AT B T BA. 26. Symmetric, since (B(A + AT )B T )T = ((A + AT )B T )T B T = B(A + AT )T B T = B(AT + A)T B T = B((AT )T + AT )B T = B(A + AT )B T . 27. Using Facts 5.3.6 and 5.3.9a, we find that (Av) · w = (Av)T w = v T AT w = v · (AT w), as claimed. 28. Write L(x) = Ax; by Definition 5.3.1, A is an orthogonal n×n matrix, so that A T A = In , by Fact 5.3.7. Now L(v) · L(w) = (Av) · (Aw) = (Av)T Aw = v T AT Aw = v T In w = v T w = v · w, as claimed. Note that we have used Facts 5.3.6 and 5.3.9a. 256
ISM: Linear Algebra
Section 5.3
29. We will use the fact that L preserves length (by Definition 5.3.1) and the dot product (by Exercise 28): ∠(L(v), L(w)) = arccos
L(v)·L(w) L(v) L(w)
= arccos
v·w v w
= ∠(v, w).
30. If L(x) = 0, then L(x) = x = 0, so that x = 0. Therefore, ker(L) = {0}. By Fact 3.3.7, dim(im(L)) = m − dim(ker(L)) = m. Since Rn has an m-dimensional subspace (namely, im(L)), the inequality m ≤ n holds. The transformation L preserves right angles (the proof of Fact 5.3.2 applies), so that the columns of A are orthonormal (since they are L(e1 ), . . . , L(em )). Therefore, we have AT A = Im (the proof of Fact 5.3.7 applies). Since the vectors v1 , . . . , vm form an orthonormal basis of im(A), the matrix AAT represents the orthogonal projection onto im(A), by Fact 5.3.10. 1 0 x1 x1 A simple example of such a transformation is L(x) = 0 1 x, that is, L = x2 . x2 0 0 0
31. Yes! If A is orthogonal, then so is AT , by Exercise 11. Since the columns of AT are orthogonal, so are the rows of A.
1 0 32. a. No! As a counterexample, consider A = 0 1 (see Exercise 30). 0 0
b. Yes! More generally, if A and B are n × n matrices such that BA = In , then AB = In , by Fact 2.4.9c.
33. Write A = [ v1
cos(φ) , for some sin(φ) − sin(φ) or φ. Then v2 will be one of the two unit vectors orthogonal to v1 : v2 = cos(φ) sin(φ) . (See Figure 5.7.) v2 = − cos(φ) v2 ]. The unit vector v1 can be expressed as v = 257
Chapter 5
ISM: Linear Algebra
Figure 5.14: for Problem 5.3.33. cos(φ) sin(φ) − sin(φ) cos(φ)
Therefore, an orthogonal 2 × 2 matrix is either of the form A = A=
or
cos(φ) sin(φ) , representing a rotation or a reflection. Compare with Exercise sin(φ) − cos(φ) 2.2.24. See Figure 5.14. a b e f is an orthogonal 2 × 2 matrix; By Exercise 33, the 3 × 3 matrix A is either of the form cos(φ) − sin(φ) 0 cos(φ) sin(φ) 0 A= 0 0 1 or A = 0 0 1 . sin(φ) cos(φ) 0 sin(φ) − cos(φ) 0 2
3 1 3
34. Since the first two columns are orthogonal to the third, we have c = d = 0. Then
35. Let us first think about the inverse L = T −1 of T . Write L(x) = Ax = [ v1 v2
Furthermore, vectors v1 , v2 , v3 must form an orthonormal basis of R3 . By inspection, the −2 3 we find v1 = 1 . 3
2 3
v3 ] x. It is required that L(e3 ) = v3 = 2 . 3
258
ISM: Linear Algebra
1 −3 3
Section 5.3
Since the matrix of L is orthogonal, the matrix of T = L−1 is the transpose of the matrix of L: −2 1 2 1 T (x) = 3 −1 2 −2 x. 2 2 1 There are many other answers (since there are many choices for the vector v 1 above). 2 1 1
3 1 3 √ 2 1 − √2 √ 18 √1 18 − √4 18
−2 −1 2 2 2 x. Then v2 = v1 ×v3 = 2 does the job. In summary, we have L(x) = 1 1 3 3 2 2 −2 1 −
36. Let the third column be the cross product of the first two: A = 2 3
0
There is another solution, with the signs in the last column reversed. 2 −3 3 2 37. No, since the vectors 3 and 2 are orthogonal, whereas 0 and −3 are not 0 0 2 0 (see Fact 5.3.2).
.
0 38. a. The general form of a skew-symmetric 3 × 3 matrix is A = −a −b 2 −a − b2 −bc ac A2 = −bc −a2 − c2 −ab , a symmetric matrix. ac −ab −b2 − c2
a b 0 c , with −c 0
b. By Fact 5.3.9.a, (A2 )T = (AT )2 = (−A)2 = A2 , so that A2 is symmetric.
39. By Fact 5.3.10, the matrix of the projection is uu T ; the ij th entry of this matrix is ui uj . 0.5 −0.1 0.5 0.7 40. An orthonormal basis of W is u1 = , u2 = (see Exercise 5.2.9). 0.5 −0.7 0.5 0.1 By Fact 5.3.10, the matrix of the projection onto W is QQT , where Q = [ u1 259 u2 ].
Chapter 5 26 18 32 24 74 −24 32 1 18 QQT = 100 32 −24 74 18 24 32 18 26
ISM: Linear Algebra
1 . 1 √ . . 41. A unit vector on the line is u = n . 1
The matrix of the orthogonal projection is uuT , the n × n matrix whose entries are all (compare with Exercise 39).
1 n
42. a. Suppose we are projecting onto a subspace W of Rn . Since Ax is in W already, the orthogonal projection of Ax onto W is just Ax itself: A(Ax) = Ax, or A2 x = Ax. Since this equation holds for all x, we have A2 = A. b. A = QQT , for some matrix Q with orthonormal columns u1 , . . . , um . Note that QT Q = Im , since the ij th entry of QT Q is ui · uj . Then A2 = QQT QQT = Q(QT Q)QT = QIm QT = QQT = A.
43. Examine how A acts on u, and on a vector v orthogonal to u: Au = (2uuT − I3 )u = 2uuT u − u = u, since uT u = u · u = u Av = (2uuT − I3 )v = 2uuT v − v = −v, since uT v = u · v = 0. Since A leaves the vectors in L = span(u) unchanged and reverses the vectors in V = L ⊥ , it represents the reflection about L. Note that B = −A, so that B reverses the vectors in L and leaves the vectors in V unchanged; that is, B represents the reflection about V . 44. Note that AT is an m × n matrix. By Facts 3.3.7 and 5.3.9c we have dim(ker(AT )) = n − rank(AT ) = n − rank(A). By Fact 3.3.6, dim(im(A)) = rank(A), so that dim(im(A)) + dim(ker(AT )) = n. 45. Note that AT is an m × n matrix. By Facts 3.3.7 and 5.3.9c, we have dim(ker(A)) = m − rank(A) and dim(ker(AT )) = n − rank(AT ) = n − rank(A), so that dim(ker(A)) = dim(ker(AT )) if (and only if) A is a square matrix. 260
2
= 1.
ISM: Linear Algebra
Section 5.3
46. By Fact 5.2.2, the columns u1 , . . . , um of Q are orthonormal. Therefore, QT Q = Im , since the ij th entry of QT Q is ui · uj . If we multiply the equation M = QR by QT from the left then QT M = QT QR = R, as claimed. 47. By Fact 5.2.2, the columns u1 , . . . , um of Q are orthonormal. Therefore, QT Q = Im , since the ij th entry of QT Q is ui · uj . By Fact 5.3.9a, we now have AT A = (QR)T QR = RT QT QR = RT R. 48. As suggested, we consider the QR factorization AT = P R of AT , where P is orthogonal and R is upper triangular with positive diagonal entries. By Fact 5.3.9a, A = (P R)T = RT P T . Note that L = RT is lower triangular and Q = P T is orthogonal. 49. Yes! By Exercise 5.2.45, we can write AT = P L, where P is orthogonal and L is lower triangular. By Fact 5.3.9a, A = (P L)T = LT P T . Note that R = LT is upper triangular, and Q = P T is orthogonal (by Exercise 11).
50. a. If an n × n matrix A is orthogonal and upper triangular, then A−1 is both lower triangular (since A−1 = AT ) and upper triangular (being the inverse of an upper triangular matrix; compare with Exercise 2.3.35c). Therefore, A−1 = AT is a diagonal matrix, and so is A itself. Since A is orthogonal with positive diagonal entries, all the diagonal entries must be 1, so that A = In . b. Using the terminology suggested in the hint, we observe that Q−1 Q1 is orthogonal (by 2 −1 Fact 5.3.4) and R2 R1 is upper triangular with positive diagonal entries. By part a, −1 the matrix Q−1 Q1 = R2 R1 is In , so that Q1 = Q2 and R1 = R2 , as claimed. 2 51. a. Using the terminology suggested in the hint, we observe that Im = QT Q1 = (Q2 S)T Q2 S = 1 S T QT Q2 S = S T S, so that S is orthogonal, by Fact 5.3.7. 2
−1 b. Using the terminology suggested in the hint, we observe that R2 R1 is both orthogonal −1 (let S = R2 R1 in part a) and upper triangular, with positive diagonal entries. By −1 −1 Exercise 50a, we have R2 R1 = Im , so that R1 = R2 . Then Q1 = Q2 R2 R1 = Q2 , as claimed.
261
Chapter 5
ISM: Linear Algebra
53. Applying the strategy outlined in Summary 0 1 0 0 of V , we find the basis −1 0 0 , 0 0 0 0 −1
a 52. Applying the strategy outlined in Summary 4.1.6 to the general element b c 1 0 0 0 1 0 0 0 1 0 0 0 0 0 we find the basis 0 0 0 , 1 0 0 , 0 0 0 , 0 1 0 · 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 so that dim(V ) = 6.
b c d e of V , e f 0 0 0 0 1 , 0 0 0 , 0 0 0 1
54. To write the general form of a skew-symmetric n × n matrix A, we can place arbitrary constants above the diagonal, the opposite entries below the diagonal (aij = −aji ), and zeros on the diagonal (since aii = −aii ). See Exercise 53 for the case n = 3. Thus the dimension of the space equals the number of entries above the diagonal of an n×n matrix. In Exercise 55 we will see that there are (n2 −n)/2 such entries. Thus dim(V ) = (n2 −n)/2. 55. To write the general form of a symmetric n×n matrix A, we can place arbitrary constants on and above the diagonal, and then write the corresponding entries below the diagonal (aij = aji ). See Exercise 52 for the case n = 3. Thus the dimension of the space equals the number of entries on and above the diagonal of an n × n matrix. Now there are n2 entries in the matrix, n2 − n off the diagonal, and half of them, (n2 − n)/2, above the diagonal. Since there are n entries on the diagonal, we have dim(V ) = (n2 − n)/2 + n = (n2 + n)/2. 56. Yes and yes (see Exercise 57). 57. Yes, L is linear, since L(A + B) = (A + B)T = AT + B T = L(A) + L(B) and L(kA) = (kA)T = kAT = kL(A). Yes, L is an isomorphism; the inverse is the transformation R(A) = AT from Rn×m to Rm×n . 58. Adapting the solution of Exercise 59, we see that the kernel consists of all skew-symmetric matrices, and the image consists of all symmetric matrices. 59. The kernel consists of all matrixes A such that L(A) = 1 (A − AT ) = 0, that is, AT = A; 2 those are the symmetric matrices. Following the hint, let’s apply L to a skew-symmetric matrix A, with AT = −A. Then L(A) = (1/2)(A − AT ) = (1/2)2A = A, so that A is in the image of L. Conversely, if A is any 2 × 2 matrix, then L(A) will be skew-symmetric, since (L(A))T = (1/2)(A − AT )T = (1/2)(AT − A) = −L(A). 262
0 b c 0 e 4.1.6 to the general element −b −c −e 0 0 1 0 0 0 0 0,0 0 1 , so that dim(V ) = 3. 0 0 0 −1 0
ISM: Linear Algebra In conclusion: The kernel of L consists of of all skew-symmetric matrices. 1 0 60. Using Fact 4.3.2, we find the matrix 0 0
Section 5.3 all symmetric matrices, and the image consists 0 1 0 0 0 0 0 0 . 1 0 0 −1
61. Note that the first three matrices of the given basis B are symmetric, so that L(A) = A − AT = 0, and the coordinate vector [L(A)]B is 0 for all three of them. The last matrix of the basis is skew-symmetric, so that L(A) = 2A, and [L(A)]B = 2e4 . Using Fact 4.3.2, 0 0 0 0 0 0 0 0 we find that the B-matrix of L is . 0 0 0 0 0 0 0 2 1 3 2 1 0 0 1 0 0 62. By Fact 5.3.9a, AT = 1 1 0 0 −1 0 0 1 0 . 0 0 1 0 0 2 −1 2 1 63. By Exercise 2.4.62b, the given LDU factorization of A is unique. By Fact 5.3.9a, A = AT = (LDU )T = U T DT LT = U T DLT is another way to write the LDU factorization of A (since U T is lower triangular and LT is upper triangular). By the uniqueness of the LDU factorization, we have U = LT (and L = U T ), as claimed.
64. a.
A −B T C + B AT D the required form. A B −B T AT =
−DT CT kA kB
=
A+C B+D kA kB
−B T − DT AT + C T −(kB)T (kA)T
=
A+C B+D
−(B + D)T (A + C)T
is of
b. k
−kB T kAT
=
is of the required form.
p −q −r −s p s −r q c. The general element of H is M = , with four arbitrary constants, r −s p q s r −q p r,s,p, and q. Thus dim(H) = 4; use the strategy outlined in Summary 4.1.6 to construct a basis. d. C −DT A −B T T D CT B A is of the required form. = AC − B T D BC + AT D −ADT − B T C T −BDT + AT C T = AC − B T D BC + AT D −(BC + AT D)T (AC − B T D)T
263
Chapter 5
ISM: Linear Algebra
Note that A, B, C, D, and their transposes are rotation-dilation matrices, so that they all commute. e. A B −B T AT
T
=
AT −B
BT (AT )T
is of the required form.
f. Note that the columns v1 , v2 , v3 , v4 or M are orthogonal, and they all have length p2 + q 2 + r2 + s2 . Now M T M is the 4 × 4 matrix whose ij th entry is vi · vj , so that M T M = (p2 + q 2 + r2 + s2 )I4 . g. If M = 0, then k = p2 + q 2 + r2 + s2 > 0, and with M −1 =
1 T p2 +q 2 +r 2 +s2 M . 1 T kM
M = I4 , so that M is invertible,
By parts b and e, M −1 is 0 −1 0 0 0 1 h. No! A = 0 0 0 0 0 −1 −BA).
in H as well. 0 0 0 −1 0 0 0 −1 0 0 and B = do not commute (AB = 1 1 0 0 0 0 0 1 0 0
65. Write 10A =
b a a b ; it is required that a, b, c and d be integers. Now A = 10 10 c d c d 10 10 c a must be an orthogonal matrix, implying that ( 10 )2 +( 10 )2 = 1, or a2 +c2 = 100. Checking the squares of all integers from 1 to 9, we see that there are only two ways to write 100 as a sum of two positive perfect squares: 100 = 36 + 64 = 64 + 36. Since a and c are required to be positive, we have either a = 6 and c = 8 or a = 8 and c = 6. In each case we have two options for the second column of A, namely, the two unit vectors perpendicular to the first column vector. Thus we end up with four solutions:
A=
.6 −.8 .6 .8 .8 −.6 , , .8 .6 .8 −.6 .6 .8
or
.8 .6 . .6 −.8
66. One approach is to take one of the solutions from Exercise 65, say, the rotation matrix 0.8 −0.6 0.28 −0.96 B= , and then let A = B 2 = . Matrix A is orthogonal by 0.6 0.8 0.96 0.28 Fact 5.3.4a.
67. a. We need to show that AT Ac = AT x, or, equivalently, that AT (x − Ac) = 0. But AT (x − Ac) = AT (x − c1 v1 − · · · − cm vm ) is the vector whose ith component is 264
ISM: Linear Algebra
Section 5.4
(vi )T (x − c1 v1 − · · · − cm vm ) = vi · (x − c1 v1 − · · · − cm vm ), which we know to be zero. b. The system AT Ac = AT x has a unique solution c for a given x, since c is the coordinate vector of projV x with respect to the basis v1 , . . . , vm . Thus the coefficient matrix AT A must be invertible, so that we can solve for c and write c = (AT A)−1 AT x. Then projV x = c1 v1 + · · · + cm vm = Ac = A(AT A)−1 AT x. 68. If A = QR, then A(AT A)−1 AT = QR(RT QT QR)−1 RT QT = QR(RT R)−1 RT QT = QRR−1 (RT )−1 RT QT = QQT , as in Fact 5.3.10. The equation QT Q = Im holds since the columns of Q are orthonormal.
5.4
1. A basis of ker(AT ) is −3 . (See Figure 5.15.) 2
Figure 5.15: for Problem 5.4.1. 1 2. A basis of ker(AT ) is −2 . im(A) is the plane perpendicular to this line. 1 265
Chapter 5
ISM: Linear Algebra
3. We will first show that the vectors v1 , . . . , vp , w1 , . . . , wq span Rn . Any vector v in Rn can be written as v = v + v⊥ , where v is in V and v⊥ is in V ⊥ (by definition of a projection, Fact 5.1.4). Now v is a linear combination of v1 , . . . , vp , and v⊥ is a linear combination of w1 , . . . , wq , showing that the vectors v1 , . . . , vp , w1 , . . . , wq span Rn . Note that p + q = n, by Fact 5.1.8c; therefore, the vectors v1 , . . . , vp , w1 , . . . , wq form a basis of Rn , by Fact 3.3.4d. 4. By Fact 5.4.1, the equation (im B)⊥ = ker(B T ) holds for any matrix B. Now let B = AT . Then (im(AT ))⊥ = ker(A). Taking transposes of both sides and using Fact 5.1.8d we obtain im(AT ) = (kerA)⊥ , as claimed. 5. V = ker(A), where A = 1 1 1 1 . 1 2 5 4
Then V ⊥ = (kerA)⊥ = im(AT ), by Exercise 4. The two columns of AT form a basis of V ⊥ : 1 1 1 2 , 1 5 1 4 im(A)
6. Yes! For any matrix A, = (ker(AT ))⊥ ↑ = (ker(AAT ))⊥ ↑ = (ker(AAT )T )⊥ = im(AAT ). ↑
Fact 5.4.1
Fact 5.4.2a
Facts 5.4.1 and 5.1.8d.
7. im(A) and ker(A) are orthogonal complements by Fact 5.4.1: (imA)⊥ = ker(AT ) = ker(A)
8. a. By Fact 5.4.6, L+ (y) = (AT A)−1 AT y. The transformation L+ is linear since it is “given by a matrix,” by Definition 2.1.1. b. If L (and therefore A) is invertible, then L+ (y) = A−1 (AT )−1 AT y = A−1 y = L−1 y, so that L+ = L−1 . c. L+ (L(x)) = the unique least-squares solution u of L(u) = L(x) = x. 266
ISM: Linear Algebra d. L(L+ (y)) = A(AT A)−1 AT y = projV y, where V = im(A), by Fact 5.4.7. 1 0 1 0 0 e. Here A = 0 1 . Then L+ (y) = (AT A)−1 AT y = y. 0 1 0 0 0
Section 5.4
Figure 5.16: for Problem 5.4.9. 9. x0 is the shortest of all the vectors in S. (See Figure 5.16.)
10. a. If x is an arbitrary solution of the system Ax = b, let xh = projV x, where V = ker(A), and x0 = x − projV x. Note that b = Ax = A(xh + x0 ) = Axh + Ax0 = Ax0 , since xh is in ker(A). b. If x0 and x1 are two solutions of the system Ax = b, both from (kerA)⊥ , then x1 − x0 is in the subspace (kerA)⊥ as well. Also, A(x1 − x0 ) = Ax1 − Ax0 = b − b = 0, so that x1 − x0 is in ker(A). By Fact 5.1.8b, it follows that x1 − x0 = 0, or x1 = x0 , as claimed. c. Write x1 = xh + x0 as in part a; note that xh is orthogonal to x0 . The claim now follows from the Pythagorean Theorem (Fact 5.1.9). 11. a. Note that L+ (y) = AT (AAT )−1 y; indeed, this vector is in im(AT ) = (kerA)⊥ , and it is a solution of L(x) = Ax = y. 267
Chapter 5 b. L(L+ (y)) = y, by definition of L+ .
ISM: Linear Algebra
c. L+ (L(x)) = AT (AAT )−1 Ax = projV x, where V = im(AT ) = (kerA)⊥ , by Fact 5.4.7. d. im(L+ ) = im(AT ), by part c, and ker(L+ ) = {0} (if y is in ker(L+ ), then y = L(L+ (y)) = L(0) = 0, by part b). 1 0 1 0 0 e. Let A = ; then the matrix of L+ is AT (AAT )−1 = AT = 0 1 . 0 1 0 0 0 12. By Fact 5.4.5, the least-squares solutions of the linear system Ax = b are the exact solutions of the (consistent) system AT Ax = AT b. The minimal solution of this normal equation (in the sense of Exercise 10) is called the minimal least-squares solution of the system Ax = b. Equivalently, the minimal least-squares solution of Ax = b can be defined as the minimal solution of the consistent system Ax = projV b, where V = im(A).
13. a. Suppose that L+ (y1 ) = x1 and L+ (y2 ) = x2 ; this means that x1 and x2 are both in (kerA)⊥ = im(AT ), AT Ax1 = AT y1 , and AT Ax2 = AT y2 . Then x1 + x2 is in im(AT ) as well, and AT A(x1 + x2 ) = AT (y1 + y2 ), so that L+ (y1 + y2 ) = x1 + x2 . The verification of the property L+ (ky) = kL+ (y) is analogous. b. L+ (L(x)) is the orthogonal projection of x onto (kerA)⊥ = im(AT ). c. L(L+ (y)) is the orthogonal projection of y onto im(A) = (kerAT ))⊥ . d. im(L+ ) = im(AT ) and ker(L+ ) = ker(AT ), by parts 1 y1 2 2 y1 = 0 , so that the matrix of L+ is 0 e. L+ y2 0 0 b and c. 0 0 . 0
14. L+ (w1 ) is the minimal solution of the system L(x) = w1 . The line S in Figure 5.17 shows all solutions of the system L(x) = w1 (compare with Exercise 9). The minimal solution, L+ (w1 ), is perpendicular to ker(L). L+ (w2 ) = L+ (projim(L) w2 ) = L+ (0) = 0 268
ISM: Linear Algebra
Section 5.4
Figure 5.17: for Problem 5.4.14. L+ (w3 ) = L+ (projim(L) w3 ) ≈ L+ (0.55w1 ) = 0.55L+(w1 ) 15. Note that (AT A)−1 AT A = In ; let B = (AT A)−1 AT . 16. If A is an m × n matrix, then dim(imA)⊥ = m − dim(imA) = m − rank(A) ↑ ↑ Fact 3.3.6
Fact 5.1.8c and dim(ker(AT ))
= m − rank(AT ). ↑
Fact 3.3.7 It follows that rank(A) = rank(AT ), as claimed. 17. Yes! By Fact 5.4.2, ker(A) = ker(AT A). Taking dimensions of both sides and using Fact 3.3.7, we find that n − rank(A) = n − rank(AT A); the claim follows. 18. Yes! By Exercise 17, rank(A) = rank(AT A). Substituting AT for A in Exercise 17 and using Fact 5.3.9c, we find that rank(A) = rank(AT ) = rank(AAT ). The claim follows. 19. x∗ = (AT A)−1 AT b = 1 , by Fact 5.4.6. 1 2 2 −1 and b − Ax∗ = 1 . 1
20. Using Fact 5.4.6, we find x∗ =
Note that b − Ax∗ is perpendicular to the two columns of A. −12 −1 21. Using Fact 5.4.6, we find x∗ = and b − Ax∗ = 36 , so that b − Ax∗ = 42. 2 −18 269
Chapter 5 22. Using Fact 5.4.6, we find x∗ =
ISM: Linear Algebra
3 and b − Ax∗ = 0. This system is in fact consistent −2 and x∗ is the exact solution; the error b − Ax∗ is 0.
23. Using Fact 5.4.6, we find x∗ = 0; here b is perpendicular to im(A). 24. Using Fact 5.4.6, we find x∗ = [2]. 25. In this case, the normal equation AT Ax = AT b is 5 , which sim15 1 − 3t plifies to x1 + 3x2 = 1, or x1 = 1 − 3x2 . The solutions are of the form x∗ = , t where t is an arbitrary constant. 66 78 90 1 x1 26. Here, the normal equation AT Ax = AT b is 78 93 108 x2 = 2 , with solux3 90 108 126 3 7 t− 6 tions x∗ = 1 − 2t , where t is an arbitrary constant. t 5 15 15 45 x1 x2 = 27. The least-squares solutions of the system SAx = S b are the exact solutions of the normal equation (SA)T SAx = (SA)T S b. Note that S T S = In , since S is orthogonal; therefore, the normal equation simplifies as follows: (SA)T SAx = AT S T SAx = AT Ax and (SA)T S b = AT S T S b = AT b, so that the normal equation is AT Ax = AT b, the same as the normal equation of the system Ax = b. Therefore, the systems Ax = b and SAx = S b have the same least-squares 7 . solution, x∗ = 11 28. The least-squares solutions of the system Ax = un are the exact solutions of Ax = projim(A) un . Note that un is orthogonal to im(A), so that projim(A) un = 0, and the unique least-squares solution is x∗ = 0. 29. By Fact 5.4.6, x∗ = (AT A)−1 AT b = where ε = 10−20 . 1+ε 1 1 1+ε
−1
1+ε 1+ε
=
1 2+ε
1+ε 1+ε
≈
1 2 1 2
,
If we use a hand-held calculator, due to roundoff errors we find the normal equation 1 1 1 x1 = , with infinitely many solutions. 1 1 x2 1 270
ISM: Linear Algebra 30. We attempt to solve the system 1 0 c0 + 0c1 = 0 c c0 + 0c1 = 1 , or 1 0 0 c1 1 1 c0 + 1c1 = 1 0 = 1. 1 c∗ 0 c∗ 1 =
Section 5.4
This system cannot be solved exactly; the least-squares solution is line that fits the data points best is f ∗ (t) =
1 2
+ 1 t. 2
1 2 1 2
. The
Figure 5.18: for Problem 5.4.30. The line goes through the point (1, 1) and “splits the difference” between (0, 0) and (0, 1). See Figure 5.18. 31. We want c0 c1 such that 1 0 3 = c0 + 0c1 c 3 = c0 + 1c1 or 1 1 1 c2 1 1 6 = c0 + 1c1 1 0 c Since ker 1 1 = {0}, 0 c1 1 1 = 3 2 2 2
−1
3 = 3. 6
∗
−1 1 0 3 1 1 1 1 1 1 = 1 1 3 0 1 1 0 1 1 1 1 6
12 = 9
3
3 2
3 so f ∗ (t) = 3 + 2 t. (See Figure 5.19.)
c0 32. We want c1 of f (t) = c0 + c1 t + c2 t2 such that c2 271
Chapter 5
ISM: Linear Algebra
Figure 5.19: for Problem 5.4.31. 1 27 = c0 + 0c1 + 0c2 0 = c0 + 1c1 + 1c2 1 or 1 0 = c0 + 2c1 + 4c2 0 = c0 + 3c1 + 9c2 1 0 1 2 3 0 27 c0 1 0 c1 = 4 0 c2 9 0
If we call the coefficient matrix A, we notice that ker(A) = {0} so ∗ 27 c0 25.65 c1 = (AT A)−1 AT 0 = −28.35 so f ∗ (t) = 25.65 − 28.35t + 6.75t2. 0 c2 6.75 0 c0 33. We want c1 such that c2 0 = c0 + sin(0)c1 + cos(0)c2 1 0 1 0 c1 1 = c0 + sin(1)c1 + cos(1)c2 1 sin(1) cos(1) 1 or c2 = . 2 = c0 + sin(2)c1 + cos(2)c2 1 sin(2) cos(2) 2 c3 3 = c0 + sin(3)c1 + cos(3)c2 1 sin(3) cos(3) 3
∗ c0 1.5 c1 ≈ 0.1 so f ∗ (t) ≈ 1.5 + 0.1 sin t − 1.41 cos t. c2 −1.41 c0 c1 34. We want c2 such that c3 c4 272
∗ c1 Since the coefficient matrix has kernel {0}, we compute c2 using Fact 5.4.6, obtaining c3
ISM: Linear Algebra 1 1 1 1 1 1 1 0 0 1 0 1 sin(0.5) cos(0.5) sin(1) cos(1) c0 0.5 sin(1) cos(1) sin(2) cos(2) c1 1 sin(1.5) cos(1.5) sin(3) cos(3) c2 = 1.5 sin(2) cos(2) sin(4) cos(4) c3 2 2.5 c4 sin(2.5) cos(2.5) sin(5) cos(5) 3 sin(3) cos(3) sin(6) cos(6)
Section 5.4
Since the columns of the coefficient matrix are linearly independent, its kernel is {0}. c0 1.5 c1 0.109 We can use Fact 5.4.6 to compute c2 ≈ −1.537 so f ∗ (t) ≈ 1.5 + 0.109 sin(t) − c3 0.303 c4 0.043 1.537 cos(t) + 0.303 sin(2t) + 0.043 cos(2t).
35. a. The ij th entry of AT An is the dot product of the ith row of AT and the jth column n n of An , i.e.
n n sin ai AT An = n i=1 n cos ai
i=1
n
n
sin ai
i=1 n i=1 n
cos ai
sin2 ai
i=1 n
sin ai cos ai
i=1
i=1
i=1 n g(ai ) sin ai . sin ai cos ai and AT b = n i=1 i=1 n n 2 g(ai ) cos ai cos ai
i=1 2π
n
g(ai )
2π 2π 2π T b. lim n An An = sin t dt n→∞ 0 2π cos t dt
0
2π
sin t dt
0 2π 0 2π
0
sin2 t dt sin t cos t dt
0
0
2π 2π sin t cos t dt = 0 0 0 2π 2 cos t dt
cos t dt
0 π 0
0 0 π
273
Chapter 5
2π
ISM: Linear Algebra
2π T and lim n An b = n→∞ ( Here
2π n
0
0
2π g(t) sin t dt 0 2π g(t) cos t dt
n
g(t) dt
n n→∞
2π
= ∆t so lim
2π n→∞ n
cos(ti ) = lim
i=1
cos(ti )∆t =
i=1 0
cos t dt for instance.
1 g(t) dt 2π −1 0 2π c cn 2π 0 0 1 c. p = lim pn = 0 π 0 g(t) sin t dt = π n→∞ 0 q qn 0 0 π 2π 1 g(t) cos t dt π 0 and f (t) = c + p sin t + q cos t, where c, p, q are given above.
All other limits are obtained similarly. )
2π
2π
0
g(t) sin t dt 0 2π g(t) cos t dt
0 2π
g(t) dt
a 36. We want b such that c
2π a + b sin 365 28 2π a + b sin 365 77 2π a + b sin 365 124 2π a + b sin 365 168
+ c cos + c cos + c cos + c cos
2π 365 28 2π 365 77 2π 365 124 2π 365 168
2π 365 28 2π 365 77 2π 365 124 2π 365 168
= 10 = 12 = 14 = 15. ∗ 10 a 12 and b = we compute b = (AT A)−1 AT b ≈ 14 c 15
2π 365 t
cos 1 sin 12.25 0.394 and f ∗ (t) ≈ 12.25 + 0.394 sin −2.726 37. a. We want c0 , c1 such that
1 sin Using A = 1 sin
1
sin
cos cos cos
2π 365 28 2π 365 77 2π 365 124 2π 365 168
− 2.726 cos
2π 365 t
.
274
ISM: Linear Algebra 1 35 c0 + c1 (35) = log 35 c0 + c1 (46) = log 46 1 46 c0 or c1 1 59 c0 + c1 (59) = log 77 1 69 c0 + c1 (69) = log 133 ↑ A so c0 c1
∗
Section 5.4 log 35 log 46 = log 77 log 133 ↑ b so log(d) ≈ 0.915 + 0.017t.
= (AT A)−1 AT b ≈
0.915 0.017
b. d ≈ 100.915 · 100.017t ≈ 8.22 · 100.017t c. If t = 88 then d ≈ 258. Since the Airbus has only 93 displays, new technologies must have rendered the old trends obsolete.
c0 38. We want c1 such that c2 2 1 110 12 0 c0 180 5 1 c1 = 120 . 11 1 c2 160 6 0 160 ∗ 125 c0 5 , so that w∗ = 125 + 5h − 25g. The least-squares solution is c1 = −25 c2 110 = c0 + 2c1 + c2 1 180 = c0 + 12c1 + 0c2 1 120 = c0 + 5c1 + c2 or 1 160 = c0 + 11c1 + c2 1 160 = c0 + 6c1 + 0c2 1
For a general population, we expect c0 and c1 to be positive, since c0 gives the weight of a 5 male, and increased height should contribute positively to the weight. We expect c2 to be negative, since females tend to be lighter than males of equal height.
39. a. We want
c0 c1
such that
log(250) = c0 + c1 log(600, 000) log(60) = c0 + c1 log(200, 000) log(25) = c0 + c1 log(60, 000) 275
Chapter 5 log(12) = c0 + c1 log(10, 000) log(5) = c0 + c1 log(2500) The least-squares solution to the above system is −1.616 + 0.664 log g. c0 c1
∗
ISM: Linear Algebra
≈
−1.616 0.664
so log z ≈
b. Exponentiating both sides of the answer to a, we get z ≈ 10−1.616 · g 0.664 ≈ 0.0242 · g 0.664 . √ c. This model is close since g = g 0.5 .
40. First we look for
c0 c1
such that log D = c0 + c1 log a. c0 c1
∗
Proceeding as in Exercise 39, we get 101.5 log a = a1.5 .
≈
0 , i.e. log D ≈ 1.5 log a, hence D ≈ 1.5
Note that the formula D = a1.5 is Kepler’s third law of planetary motion.
41. a. We want log 370 log 533 log 908 log 1, 823 log 3, 233 log 4, 871
c0 c1
such that log D = c0 + c1 t (t in years since 1970), i.e.
= c0 + c1 (0) = c0 + c1 (5) = c0 + c1 (10) = c0 + c1 (15) = c0 + c1 (20) = c0 + c1 (25) c0 c1
∗
The least-squares solution to the system is or D ≈ 102.53 · 100.047t = 339 · 100.047t .
≈
2.53 , i.e. log D ≈ 2.53+0.047t 0.047
b. In the year 2000, t = 30 and D ≈ 8, 700. The formula predicts a debt of about 8.7 trillion dollars. For the year 2010, the formula predicts a debt of about 25.7 trillion dollars. c. In the late 1990’s, the United States managed to balance their budget.
276
ISM: Linear Algebra
Section 5.5
42. Clearly, L is a linear transformation. We will use Fact 4.2.4a and show that ker(L) = { 0} and im(L) = im(A). Now ker(L) = ker(A) ∩ im(AT ) = {0}, by Facts 5.4.1 and 5.1.8b. Also, im(L) = {Av : v in im(AT )} = im(AAT ) = im(A), by Exercise 6.
5.5
1. Since f is nonzero, there is a c on [a, b] such that f (c) = d = 0. By continuity, there is an ε > 0 such that |f (x)| > |d| for all x on the open interval (c − ε, c + ε) where f is defined 2 (see any good Calculus text).
Figure 5.20: for Problem 5.5.1. We assume that c − ε ≥ a and c + ε ≤ b and leave the other cases to the reader. Then 2 b c+ε d2 ε d f, f = > 0, as claimed. (See Figure 5.20.) dt = (f (t))2 dt ≥ 2 2 a c−ε 2. We perform the following operations: f, g + h = ↑ g + h, f = ↑ g, f + h, f = ↑ f, g + f, h
Definition 5.5.1a
Definition 5.5.1b
Definition 5.5.1a
3. Note that x, y = (Sx)T Sy = Sx · Sy. a. We will check the four parts of Definition 5.5.1 α. β. x, y = Sx · Sy = Sy · Sx = y, x x + y, z = S(x + y) · Sz = (Sx + Sy) · Sz = (Sx · Sz) + (Sy · Sz) = x, z + y, z 277
Chapter 5 γ. cx, y = S(cx) · Sy = c(Sx) · Sy = c x, y
2
ISM: Linear Algebra
δ. If x = 0, then x, x = Sx · Sx = Sx Answer: S must be invertible.
is positive if Sx = 0, that is, if x is not in the
kernel of S. It is required that Sx = 0 whenever x = 0, that is, ker(S) = {0}. b. It is required that x, y = (Sx)T Sy = xT S T Sy equal x · y = xT y for all x and y. This is the case if and only if S T S = In , that is, S is orthogonal.
4. a. For column vectors v, w, we have v, w = trace(v T w) = trace(v · w) = v · w, the dot product. b. For row vectors v, w, the ij th entry of v T w is vi wj , so that v, w = trace(v T w) =
m i=1
vi wi = v · w, again the dot product.
5. a. b.
A, B = tr(AB T ) = tr((AB T )T ) = tr(BAT ) = B, A In the second step we have used the fact that tr(M ) = tr(M T ), for any square matrix M . A + B, C = A, C = tr((A + B)C T ) = tr(AC T + BC T ) = tr(AC T ) + tr(BC T ) + B, C = tr(cAB T ) = ctr(AB T ) = c A, B
n
c. d.
cA, B A, A
= tr(AAT ) =
i=1
vi ·, ·
m
2
> 0 if A = 0, where vi is the ith row of A.
We have shown that
does indeed define an inner product.
n m
6. a. The iith entry of PQ is
k=1 m n
pik qki , so that tr(P Q) =
i=1 k=1
pik qki .
Likewise, tr(QP ) =
i=1 k=1
qik pki .
Reversing the roles of i and k and the order of summation we see that tr(P Q) = tr(QP ), as claimed. b. Using part a and the fact that tr(M ) = tr(M T ) for any square matrix M , we find that A, B = tr(AT B) = tr(BAT ) = tr((BAT )T ) = tr(AB T ) = A, B 278
ISM: Linear Algebra
Section 5.5
7. Axioms a, b, and c hold for any choice of k (check this!). Also, it is required that v, v = k v, v be positive for nonzero v. Since v, v is positive, this is the case if (and only if) k is positive. 8. By parts b and c of Definition 5.5.1, we have T (u + v) = u + v, w = u, w + v, w = T (u) + T (v) and T (cv) = cv, w = c v, w = cT (v), so that T is linear. If w = 0, then im(T ) = {0} and ker(T ) = V . If w = 0, then im(T ) = R and ker(T ) consists of all v perpendicular to w.
1
9. If f is even and g is odd, then fg is odd, so that f, g =
−1
f g = 0.
10. A function g(t) = a + bt + ct2 is orthogonal to f (t) = t if
1
f, g =
−1
b c (at + bt2 + ct3 ) dt = [ a t2 + 3 t3 + 4 t4 ]1 = −1 2
2 b = 0, that is, if b = 0. 3
Thus, the functions 1 and t2 form a basis of the space of all functions in P2 orthogonal to f (t) = t. To find an orthonormal basis g1 (t), g2 (t), we apply Gram-Schmidt. Now
1
1 =
√ 5 2 2 (3t
1 2
1 dt = 1, so that we can let g1 (t) = 1. Then g2 (t) =
−1 √ 5 2 2 (3t
t2 − 1,t2 1 t2 − 1,t2 1
=
t2 − 1 3 t2 − 1 3
=
− 1)
Answer : g1 (t) = 1, g2 (t) =
− 1)
11. f, g = cos(t), cos(t + δ) = cos(t), cos(t) cos(δ) − sin(t) sin(δ) = cos(δ) cos(t), cos(t) − sin(δ) cos(t), sin(t) = cos(δ), by Fact 5.5.4. Also, g, g = 1 (left to reader). Thus, ∠(f, g) = arccos
f,g f g
= arccos(cos δ) = δ.
π π −π √1 2π
1 12. By Fact 5.5.5, a0 = |t|, √2
π 1 |t| dt = √ , bk = |t|, sin(kt) = π 2 −π 0, since the integrand is an odd function. =
π
|t| sin(kt) dt =
π π 0
ck
= |t|, cos(kt) =
1 π
−π
|t| cos(kt) dt = =
2 π
π
t cos(kt) dt =
0
= Summary:
2 1 π π k2 [cos(kt)]0
2 πk2 (cos(kπ)
− 1) =
2 1 t sin(kt) − π k 0 0 if k is even 4 − k2 π if k is odd
1 sin(kt) dt k
279
Chapter 5 a0 bk ck =0 =0 = 0 4 − k2 π if k is even if k is odd
ISM: Linear Algebra
13. The sequence (a0 , b1 , c1 , b2 , c2 , . . .) is “square-summable” by Fact 5.5.6, so that it is in 2 . Also, (a0 , b1 , c1 , b2 , c2 , . . .) 2 = a2 + b2 + c2 + b2 + c2 + · · · = f 2 , by Fact 5.5.6, so that 0 1 1 2 2 the two norms are equal.
14. a. This is not an inner product since there are nonzero polynomials f (t) in P2 with f (1) = f (2) = 0, so that f, f = (f (1))2 +(f (2))2 = 0. (For example, let f (t) = (t−1)(t−2).) b. This is an inner product. We leave it to the reader to check axioms a to c. As for d: A nonzero polynomial f in P2 has at most two zeros, so that f (1) = 0 or f (2) = 0 or f (3) = 0, and f, f = (f (1))2 + (f (2))2 + (f (3))2 > 0.
15. First note that b =
1 0 0 1 , = , = c, by part a of Definition 5.5.1, 0 1 1 0 so that b = c. Check that if b = c then v, w = w, v for all v, w in R2 . Check that parts (b) and (c) of Definition 5.5.1 are satisfied for all values of constants c,d. Next, we x x1 = , 1 need to worry about part (d) of that definition. It is required that x2 x2 x1 x2 + 2bx1 x2 + dx2 be positive for all nonzero . We can complete the square and 1 2 x2 write x2 + 2bx1 x2 + dx2 = (x1 + bx2 )2 + (d − b2)x2 ; this quantity is positive for all nonzero 1 2 2 x1 2 2 if (and only if) d − b > 0, or, d > b . In summary, the function is an inner product x2 if (and only if) b = c and d > b2 .
16. a. We start with the standard basis 1, t and use the Gram-Schmidt process to construct an orthonormal basis g1 (t), g2 (t).
t− 1,t dt = 1, so that we can let g1 (t) = 1. Then g2 (t) = t− 1,t 1 = 0 √ 3(2t − 1). √ Summary: g1 (t) = 1 and g2 (t) = 3(2t − 1) is an orthonormal basis. 1 1 1
=
1 t− 2 t− 1 2
=
b. We are looking for projP1 (t2 ) = g1 (t), t2 g1 (t) + g2 (t), t2 g2 (t), by Fact 5.5.3. 280
ISM: Linear Algebra √ 1 t dt = and g2 (t), t2 = 3 We find that g1 (t), t = 3 0 1 that projP1 t2 = 1 + 2 (2t − 1) = t − 1 . See Figure 5.21. 3 6
2 2 1 1 0
Section 5.5 √ 3 (2t − t ) dt = , so 6
3 2
Figure 5.21: for Problem 5.5.16b.
17. We leave it to the reader to check that the first three axioms are satisfied for any such T . As for axiom d: It is required that v, v = T (v) · T (v) = T (v) 2 be positive for any nonzero v, that is, T (v) = 0. This means that the kernel of T must be {0}. c1 c2 18. Let the orthonormal basis be f1 , . . . , fn and f = c1 f1 +· · ·+cn fn ; then [f ]B = . . The . . Pythagorean Theorem tells us that f
2
= c2 + · · · + c2 = [f ]B n 1
2
cn , so that f = [f ]B .
19. If we write v = = [x1 x2 ] p q r s
p q y x1 , w = 1 , and A = , then y2 r s x2 y1 y2
y x1 , 1 y2 x2
= px1 y1 + qx1 y2 + rx2 y1 + sx2 y2 . Note that in Exercise 15 we 1 1 , 0 0 be positive.
considered the special case p = 1. First it is required that p = Now we can write y x1 , 1 y2 x2
q r s = p x1 y1 + p x1 y2 + p x2 y1 + p x2 y2 and use our work
q r s in Exercise 15 with b = p , c = p , d = p to see that the conditions q = r and q 2 < ps must hold. In summary, the function is an inner product if (and only if) the entries of p q matrix A = satisfy the conditions p > 0, q = r and det(A) = ps − q 2 > 0. r s
281
Chapter 5
ISM: Linear Algebra
20. a.
x1 1 , x2 0
= [x1 x2 ]
1 2 2 8
1 = x1 + 2x2 = 0 when x1 = −2x2 . This is the line 0
spanned by vector b. Since vectors
−2 . 1
1 −2 and are orthogonal, we merely have to multiply each of them 0 1 1 0
2
with the reciprocal of its norm. Now is a unit vector, and Thus −1 1 , 1 0 2 −2 1
2
= [1 0] 1 2 2 8 −2 1
1 2 2 8
1 = 1, so that 0 −2 1
1 0 = 2.
= [ −2 1 ]
= 4, so that
is an orthonormal basis.
21. Note that v, w = (v + w) · (v + w) − v · v − w · w = 2(v · w), double the dot product. Thus it’s an inner product, by Exercise 7. 22. Apply the Cauchy-Schwarz inequality to f (t) and g(t) = 1; note that g = 1:
1 2
| f, g | ≤ f
g = f or f, g
2
≤ f
2
or
0
f (t) dt
≤
1 2 0 (f (t))
dt.
23. We start with the standard basis 1, t of P1 and use the Gram-Schmidt process to construct and orthonormal basis g1 (t), g2 (t). 1 =
t− 1 2 t− 1 2 1 2 (1
· 1) + 1 · 1 = 1, so that we can let g1 (t) = 1. Then g2 (t) =
t− 1,t 1 t−(1,t)1
=
= 2t − 1.
Summary: g1 (t) = 1 and g2 (t) = 2t − 1 is an orthonormal basis.
24. a. f, g + h = f, g + f, h = 0 + 8 = 8 b. g+h = g + h, g + h = g, g + 2 g, h + h, h = √ √ 1 + 6 + 50 = 57 is an orthonormal basis of
c. Since f, g = 0, g = 1, and f = 2, we know that span (f, g). Now projE h =
f 2,h f 2
f 2,g
+ g, h g =
1 4
f, h f + g, h g = 2f + 3g. 282
ISM: Linear Algebra
Section 5.5
d. From part c we know that 1 f, g are orthonormal, so we apply Fact 5.2.1 to obtain 2 the third polynomial in an orthonormal basis of span(f, g, h):
h−projE h h−projE h
=
h−2f −3g h−2f −3g
=
h−2f −3g 5
2 3 = −5f − 5g + 1h 5
Orthonormal basis:
1 2 2 f, g, − 5 f
1 − 3g + 5h 5
25. Using the inner product defined in Example 2, we find that x = 5.5.6). 26. a0 = bk =
√1 2π π 1 π
x, x =
π
1+
1 4
+
1 9
+···+
1 n2
+··· =
π2 6
=
π √ 6
(see the text right after Fact
f (t) dt = 0
−π
f (t) sin(kt) dt =
−π
1 π
0
π
−
sin(kt) dt +
−π 0
sin(kt) dt
=
2 π
π
sin(kt) dt
0
2 = − kπ [cos(kt)]π = 0 π
0
4 πk
if k is even if k is odd
ck =
1 π
f (t) cos(kt) dt = 0, since the integrand is odd. See Figure 5.22.
−π 4 π 4 π 4 π
f1 (t) = f2 (t) = f3 (t) = f4 (t) = f5 (t) = f6 (t) =
π
sin(t). See Figure 5.23. sin(t) + sin(t) +
4 3π 4 3π
sin(3t). See Figure 5.24. sin(3t) +
4 5π
sin(5t)
27. a0 = bk = ck =
√1 2π π 1 π
1 f (t) dt = √ . 2 −π f (t) sin(kt) dt = 1 π 1 π
π 0 π
−π π
sin(kt) dt = − cos(kt) dt =
1 [cos(kt)]π = 0 kπ
0
2 kπ
if k is even if k is odd
1 π
f (t) cos(kt) dt =
−π 1 2
0 1 2
1 [sin(kt)]π = 0 0 kπ +
2 π
f1 (t) = f2 (t) = . . .
+
2 π
sin(t), f3 (t) = f4 (t) =
sin(t) +
2 3π
sin(3t)
283
Chapter 5
ISM: Linear Algebra
Figure 5.22: for Problem 5.5.26.
Figure 5.23: for Problem 5.5.26.
π
28. f
2
= f, f =
1 π
(f (t))2 dt =
−π
1 π
π
1 dt = 2
−π
284
ISM: Linear Algebra
Section 5.5
Figure 5.24: for Problem 5.5.26. 1 k2 1 = k2
Now Fact 5.5.6 tells us that 1 1 1 π2 1+ + + +··· = . 9 25 49 8
π
16 π2
+
16 9π 2
+
16 25π 2
+ ··· =
16 π2 k odd
= 2, or
k odd
29. f
2
= f, f =
1 π
(f (t))2 dt =
−π 1 2
1 π +
π
1 dt = 1
0 4 9π 2 4 25π 2 1 2 4 π2 k odd
Now Fact 5.5.6 tells us that
+
4 π2
+
+ · · · = 1, or
+
1 k2
= 1 or
k odd
1 1 1 1 =1+ + + +··· = 2 k 9 25 49
1 2 4 π2
=
π2 . 8
30. There is an invertible linear transformation T (x) = Ax from R2 to R2 that transforms E into the unit circle. (If E is parametrized by cos(t)w1 + sin(t)w2 , let A = [ w1 w2 ]−1 . Compare with Exercises 2.2.50 and 2.2.51.) Now x is on E if T (x) = 1, or T (x) · T (x) = 1. This means that the inner product x, y = T (x) · T (y) does the job (see Exercise 17). (There are other, very different approaches to this problem.) 31. An orthonormal basis of P2 of the desired form is f0 (t) =
1 2 5 2 2 (3t 1 √ , f1 (t) 2
=
3 2 t, f2 (t)
=
1 − 1) (compare with Exercise 10), and the zeros of f2 (t) are a1,2 = ± √3 .
285
Chapter 5
1 2
ISM: Linear Algebra
Next we find the weights w1 , w2 such that
−1
f (t) dt =
i=1
wi f (ai ) for all f in P1 . We
need to make sure that the equation holds for 1 and t. 2 = w 1 + w2 1 0 = √3 w1 −
1 √ w2 3
, with solution w1 = w2 = 1.
1
It follows that the equation
1 1 √ +f −√ holds for 3 3 −1 all polynomials f in P1 . We can check that it holds for t2 and t3 as well, that is, it holds in fact for all cubic polynomials. f (t) dt = f (a1 ) + f (a2 ) = f
1 −1
t2 dt =
2 equals 3
1 √ 3
2
1 + − √3
2
1
= 2 , and 3
t3 dt = 0 equals
0.
−1
1 √ 3
3
1 + − √3
3
=
True or False
1. T, by Fact 5.3.4a 2. F. We have (AB)T = B T AT , by Fact 5.3.9a 3. T, since (A + B)T = AT + B T = A + B 4. T, by Fact 5.3.4 5. F. Consider 1 1 . 1 1
6. T. First note that AT = A−1 , by Fact 2.4.9. Thus A is orthogonal, by Fact 5.3.7 7. F. The correct formula is projL (x) = (x · u)u, by Definition 2.2.1. 8. T, since (7A)T = 7AT = 7A. 9. F. Consider T (x) = 10. T, by Fact 5.3.9.b 11. T, since (ABBA)T = AT B T B T AT = ABBA, by Fact 5.3.9a 12. T, since AT B T = (BA)T = (AB)T = B T AT , by Fact 5.3.9a 286 1 1 x. 0 0
ISM: Linear Algebra
True or False
13. F. dim(V ) + dim(V ⊥ ) = 5, by Fact 5.1.8c. Thus one of the dimensions is even and the other odd. 14. T. Consider the QR factorization (Fact 5.2.2) 15. F. The Pythagorean Theorem holds for orthogonal vectors x, y only (Fact 5.1.9) 16. T. det a b a c = ad − bc = det c d b . d
17. T. If A is orthogonal, then AT = A−1 , and A−1 is orthogonal by Fact 5.3.4b. 18. F. Consider 0 1 1 0 0 −1 1 0 . = 1 0 0 −1 0 1 1 0 . Then AB T = 0 0 0 0 isn’t equal to B T A = 0 0 . 0 1
19. F. Consider A = B =
20. F. It is required that the columns of A be orthonormal (Fact 5.3.10). As a counterexample, 2 4 0 consider A = , with AAT = . 0 0 0 21. T, since the columns are unit vectors. 22. T. Use the Gram-Schmidt process to construct such a basis (Fact 5.2.1) 23. F. The columns fail to be unit vectors (use Fact 5.3.3b) 24. T, by definition of an orthogonal projection (Fact 5.1.4). 25. F. As a counterexample, consider 26. T, by Fact 5.4.1. 27. T, by Fact 5.4.2a. 28. F. Consider A = 29. F. det 30. T. 2 0 , or any other symmetric matrix that fails to be orthogonal. 0 2 −1 0 0 1 is orthogonal.
1 2 (A
1 0 0 −1
and
0 −1 . 1 0
−1 0 = −1 − 0 = −1, yet 0 1 − AT )
T
1 2 (A
1 = 1 (A − AT )T = 2 (AT − A) = − 2
− AT ) .
31. F. Consider A =
−1 1 . 0 1 287
Chapter 5 32. T. By Definition 5.1.12, quantity cos(θ) =
v·w v w
ISM: Linear Algebra is positive, so that θ is an acute angle.
33. T. In Fact 5.4.1, let A = B T to see that (im(B T ))⊥ = ker(B). Now take the orthogonal complements of both sides and use Fact 5.1.8d. 34. T, since (AT A)T = AT (AT )T = AT A, by Fact 5.3.9a. 35. F. Verify that matrices A = 36. F. Consider B = 5.4.1 and 5.4.2. 37. T. We know that AT = A and S −1 = S T . Now (S −1 AS)T = S T AT (S −1 )T = S −1 AS, by Fact 5.3.9a. 38. T. By Fact 5.4.2, we have ker(A) = ker(AT A). Replacing A by AT in this formula, we find that ker(AT ) = ker(AAT ). Now ker(A) = ker(AT A) = ker(AAT ) = ker(AT ). 39. T. Try A = 1 + cos θ − sin θ − sin θ . , so that A + B = sin θ 1 + cos θ cos θ 1 + cos θ − sin θ It is required that and be unit vectors, meaning that 1 + 2 cos θ + sin θ 1 + cos θ √ 1 0 cos2 θ + sin2 θ = 2 + 2 cos θ = 1, or cos θ = − 1 , and sin θ = ± 23 . Thus A = and 2 0 1 1 0 0 1 and B = cos θ sin θ B=
1 −2 √ 3 2
1 0 0 −1
and B =
1 1 0 −1
are similar.
0 1 . The correct formula im(B) = im(BB T ) follows from Facts 0 0
−
√ 3 2 −1 2
40. F. Consider A =
0
1 2
is a solution.
−1 2 , for example, representing a rotation combined with a scaling. 0
41. T. We attempt to write A = S +Q, where S is symmetric and Q is skew-symmetric. Then AT = S T + QT = S − Q. Adding the equations A = S + Q and AT = S − Q together 1 1 gives 2S = A + AT and S = 2 (A + AT ). Similarly we find Q = 2 (A − AT ). Check that 1 1 T T the decomposition A = S + Q = ( 2 (A + A )) + ( 2 (A − A )) does the job. x1 42. T. Apply the Cauchy-Schwarz inequality (squared), (x · y)2 ≤ x 2 y 2 , to x = . . . xn 1 and y = . . . (all n entries are 1). 1 288
ISM: Linear Algebra
True or False
43. T. Let A =
x y x2 + y 2 xz + yt x2 + yz xy + yt . We know that AAT = A2 , or = . 2 2 z t xz + yt z + t zx + tz yz + t2 We need to show that y = z. If y = 0, this follows from the equation x2 + y 2 = x2 + yz; if z = 0, it follows from z 2 + t2 = yz + t2 ; if both y and z are zero, we are all set.
2
44. T, since x · (projV x) = (projV x + (x − projV x)) · projV x = projV x x − projV x is orthogonal to projV x, by the definition of a projection.
1 45. T. Note that 1 = A x x x . See Definition 5.3.1.
≥ 0. Note that
=
1 x
Ax =
1 x
Ax for all nonzero x, so that Ax =
46. T. If A =
We use the quadratic formula to find the (real) solutions x = 2 √ a+c± (a−c)2 +4b2 . Note that the discriminant (a − c)2 + 4b2 is positive or zero. 2 47. T; one basis is: 1 0 1 0 0 1 0 −1 , , , . 0 1 0 −1 1 0 1 0
1 √ 5
a b a−x b is a symmetric matrix, then A − xI2 = . This mab c b c−x trix fails to be invertible if (and only if) det(A − xI2 ) = (a − x)(c − x) − b2 = 0. √
a+c± (a+c)2 −4ac+4b2
=
48. F; A direct computation or a geometrical argument shows that Q = resenting a reflection, not a rotation.
1 2 , rep2 −1
49. F; dim(R3×3 )= 9, dim(R2×2 )= 4, so dim(ker(L))≥ 5, but the space of all 3 × 3 skewsymmetric matrices has dimension of 3. 0 −1 0 0 0 −1 0 0 0 A basis is 1 0 0 , 0 0 0 , 0 0 −1 . 0 0 0 1 0 0 0 1 0
50. T; Consider an orthonormal basis v1 , v2 of V, and a unit vector v3 perpendicular to V, 1 0 0 and form the orthogonal matrix S = [v1 v2 v3 ]. Now AS = v1 v2 0 = S 0 1 0 . 0 0 0 1 0 0 Since S is orthogonal, we have S T AS = S −1 AS = 0 1 0 , a diagonal matrix. 0 0 0
289