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Unformatted text preview: HW3
2.13.7515, 24,3}, 93 3. Not linear, since y2 = $1933 is nonlinear. 131
7. Note that $1171 +    + mmﬁ'm 2 [51...ﬁ'm] mm , so that T is indeed linear, with matrix [171 172 17ml 15. By Exercise 133., the matrix [3 ‘2] is invertible if (and only if) a2 + b2 75 0, which
. ——b . . . . . .
13 the case unless a = b x 0. If a] is invertible, then its inverse is 391:? [ _: 2] , by Exercise 13b. 24. Compare with Example 5. Figure 2.5: for Problem 2.1.24. 37. Since f 2 '17 +  17), we have T(.i:') = T (17+ Mu? — = T(ﬁ') + k(T(u7) — T073), by
Fact 2.1.3 Since is is between 0 and 1, the tip of this vector T(.f) is on the line segment connecting
the tips ofT(17) and TUE). (See Figure 2.14.) Til?) — T6"). translach Ital?)  T50). translmed Figure 2.14: for Problem 2.1.37. 2 1:1 1'1
43. a. T(:E) = 3 3:2 = 23:1 + 31:2 + 41:3 : [2 3 4] 3:2
4 1‘3 133 The transformation is indeed linear1 with matrix [2 3 4]. '01
b. If 17 = '02 , then T is linear with matrix ['01 '02 1:3], as in part (a).
.US
33] 1'1
c. Let [a b c] be the matrix of T Then T 1'2 = [a b c] 352 = ax; + (J32 + (11:3 =
$3 $3
a $1 (I.
b  3:2 , so that 17 = I) does the job.
c :63 c 2.2.: 14.9, 2"!» 33,38,945 14. a. Proceeding as on Page 58 of the text, we ﬁnd that A is the matrix whose ijth entry is ui'UjZ ‘ 2 . ul “1112 u1u3
A: mm 11% ugug . 2 b. The sum of the diagonal entries is off + v.52, + u§ = 1, since u is a umt vector. (1. b ’Ul ._ 0/01 +1302 __ 'U1
17' we want‘ [b —u.: [v2] _ [be] —av2] _ [112} Now, (a — 1)‘u1 + lrUz = 0 and (ml — (a + U112, which is a system with solutions of the form [ (1 giant] , where t is an arbitrary constant. Let’s choose it = 1, making 17 = [I E a]. Similarly, we want Adi : —’d)'. We perform a computation as above to reveal ti} =
— l . . . _. _. . [a b ] as a possrble chorce. A qurck check of v  w = 0 reveals that they are indeed perpendicular. Now, any vector i" in IR can be written in terms of components with respect to L :
spanw) as 55 = :r" +53i = ca+ do. Then, T(s) 2 A55 2 A(ca+ (in?) : A(cU)+A(d¢D’) :
cAﬁ+ dAin = (317— do? = "H — 5:“ : refL(:i:'), by Deﬁnition 2.2.2. (The vectors '5 and iii constructed above are both zero in the special case that a = 1
and b = 0. In that case, we can let 13' = é] and 15 : 62 instead.) 0 24. a. A=[17 11?],soA[(1)] =‘UandA[1 = 13. Since A preserves length, both if and iii . . 1 0
must be unit vectors. Furthermore, since A preserves angles and [0] and [I] are clearly perpendicular, 13' and iii must also be perpendicular. b. Since 23 is a unit vector perpendicular to 17, it can be obtained by rotating 6 through 90
degrees, either in the counterclockwise or in the clockwise direction. Using the corre .. —1 _. —b _. 0 1 _.
spending rotation matrices, we see that w = [0 ] v = [ ] or w = [_1 0] v = 1 0 a.
b
ma '
. . . a —b . .
c. Followmg part b, A 13 either of the form b a , representing a rotation, or A = b —a
33. Geometrically, we can ﬁnd the representation 1'3 2 {1'1 + 272 by means of a parallelogram,
as shown in Figure 2.26. [a b ] , representing a. reﬂection. To show the existence and uniqueness of this representation algebraically, choose a nonzero vector 1171 in L1 and a nonzero 162 in L2. Then the system $11171 + $2132 : 0 or
.. _. m — . . _. _. — [ml 102] = D has only the solution 3:1 = $2 = 0 (1f $1101 + $2102 = 0 then
321161 = —a:2'u72 is both in L1 and in L2, so that it must be the zero vector). Figure 2.26: for Problem 2.2.33. 1'1
:1‘2
for all 17 in R2 (by Fact 1.3.4). Now set 171 : 11th and 172 : 12152 to obtain the desired
representation if = 171 +172. (Compare with Exercise 1.3.57.) Therefore, the system 1: “El + $2152 2 13' or [llll n72] [ ] = 17 has a unique solution 2:1, 1:2 To show that the transformation T07) = '61 is linear, we will verify the two parts of Fact
2.1.3. 0 Let ‘U = 171+ 172, '10 = 1171 +1172, SO that 174117 = (171 + 131) + (172 + and 16317 = A117] + A7172. TTTT T 1 1T in L1 in L2 in L1 in L2 in L, in L2 in L1 in L2 a. T(17+ u?) = 171 +131 2 + T(1L"), and
b. CHM?) 2 kn] = kT(ﬁ), as claimed. .22] , so det(A) = 11qu —u;u2u1u2 : 0. b. A 2 [z _a] , so det(A) = —a2 — b2 = —(a2 + b2) = —1. (l. A = or , both of which have determinant equal to 12 — 0 = 1. 2 2 i1
46. We want to write A = k[ k J , where the matrix B = kn] represents a a‘lc'a‘ln k E
7%)2 = 1, meaning that a2+lb2 2 k2, or, k = x/a2 + b2. Now A—1 : 39%? (I: b ] = 51514 = ﬁB, for the reﬂection matrix B and the scaling He reﬂection. It is required that (%)2+ A —a
factor I: introduced above. In summary: If A represents a reﬂection combined with a
scaling by k, then 14—1 represents the same reﬂection combined with a scaling by 2.3: 1,6,H, lull): ‘13, LII[J L13 —1
1 Z __ 8 _3
1“My? 3. 1 a]: 1 0. 8 3 35061342 3] =[_ . . 2
5 8: 0 1 G 1: —5 2 5 8 5
1 2 1
6. Use Fact 2.3.5; the inverse is 0 1 —2
0 0 l 19. Solving for $1,332, and $3 in terms of yhyg, and y3, we ﬁnd that 331 = 3y1 ‘ 3312 + 113.113
$2 = —3y1 + 492  ya
173 = .111 — '3'sz + £93 41. a. Invertible: the transformation is its own inverse. b. Not invertible: the equation T(:E) = b has infinitely many solutions if I; is on the plane,
and none otherwise. 43. We make an attempt to solve the equation 37 = A(B:E) for :E': Bi 2 14137, so that :f = B‘1(A_1g'). 1 0 —1 —2
I 0 1 2 3
44. a. rrerl/I4) = 0 0 0 O , so that rankﬂth) = 2.
0 0 0 O
b. To simplify the notation, we introduce the row vectors 1? = [l 1 1] and 15 2 [011211 (n—1)n] with 71 components. Then we can write Afn in terms of its rows as A1,, 2 Applying the GaussJordan algorithm to the ﬁrst column we get ~2w
—(n * 1)1I}' All the rows below the second are scalar multiples of the second; therefore, rankMI") = 2. o. By part (b), the matrix AL; is invertible only if n = 1 or n. : 2. 0 1 O 0 _‘ 0
48. Let A r— 0 I] and I; = 0 . The equation A5 = b has the unique solution 57' = [ O].
0 (J
t 2.3.4 applies to square matrices only. Note that Fae ...
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This homework help was uploaded on 02/25/2008 for the course PL PA 201 taught by Professor G.w.hudler during the Spring '07 term at Cornell.
 Spring '07
 G.W.HUDLER

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