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Unformatted text preview: 8. 24. Note that A is symmetric and orthogonal, so that the eigenvalues are 1 and —1 (see 28. 2 1 1 1
. f [ 1] ,ﬁ [—4] , ”Cl—5 [I] is an orthonormal eigenbasis. 1
x/T5 W 3 —1 4 0
ﬁt 3] andD=[0 —6l' Exercise 23). 1 0 1 0
E1 = span 3 , i and E_1 = span 3 , _: , so that
1 0 —1 0
1 0 1 O
l 0 1 1 1 0 1 1 . . ‘.
73 0 , 7,: 1 , Ti 0 , W *1 is an orthonormal eigenbabls.
1 O ~1 0
For A 71 0
—A 0 1 —A 0
—A 1 —A
fA()\)=d€t " : =§det
O —~A 1 0 —)\
1 1 1 1  A A A A
A 0 1
A 1
= ﬁdet
0 —A 1
0 0 ~ 0 —A2 + A + 12 z 1119?  A ~ 12) = —A”(A  4W + 3) Eigenvalues are 0 (with multiplicity 11), 4 and —3. . wait/S
Eigemﬁlue’s for 0 are 51 — é'¢(i = 2,. .. , 12),
1 1
1 ' 1
E4 = span 2 (12 ones), 133 = span 3 (12 ones)
1 L 1
4 ——3 3], 1 [:13] is an orthonormal eigenbasis, with A1 = 4 and A2 = ‘6’ 30 S = S 1 1 1 1 1 1 1 1 1 1 1 1 1
—1 0 0 O 0 0 0 O O 0 0 1 1
0 1 0 0 O 0 0 O O O 0 1 1
O 0 ~—1 0 0 O 0 O 0 0 O 1 1
0 0 0 —1 O 0 0 0 0 0 0 1 1
0 0 0 0 —1 0 O 0 O 0 0 1 1
= O 0 O 0 O —1 0 0 O 0 0 1 1
0 O 0 0 0 0 —1 0 0 O 0 1 1
0 0 0 0 0 O O —1 0 0 0 1 1
O 0 O 0 O 0 0 0 —1 0 0 1 1
0 0 0 O 0 0 0 0 0 —1 0 1 1
0 0 0 O 0 0 0 0 0 0 ~1 1 1
O 0 0 0 0 O 0 0 O O 0 4 —3 diagonalizes A, and D = S‘IAS will have all zeros as entries except ([12, 12 = 4 and £113, 13 == ”3. 81} W 14. 17. 24. _. q 2 "‘
28. /\1=2, A2=10;171 =[ 3],172= [é]; C1 3 ”#62: 23330 that x(t)=—1§e t[ 2] + 4 6
ATA=l6 13l§A1=16,x\2=1;01=4,02=1 eigeuvectors of ATA 2 171 = El“ 1 "‘ 1 —2 4 a 2 .g .,
[2]’v2~':zr;[ 1],UI=;1;A 1=§§[1],u2=3§;Av2 H f ["5], so that
1 2 ‘1 _ 4 0 1 1 2
75L 2l’2‘lo ll’V=7%‘[2 ll. We need to check that A (%‘}61 +   ' + ngﬁm) 4: projimAE (see page 221 of the text). C:
H 0‘! 'ﬁ1m++wvm) = 93fA171+m+9§7§A27m :(bﬁ1)a'1+~+(bﬂm)ﬁm 0' 0m ButA( = projimAE, since 1731, . . . ,ﬁm is an orthonormal basis of ‘un(A) (see Fact 5.1.5). 1 l 1
~ 2 1 1 a 1 a 3 1 4 ~
4 1 —1 We are told that E}? = A53. Let E'(t) =: ek‘ﬂt). Then a? = ~dd—t(ektzif) = (3456“) 55+ emit3E =
kek‘a’f + ek‘As'c’ = (A + kIn)(e’"5c') = (A + kIn)é', as claimed. m mwm 32. See. Exercise 26 and Figure. 9.7. Figure 9.7: for Problem 9.1.32. 26. ,\1 =3,,\2 =—2; 771 = [3,172 =[ *2 3
,C1 = 5,02 = —1, so that 53(t) = 5e3‘[ Figure 9.20: for Problem 9.1.46. 3‘2. ’2 .jzspan [3+1] and E_.; =span [3:64]. 6. A1,2=ii?Ei:ker[ 5 ~3—i
General solution: a .1. 3+" ,1. 3—4
a:(t)=cle‘[ 5L]+cze ‘[ 5’] If C] = 62 = 1 then f(t) = (cos(t)+isin(t))[ 6 0030‘) — 2 sin(t)
[ 10 cos(t) ] ' 5 3+'i + (cos(t) ~ isi11(t)) [3:] : 1
1
_ 12. We will show that the real parts of all the eigenvalues are negative, so that the zero
state is a stable equilibrium solution. Now the characteristic polynomial of A is f A()\) =
—)\3 — 2A2 — A — 1. It is convenient to get rid of all these minus signs: The eigenvalues
are the solutions of the equation g()\) = A3 + 2A2 + A + 1 = 0. Since g(—1) = 1 and
g(——2) = —1. there will be an eigenvalue A; between —2 and ~1. Using calculus (or a
graphing calculator), we see that the equation g()\) = 0 has no other real solutions. Thus
there must be two complex conjugate eigenvalues p :l: iq. Now the sum of the eigenvalues
is A1 + 210 = tr(A) = —2, and p = ig—Ai will be negative , as claimed. The graph of 90‘).
is shown in Figure 9.33. g(7t)=7t3+27»2+)t+1 Figure 9.33: for Problem 9.2.12. 1
—b ——c —c:l: Vci —4b
————2 . 36. A = [ 1 and fA()\) = A2 + c/\ + b, with eigenvalues Am = a. If c z; 0 then M; = iix/E. The trajectories are ellipses. See Figure 9.40. The block oscillates harmonically, with period %. The zero state fails to be asymp—
totically stable. b. A13 3 Feiixz/lib—Ez V Figure 9.40: for Problem 9.2.36a. The trajectories spiral inwards, since Re()\1) = ReO‘z) = ~§ < 0. This is the case of
a damped oscillation. The zero state is asymptotically stable. See Figure 9.41. Figure 9.41: for Problem 9.2.36b. _ A .. n _ rru . I , ,L1L1 LA.. ...
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 Fall '05
 HUI
 Math, Linear Algebra, Algebra

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