*This preview shows
pages
1–3. Sign up
to
view the full content.*

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **Math 26", HW IL, -‘ 5 IOJIéqu/lolzz' 3;! ’5, '5‘, 2‘1, 3:, 57.
_-I i 20, 2v, 26
8.1
10. ,\1 = A2 = 0 and A3 = 9.
2 1
171 = —V}—g 1 is in E0 and '52 = g —2 is in E9.
0 2
. 2
Let 173 2 171 x 172 = 3—5—5 —4 ; then '51, '52, 173 is an orthonormal eigenbasis.
i l L
V5 3 W5 0 0 0
S = f —% —34ﬁ and D = 0 9 0 15. Yes, if Ali 2 A17, then A’lﬁ' = @7, so that an orthonormal eigenbasis for A is also an
orthonormal eigenbasis for A‘1 (with reciprocal eigenvalues). 16. a. ker(A) is four-dimensional, so that the eigenvalue 0 has multiplicity 4, and the remain-
ing eigenvalue is tr(A) = 5. l). B = A + 215, so that the eigenvalues are 2, 2, 2, 2, 7. c. det(B) = 24 -7 = 112 (product of eigenvalues) 17. If A is the n x in matrix with all 1’s, then the eigenvalues of A are 0 (with multiplicity
n — 1) and n. Now B = qA + (p — q)In, so that the eigenvalues of B are p — q (with multiplicity n — 1) and qn + p — (1. Thus det(B) = (p — q)"‘1(qn + p — q). 19. Let L-(f) = A55. Then ATA is symmetric, since (ATA)T : AT(AT)T : ATA so that
there an orthonormal eigenbasis 171, . . . ,ﬁm for ATA. Then the vectors A171 , Affm
are or Ogona,since iii-Aﬁ- = (A5.)TA5. = ﬁrArA—g : 1. T «I _ 1 am;
Aj(ﬁi'ﬁj)=0ifi7§j. J z ‘7 1 “J “t (A Avg) —vg.(AJ-vj) = 20. By Exercise 19, there is an orthonormal basis 17;, . . . ,ﬁm of R“ such that T(-D'1 ), . . . ,T('J,,.) are orthogonal. Suppose that T(171),. . . ,T(17r) are nonzero and T(1‘)’,+1),... ,Twm) are
zero. Then let 1:71- : WTﬁﬁ) for 3' = 1,...,r and choose an orthonormal basis
1HT+1,...,1iin of[span(1D’1,...,w'r)]i. Then 131,...nfiﬂ does the job. 22. a. If we let I: = 2 then A is symmetric and therefore (orthogonally) diagonalizable. b. If we let I: = 0 then 0 is the only eigenvalue (but A ¢ 0), so that A fails to be
diagonalizable. 29. By Fact 5.4.1 (im AH = ker(AT) = ker(A), so that 13' is orthogonal to if}. 31. True; A is diagonalizable, that is, A is similar to a diagonal matrix D; then A2 is similar
to D2. Now rank(D) = ram-((132) is the number of nonzero entries on the diagonal of D (and D2). Since similar matrices have the same rank (by Fact 7.3.6b) we can conclude
that rank(A) = rank(D) : rank(D2) 2 rank(A2). 32. By Exercise 17, det(A) = (1 — q)“‘l(qn + 1 — q). A is invertible if det(A) 74 0, that is, if
q % 1 and q 5% ‘11 ~_ 0 —1 1:] _ —J,'2 ._
20. Ax—[l 0][ J-[ x1].SeeF1gure9.5. 1 It appears that the trajectories will be circles. If we start at [0 we will trace out the unit Circle :E(t) = ] . Figure 9.5: for Problem 9.1.20. 24. We are told that ﬁg: 2 Ai Let 5(t) = ek‘ﬁt). Then % = ﬁ(ek‘§:’) = (d M) 5+ ek‘df -—
kek‘f-I- ek‘Aa? = (A + kIn)(ek‘:E) 2' (A + kIn)é', as claimed. _ .. 1 _. —2
26' A! — 3: A2 = ‘2; '01 =[ JVU2 =[ 3] ,C1 = 5,02 = —1, so that 55(15): 5e3‘[1J‘ ...

View
Full
Document