This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 7
Dimension and Structure EXERCISE SET T21
2. {a} v1 = in {h} V3 = 5v; — 5v; [1:] v4 = {1191+ 41m + 8?;
6. {a} Any one of the vectors {1,3}, {2.6L (—1,—3} forms a basis for the line I = :1 y = at. {b} Any two of the vectors [1,1,3], {1,—1‘2L {2,0,5} form a. basis for the plate .7; =r1+r«2,
y =[1—tg1 Z: 3t1+2t3. 1 2—2 1 3H]
“‘3'1‘1'u dhedoed hlt‘
‘ ‘ nﬂ ter u rower: eon mm
It]. The augmented rnntrnt ofthe eyetemle 2 3 4 3 Lien
U l 'D—l III}
1 o o e—gio
_ I _ I
of this matrix is n 1 ﬂ 1 1:0 . Thus the general solutionis
u o 1 o—§.n
o o n n nio x= 35+ genr 43mm} = s[—31l,[},l.ﬂ: +n[%1 e1,§,n,1) where —oo =: 3,: a: oo. The solution spnee is 2dimensionsl 1with canonical basis {vhvn} where v1=E—3rllﬂ1llﬂ}m'd 1'? = {%r_11 grﬂllj' D2. Yen1 theyr are linearly independent. If we write the 1reetorn in the order
Iv" = {utﬂ'ﬂiﬂlllilva = {Drﬂtﬂtlttlll11v2 = [ﬂ‘ﬂl li‘!‘l‘]lvl = t t i t * iirii thEﬂ,lhECﬂ1L‘iﬁ of the positions of the lending le+ it is clear that none of these vectore can be written
no n lJnesr oomhination of the preoeding ones in the list. P31 1“: 55 '1 theﬂ {hair}: = kinx} :I] if and only iinx = ﬂ:[,h1ﬁ:ka}J 2 ae
EXERCISE SET 12
S. [a] An arbitrary vector x = (Ly, 2] can be written an
F an + [y z}n+ [r 'I‘I'iV3+UVi thus the vectors in. v2, Val and in. open R3. On the other hand, since in; = 2v: +1»;T the
vectors in, V2. v1: and v4 are. linearly dependent and do not iorm a home for H3. {h} The vector equation it = [311?], + oat'2 + cava + e41“ is equivalent to the linear ayetem l 1 l
l l D
1 U D EMILE 1
= 2
3 The augmented matrix of this system is GMTp:
“Ml—l and the reduced row eeheloo form of this matrix
3
1 is
U
2
1 l 1 {II I]
{1' 1 I]
I] I] l
TlaLuoT setting :24 = L a genﬂl'ﬂl solution of the exertion is given by C1=3,Cg=—1"211{:3=—1—tlc4=1
where—ooﬂtéoo.Thu5 V: 3V1—{1+2E}V2 —{1+E}V3+1Vq for 3113' “illIE ﬂf I. For example. mrresponding 15.3. t = Dr :2 _1r and t: _%1 “"3 have v:
3V1V2 V3.V=3v1 +192 —v., and v=3VI_ %v3_ %v4_ 16. {a} Sinoe u o n = [1H2] + {1H1} + [3][—l} = I], the vector u is orthogonal to n. Thus V iswntained
in W.
{h} The line V is parallel to u = [1,21 —5}. and the plane W has nonnal vector 11 = {312,1}. Since
1.1 n = [13(3}+[2]{2]+{5}{1}= 2 a! t}. the vector u is not orthogonal to n. Thus if is not
contained in W. 18. (a) The vector equation e1{l_. l. l] + m[1.1,ﬂ} + e3[110.[l] = (4,3,1?!) is equivalent to the linear sve tE‘III
1 l 1 {:1 a].
l I l] e; = 3
l D [I (:3 ﬂ whieh hes solution c1= n. a. z 3. c3 = 1. Thus {4. in}: = on, 1, 1) + 3(1.1,e}+1{1,e.e} and,
by linearity. we have Tl‘iaiﬂl = Ul312.0.1i+ 3(2.1.3.—1i+1i5,*2.l.ﬂ}=(11,1,1u.—3) {b} The vector equation e1{1,1.l}+ C:[l,11ﬂ} + ea{1,ﬂ,tl} = {mt,e} is equivalent to the linear sys
tem HHI—n l 1 u:
1 [I 1132 = b
0 I] 133 C
which has solution :e; = e, I: = h — c. 1:. = n — h. Thus
{01:11 a} = £111 1 + _ Elli1r + {a '— ﬂ!
and so. by linearity, we have T{a.h,e} = e[312,[}_.l}l + {h —e]{2,l,3.—1}+[e—hj{5,—2.l,ﬂ]
={ea—35H.~2u+3t+c.a+2t—3c.t+2c} 5 —3 l
(e) From the formula ehove,urehave[T]= "'3 g l
n —1 2 D1. [:1] True. Any set of more than n vectors in R" is linearly.r dependent. [b] True. Any set of less than n. vectors helmet be spanning set for R“. {1:} True. II" every.r vector in R“ can he expressed in exact];r one way as a linear combination of
the vectors in S. then S is a basis for R“ and thus must contain exactlyr as vectors. {:1} True. If Ax = I) has inﬁniter Imam},r attentions1 then detEA} = D and so the roar vectors of A are linearly dependent.
{e} True. If V E W {or W E V} and if dimll’} = diln[W], then 1’ = W. Dd. Let A be the matrix having the vectors v1 and v; as its eelumns. Then
'11'313'l4‘1i='l5‘i12 ﬂ — $1123}  {0063 a  6052.3] = ﬂoosﬁo + eos2ﬁ‘ and detliﬂ} 5% I] if and only if easier 5! (1052131 i.e.. if and only if or 75 :hﬂ + law where it = I]. i1,
i=2. For that! 1values of a and :3. the vectors v1 and v; form a heels for R2. P5. Since 13 = {V1,V2,....Vn} is a basis for R“, siren.r 1hector x in R" can be expressed as s linear
combination x = c1vl+ t:ng + r   + ens" for exactly one choice of sealers c1,e3,...,cn. Thus it
makes sense to deﬁne s transformation T: R“ —> R” by setting Tix} = Ticlvl + ‘2‘“? + ' " + Cnth = cit'1 + new; +   + there 19. “I
It is easy to check that T is linear. For eitsnnpleI it‘x : 2 131.3 and I = div? than
i=1 :1 I
11 Ttx +3} = 112.19 + djivJj = Z {cj +d3}w, =29“ + Zsjwj = Tic +rse
1:1 j=l j=1 3i=1 and so T is sdditite. Finally, the transformation T has the ::I1't:I[:.~cI't3:r that TVJ: = wj for each
3' = 1.....n. and it is clear [rem the deﬁning formula that T is Luu'quely determined by this
property. EXERCISE SET 13 8. Here W is the line ufintersecticn of the planes at + y + z = El and 2: — y + e = t}. Parametric equations
for this line are x = t, y = I]1 z = t} thus W corresponds to vectors of the form Ll = tit—1,12], 1}. It
follows that E sector w = {:1:,y.z} is in WJ if and only it's: 2 2, Le. if and oral}r ifw is of the fonn w = (135.1"): 111, 0,1) + slit], 11 [It]
Parametric equations for this plane are given by s: = r, y = s, z z r. 23. The reduced row echelon form of the matrix A is 1 e I] i —E
c 1 e t g
R: 1 3
t] ID I] 0 it Thus the vectors r1 = {1,13101 }1 —§), r3 = [D.l.ﬂ,%,§}, and r3 = {t}.ﬂ,l,§. g} form n. basis for the
row space of A. We also conclude from an inspection of R that the null space of A {solutions of
A1: = (1} consists of vectors of the form 3‘: Sir—i1 _%r_%illul +tt§r"§r—%Ti}+ll Thus the motors n1 2 —ﬁ,—%,e§,l,ﬂ}, :13 = [E,—%, §1(},1] formshssis for the null space of A.
It is teams}.? to check that r;  n}; = I] for all 1', 3'. Thus row{A} end nulI{A]I are orthogonal suhspscte of
R“.
n l I} I:
32. {a} IfA :10 I} and x= y , then Ax=ﬂ ifanrlonlp ifxzyzﬂ. Thus thenull spacecfA
I] U U z corresponds to points on the z—axis. 0n the other hand, the column space of A consists of all
sectors that are of the form H
II
tr:
Chi—n:
+
H
{SDI
+
N
as:
H
GﬂH which corresponds to points on the mgr—plane. {h} The matrix B = U 'l} U
E! l 1]] has the speciﬁed null space and column spare.
I] I] l D1. DIE+ Do. {a} {a} Pulse. This statement is true if and only it tho nonzero town of A are linsaa'lglr innhspendnnt1
to.r if there are no additional zero rows in an echelon Form. {1:} True. If E is an slantstitan”r matrix, then E is invertible and an EA; = I] if and mil},r if
A): = II. {n} Thus. If A has rank n, then there can he no zero rows in an echelon form for A; thus the
reduced row echelon form is the identity matrix. [d] False. For instantls1 if m = n and A is invertible, than row{A) = R” and nuiltA] = [e] True. This follows from the fast [Theorem 13.3} that Si is a. subspace of R". {a} Trno. If A is invertible1 then the rows of A and the columns of A each form a basis for R”;
thus l'l'IWL‘lJ = oolfA] = H“. [h] False. In Each the opposite is true: If H" is s. snhspsoe of V1 than t” is a subspace of H”
sinoe ever}! vector that is orthogonal to V 1arill also he orthogonal to W. {3} False. The speciﬁod condition implies that rowfsl} I; rowiB] from which it [011m that
nul1{A] = msEAJi ; mw[B:JJ' = nulHB) hut in general the inclusion Can he proper. is) False. For example A = i“ 1 0 ﬂ and B = [E ‘11] have the same row space but different column
spaces. {e} True This is in ﬁrst true for antr invertible matrix E. The rows of EA are linear combinations
of the rows of A; thus it is always true that row{EJl.J E tourist). If E is introstibia1 than
A = E‘1I[EA] and so we also have roar{A] Q row[EA]. if nuilﬁﬂ} is a line through the originr then row{A}I = nul]{A}J is the plane through the origin
that is perpendicular to that line.
If oohfﬂ] is a line through the origin1 than nullEAT} = oolEA'lllJ' is the plane through the origin
that is perpendicular to that line. (11} P3. If F is invertible1 then (PM.th = PIAJL} = ﬂ ifand only it Ax. = I]: thus the matrices PA and A have P4. the same null space, and so nuilityljFA} = nullitﬂAt From Theorem I13. it follows that PA and
A have also have the same rot:r spans. Thus rank[PA} = dimfnowiPA}}I = dimtrowmj} = rankL‘l]. From Theorem "(3.4 we have SJ = spanl[S}J, and [spanESJIJ'J'L =span{S] since spsn{S] is a
subspace. Thus [SJ'F' =[spanlISl1JJ‘ = spanES] ...
View
Full
Document
 Spring '08
 anony
 Linear Algebra, Vectors, Vector Space, Row echelon form

Click to edit the document details