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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] ...
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