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Unformatted text preview: EXERCISE SET 14
4. The mduoed row echelon form for A is l I] 1 2 1
ti 1 I l 2
ﬂ CI [l [I U
I] E} I] i} [1 Thus ranlttA} = 2, and o general solution of Ax = [i is given by x = r{—1.—1.1,o,o}+ ski—1.0.1.0]+t[1.—2,EI,I],1}
It follows that nullitytﬁ'l} = 3. Thue renltirl] + nullitﬂA} = 2 + 3 = 5. a. [a] If A ie a. T 5: ti matrix hearing rank 5, then its nullity must he 9 — 5 = 4. Thee there are .5 pivot
variablm and 4 free parameters in o general solution :1fo = D.
{In} If A is en 8 3: l3 matrix havin nullity 3. then its rank must he 5 — 3 = 3. Thus there are 3 pivot
variables and 3 free oermuetere in o general solution of Ax = ﬂ.
{1:} If A in e. "r' x T matrix whooe row echelon forms have 3 IlﬂIleI'D rows1 then A has rank 3 and
nullity 'i' — .'i = 4. Thus there are 3 pivot variables and 4 free parameters in a general solution
of Ax = I}. 22. If u and v are nonzero oolurnn vectors and A = uvT, then Ax = {turzj = u[vTx} = {v  xju. This
Ax = {I ifand only if tr + x = I], i.e.. if and onlyr if): is orthogonal to V. This shows that kerﬂ'} = v .
Similarly. the range of T ooneiete of all vectors of the form An = [1r  :iu. and on rentT} = spen{u}. D1. {a} True. For example. if A is m x n where m :3: n [more rows. then oolumnei then the rows oi A form it set of m 'L'ECtﬂI'S in R“ and must therefore he linen.le dependent. [in the other
hand, if m :: n then the oolunme of A must be linearlyr deperrdeot. {h} Pelee. If the additional row is a linear combination of the existing tone. then the ten]: will
not be increased. (1:) Pelee. For example. if m =1 then renkﬁﬁ.) = 1 end nullityfA} = n — 1.
{:1} True. Sueh 3. matrix must have rank two than it; thus nullity{A} = n — mnhfﬂ) = l. {ell Felee. If .43: = h is iuoooeistent for tome h then A is not invertible and on An = I] has
nontrivial eolutione; thus nuilityfA} 33 1. [ﬂ True. We must hove rnnkEA} + nullityiA) = 3; thee it is not possible to have renhM) and
nuﬂity{.ﬁl} both equal to 1. DH. Ii' .1. 9E I]. then the reduoed row echelon from of A is l I] [l [I
{1101573
1: o 1—;
o o o o Ive;
Ulii
ﬂﬂﬂﬂ
[lﬂﬂﬂ and ED renitﬂ] = 2. Thus Jr = ﬂ is the value for which the matrix A has lowest rook. P4. Since the set VLJ W contains as vectors. it suﬂices to show that VU W is a linearlyr independent
set. Suppose then that 61,123,. . “Ci. and s1,s2.....s._t are seniors with the property that 1:11?! +£2V2+hr+ﬂk1fﬁ +d1W1 +dlwl ‘l' *' ' +dn—twnk =0 Then the vector :1 2 eye] + 32?: + + I:ng = —{d1w1 +dg‘ll": + + dn_j.w.._k} belongs both
to V = weld} and to W = nuIIIEA} = MWIEAIJ'. It follows that 11 ~ n = nH2 = {l and so n = ﬂ.
Thus wesimultaneously have c1v1+ citr2 +   r + curs = I] and tiles] + rigid'2 + ~ ~ r + reﬁnery]; = II]. Since the vectors in, em. .. .w, are linearly independent it follows that c1 = c; = 2 ct = I].
and since wlrwg,...,wﬂ_y are linearly independent it follows that d1 = on = m: 14,}, = [L
This shows that V U W is a linearly independent set.
EXERCISE SET 15
1 o s g 1%
o 1 o —E —”
2. The reduced roer echelon form of A is u D 1 1; TE. and the reduced row echelon lotto
" _3
(I {l I] ll ll 1 o n 1
U I ll 1
of AT is [II D 1 ﬂ . Thus diletowl'AJJ =3 and dim{col[21} =dimi'rowaTjj = 3. It follows that
{I o o o
I] I] U [l dietitians}; = s a s = 2, and diminulltﬁlfj} = 4 — s = 1. Since dimluulltﬂii = 21 there are 2 free
parameters in a general solution of Ax = Ill. 3. [a]: Since Iank'EA} f5 millILA I b] the system is inconsistent. _
(h) Since ranh{A} = rankA  b] = 2 the system is consistent. The number of parameters In. a general
snluticmisnr=4—2=2. I a
{c} Since molds} = ranlrht  h] : 3 the system is consistent. The number of parameters in s garter
solution is n — r = 3 — 3 = D. i.e.. the system has a unique solution. I
[cl] Since real1M} = raak[A h] = 4 the system is consistent. The number ofparameters in a general
solutionisn—r=T—i=3. 1 4: s1
. . ted matrix ofﬂnzheenhereducedto IJ —3l tog3hr
14 Theaugrnen u “HI—zhI'h
is either inconsistent [it til — 262 + b; at ll}. or has exactly one solution {if b1 — + ta =I D}. The
latter includes the case by = h; = h; = I]: thus the system A]: = [I has only the tnvrsl solution. . Thus the system Ax = b D4. {a} False. The row space and column space always have the same dimension.
{h} False. It is always true that rankfﬂ = rsnltIAT]. whether A is square or not. {e} True. Under these assumptions. the system Ax: h is consistent {for any h]: and so the
matrices A and [A  b] have the same rank. Ed] True. If an m x n matrix A has full row rank and full column rank, then m. = dimfrow(A}J =
leiHRH} = din[WHEN = n. {e} Time. If ATA and AAT are both invertible then. from Theorem 7.5.11]. A has full cohunn
rank and full row rank; thus A is square. if} True. The rank ate 3 x 3 matrix is I], 11 2, or 3 and the corresponding nullity is 3.2.1.0r [L P1. FDqu Theorem T.5.B{a} we have null{ATA] = nulllel}. Thus if A is m x n. then ATA is n x n and
so Maui’s} = n — museum} 2 n — sinners) = more}. Similarly, nullHAT] = nullUlT]
and so rsnk{A.¢ir} = m. — nullityLdAT] = m — nullityfAT} = rarrlcﬂrlT} = ranﬂel). EKEHGiEE SET 15 10. Let A be the matrix hating the given vectors as its columns. The roduoecl row echelon form of A is
l [I 2 ll —1 n 1 —1 I] 3
R: o u o 1 2
o o o o t From this weconelude thstthe sectors 1"”th V4 farm abaﬁlafm w = “HAL and that “3 = 2'“  v3
and. V5 # ''1 + 3V: + 2":1 14. The reduced row echelon form of A is R = . From this we conclude that the let. 2nd. I: I: N: un 0
[I
Ill :‘DDH
GDI—‘E and 4th columns of A are the pivot columnﬂ, and so these three columns form a basis for oolfA}.
The ﬁrst. three rows of R form a basis for row[A} = row{R}. We have Ax = I] if and only if Rx = ﬂ.
and the general solution of the latter is x = (£3, —§s. 3.0} = sié, —§_. 1.0]. Thus (g, —E, LU] forms
a basis for nullIEA}. The reduced row echelon form of the partitioned matrix [A I h] is 10—; Ujt uss
u 1 s we a s—s
I
o o o 1:; o—ﬁ —%
_ _ _ q _ _ _ _ _ _ _ _ ______......_________
o o 0 ago 1 —§ § Thus the vector [(1, 1, — g1 forms a basis for nullEAT}. 11:1:+ {a} The 1st, 3rd1 and 4th columns of A are the pivot oolumns: thus these sectors form a. basis for
cclﬁA}. {h} The ﬁrst three rows of A form a basis for row[A].
{c} We have Ax = {I if and only if R): = ﬂ,i.e.1i.fa.nd only.r ifx is ofthe form x = ll—‘urtrur = Ill—4r 1:010} Thus the vector u = [—4, 1,010} forms a basis for nul][A].
[d] We have ATJL = 9 if and only if C'x = II}, i.e., if end on]:r if x is of the form 1‘: [_%tl_gslﬂla!t} =tll_irorl]rﬂrl] +3lﬂl —ilulllﬂ]
Thus the sectors V] = %, 0,0,0. 1} and V2 2 {D1 —g,ﬂ, 1,0} form a basis for n11l{AT}I. D1. {a} lb} (d) If A is a 3 r; 5 matrix, then the number of lending 1‘s in a rote echelon form of A is at most
3, the number of parameters in a general solution of An: = D is at most 4 {assuming A ?E D].
theranlroiAieatmost :t. therank min!" is at most 3. and the uullityofATisat most 2
{amuming A 5,6 0}. If A is a 5 K 3 matrix, then the number of leading 1’s in a. roar echelon form of A is at most
3, the number of parameters in a general solution of Ax = D is at most 2 {assuming A .15 U].
theranlrofAieatrnost 3,,therankofAThtat most iamlthe uullityofATisat moetri
{morning A ;E D). If A is a 4 3: 4 matrix, then the number of leading 1‘s in a rear echelon form of A is at meet
at~ the number of parameters in a general solution of Ax = D is at. most 3 {assuming A 9‘. 0],
therankofAiaatmcat I1,theranlt ofAT isatmeet 4, and thenuuityorAT isat most 3
{assuming A 79 G}.
IfAisanmxnmatﬁx.theuthenumberofleatlingl’sinaronrechelonform ofAie atmost
m, the number of parameters in a general solution of A1: = D is at most Tl  1 {assuming
A at 0}. the rank of A is at most n:irt{.'rn,rt}1 the rank of AT is at most mln{m,rt}, and the
nullityr of AT is at most tn —1 {assuming A at 0]. D2. The pivot columns of a metrix A are those columns that corrmpond to the columns of a now
echelon form R of A which contain a leading 1. For ternnnole1 E} El —2 D 'l' 12 1 2 [I 3 '1] 'f
A = 2 4 lt'lI ﬁ 12 23 + R = '3 I] l ﬂ 0 l
2‘ 4 m5 6 —5 —1 U I] I] {l l 2 and no the lst1 3rd, and 5th Domains of A are the pivot columns. ...
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