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Unformatted text preview: Dr. Doom Lin Alg MATHl850/2050 Section 5.2 Page 1 of 8 Section 5.2. Subspaces. Deﬁnition: Subspace. A subset W of V is called a subspace of V if W is
itself a vector space under the addition and scalar multiplication deﬁned on V. In general, given a subset W of a vector space V, to show that W is a subspace
of V, each of the 10 vector space axioms have to be shown to hold for W.
However, because W is a subset of V and you use the same rules of addition
and scalar multiplication, most of the axioms are “inherited” from V. That
is, most of them don’t have to be checked. The next theorem states that, in
fact, only two things have to be checked. Theorem 5.2.1: Subspaces. If W is a set of one or more vectors from
a vector space V, then W is a subspace of V if and only if the following
conditions hold. (a) If u and V are vectors in W, then u + V [email protected] (b) If k is any scalar and u is any vector in W, then ku is. Example: Let W consist of the points on the line y = a: + 1, and so, W is
a subset of R2. Show that W is NOT a subspace of R2. Rawx LIST Lemme, we surouueu \AJ “3: amen buucL Ammo». So \0 Lc NOT A $U$$9k€£ or— ML. Dr. Doom Lin Alg MATH1850/2050 Section 5.2 Page 2 of 8 Example: The set of points W that lie on a line through the origin is a
subset of R3. To show that W is a subspace of R3, we need to show the
conditions of Theorem 5.2.1 are satisﬁed. \L) = {(Atpa‘z) 6 \Q? \ 6:13.): ﬁca,‘o,c)k
“"—‘ .
= A, wiecrfon vector. FoL we Lm€ .4 3 cuecv. ($0609.; uuoEG. Abbir80»: 'u: g 2.2 eUJ ,cuecv. 5.999“). u,\_ge\U News g¢CU\,UM\1~3 2< Ea k4 V:(U\JV1\\J3) & \J—— Sol For. sou; 5,956 19., aisacwsw .. G [L Meatmm cLosoac ween. scALAo. worm . 'Lv L’LQUJ & U—élﬁ ,cqecc ‘49 G“). Mm WWWM So lkgéwi elk Dr. Doom Lin Alg MATHl850/2050 Section 5.2 Page 3 of 8 Example: The set of all points on a plane is a subset of R3. From Example
6 on page 225 of text, every plane through the origin is a vector space. Thus,
every plane through the origin is a subspace of R3. Example: In the previous section we saw that the set that contained only
the zero vector was a vector space (called the zero vector space). Because
every vector space has a zero vector, the zero vector space is a subspace of
every vector space. In addition, the vector space V is a subspace of itself. Subspaces of R2.
0 the zero subspace
0 lines through the origin
0 R2 It turns out that these are the only subspaces of R2. Subspaces of R3.
0 the zero subspace
0 lines through the origin
0 planes through the origin
0 R3 It turns out that these are the only subspaces of R2. Dr. Doom Lin Alg MATH1850/2050 Section 5.2 Page 4 of 8 Example: The set P2 of polynomials of degree g 2 (p 2 19(33) 2 co + 013: +
02332) is a subset of F (—00, 00), the set of realvalued functions on (—00, 00).
Show that P2 is a subspace of F(—oo, oo). Cue=01. cmsum: ounce. A—ou'u(oﬂ = '\F i) & 16 P1,: LHECV— er€ ‘97. $0 LET Q) lg @Z , 5° ?5 C'+C‘7<I +C'LX1 7 i
1": J~9+Axx +A1X. ”HEVE 11% Ct, £ 24.15 Y'kl: (Cotoh)? @\+J.\x ‘\ Cc1+ A act So P.\.Lé P).
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Cl"EUL (4.050% most scnuw. Mum n: \F f4 ‘93, & \aéi (’ 7, \‘V" 64934 afﬁx4:31))!" So \koépz.s e ((1 e10. 9% So PL is A SOQSDACE OF gc—QJQ) . Dr. Doom Lin Alg MATH1850/2050 Section 5.2 Page 5 of 8 Theorem 5.2.2: Solution space of homogeneous systems. If AX = 0
is a homogeneous linear system of m equations in n unknowns, then the set
of solution vectors is a subspace of R". Example: Consider AX = 0 with X\ x1— x3 7. l ‘1— L ‘LL 2. \ Z l a? \ i/z ,\ O
A : o L o 4... o \ a, _\___, o k a a
7. o 1 9. an; L o 4 0 3131341 0 o o a M X‘s“k (Folké(ﬂ) GET 36,50) x\=$t x. \
so X; t k 0
X3    s
Sownou 661’ l$ A suesmce oF l2, . Proof of 171771522 6A1 V») is THE scum'qu 561' 0(— AE: Q CLosuw upbee. ADBiTCo» . LC g 2g é UL) (q a! Ate sownaus)
TBEA Lac—xv. 91+! evJ C 092 L5 A Sownon), Qt) \lé‘k) °D A8519. JAQ'JQ ACgurg): AchA! = 9+9. =9. $0 ‘03". Gull 6L0$ULE unoev. Smut MULT'M: , A(\¢Lﬂ= lL(Akl)= \LQ =Q I$0 legend Dr. Doom Lin Alg MATH1850/2050 Section 5.2 Page 6 of 8 Deﬁnition: Linear Combination. A vector w is called a linear combi nation of the vectors V1, V2, . . Where 161,192, . . . ,Vr if it can be expressed in the form W=l€1V1+l€2V2+“‘+k7~VT . , kr are scalars. Examples: 1. In R”, with the standard unit vectors e1, e2, . . . ,en ‘ n
W 19619 , £=(x\,xz,..~.xn)= x\€;.+x;€_z+~+xu€_n 2. Consider the two vectors u = (1, —1, 1) and V = (—1,2, 1) in R3. Show
that W = (1,0,3) is a linear combination of u and V and that X = (1,0,0) is not
F09. w wAuT scams L an, (em) Sucu. THAT
9=ke*%u
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2“ 9mm; . "lLrtl‘L'L =
7.“ emu’a : k\ + k1. =
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\ \ 0 251239“ O z .\ pa stationed. Dr. Doom Lin Alg MATHl850/2050 Section 5.2 Page 7 of 8 Theorem 5.2.3. If V1, V2, . . . ,Vr are vectors in a vector space V, then: (a) The set W of all linear combinations of V1, V2, . . . ,V7n is a subspace of V. (b) W is the smallest subspace of V that contains V1, V2, . . . ,V7n in the sense
that every other subspace of V that contains V1, V2, . . . ,V7n must contain W. @ Cuecv. meson: woeo. Abbi’rfcﬂ : '\F g: 3 \g Q“) , CHECK. guy eUJ, (Aﬁﬂéw =3 uﬁll.\\lli'\LL\l1\n.+\<r¥lv gr \E_=C\_j,\+Cz\l2+...+Cf\_f_r .— 2 wag: Q¢k+c,)\_I,JrQ¢2+Cz)\_J,\u+(kr+cr3\lr so [Lynn ékﬂl cued. CLO€>UIL€ uuoea. SCALAL MULT‘Q: “3 gem] , Véw. , meat lcgéUJ norm gr As ABNé' kg, = Q¢k32‘4(kkz)gz¥u.+Ckkr)\lr so ikgégl I Deﬁnition: Spanning. If S 2 {V1, V2, . . . ,V,} is a set of vectors in a vector
space V, then the subspace W of V consisting of all the linear combinations
of the vectors in S is called the Space spanned by V1,V2, . . . ,VT, and we
say that the vectors V1, V2, . . . ,V7n span W. To indicate that W is the space
spanned by the vectors in the set S, we write W = span(S) or W = span {V1,V2, . . . ,Vr} Example: 1. R3 = span{e1, e2, e3} 2. If W denotes the asyplane in R3, W = span{e1, e2} Dr. Doom Lin Alg MATHl850/2050 Section 5.2 Page 8 of 8 Theorem 5.2.4. If S = {V1,V2, . . . ,VT} and S’ = {W1,W2, . . . ,Wk} are two
sets of vectors in a vector space V, then span {V1,V2, . . . ,VT} 2 span {W1,W2, . . . ,Wk} if and only if each vector in S is a linear combination of those in S’ and each
vector in S’ is a linear combination of those in S. Example: Say 8 = {e1,e2}, S’ = {(1,2,0), (1,1,0), (3,1,0)}. Then
\a(o,l;o) 2‘ 51,, g;
Chow) span(S) = span(Sl) if 394%: so sueay vecmz is 5’ (s A Lincone. <29
g. 23.1 9 =<~Aly. + zynoyz €1 —= l \h "' L") ‘11 *0 93 so enemy vanon. \M S (Au 3.;
€XN£6$€D As A d». CQMB. oF . I UGCt°L§ ”£8 3: Meg): Wﬁs’). ...
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 Spring '09
 MihaiBeligan
 Linear Algebra, Algebra

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