308solutionhw4 - EXAMPLtrS OF SUBSPACtrS 65 30. Since d is...

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Unformatted text preview: EXAMPLtrS OF SUBSPACtrS 65 30. Since d is in both U and I{ 0 : 0* d is in U *V. Suppose x a1d y are in U +V andwritex:ul fvr,y:rr2+v2 whereul ,u2 areinu andvl ,v2 areiny'. Thenx fy: (ur+.tz)+(.rt+v2) is inu-tv. If a isascalarthenax: aurfavr isinu-lv. It follows that tI * V is a subspace of R". 31' clearlv g is in tl ov. suppose x and y are in u nv. Then x and y are in [/ and since [/ is a srrbspace, x +y is in t/. Also for any scalar a, ax is in I/. Similarly, x fy and ax are in 7' Therefore x fy and cx are in (l a v. It follows that u o v is a subspace of R". 33' (u) Clearlvdisintluv. Srrpposex isint/Uv andleta beasca,lar. Ifx isinl/, then ax is also in U' Similarly, if x is in l/, then ax is in V. In either case, ox is inUUV. (b) Assume that u *v is in (/. Since -u is in t/ and tI is closed nncler addition we see that t- ("+v)+ (-u) is in U. This contradicts the assumption that v is not in Lr. Sirnilarly, u 1v is not in Z. 34' Since tr4l is non-empty, W contains a vector x. By (s3) the vector 0x: 0 is in tr4l. By Theorem 2,W is a subspace of R". 3.3 Examples of Subspaces 32. The vector u : are in UUV btft 1. By definition Sp(.9) , then x is in Sp(S) equationr-lA:0. is in [/ and the vector v : irr neither IJ nor V. f0.l I I I is irr V. Tlrus u arrd v Lrj f 'l | -1 | L o.J u*v is ,, 1 rz) the : ,,| _1 ] if and only 2. BydefinitionSp(S) : {rf ?l, t anyrealnumber}. If x: f "r l sp(s) tJ,i: J 3r ! 2y :9. : t any real nurnber 1. Thus if x : ifrtlrz:0. In particular Sp(S) is in R2 line with 4. is in .R2 , then ne with equation- tOl Sp(J) : tf l01,, anyrealrrumber): {e}.Sp(.9) isthepoint(0,0). Sp(s) : {t, :1' t x : klai_ kzb for scalars k7 and, k2}. For an arbitrary vector x in R2, t : I :.: I,x: ktai- keb if kj : Srt + Zrp and, ke: -rt - rs. It follows that Lrz l Sp(.9) : 62. tt0 CHAPTER 3. THE VECTOR SIPACERI{ sp(s) : {*' ; T',r*",,;iori, arkzd: x }ras augmented matnx [ -'1 0 rz ) in R2' " : L t? I -- , rt follows rtiat Tnit T*""'"ir"T' j"u o*ur"lving vields k1 : -r2' and k2: 11 + 12' lt lo r h+r2) lni'i:?' Irr ] theequationkla+kzc:x hasaugmentedmatn" L -l 2 rz ) 6. Ifx:L;; I |1 _,2 r, l,sotheequationhasasotutionifandonlvif This matrix reduces tt \ O 0 ri * *" \'"" '- ^'', *,5u, is, Sp(S) is the line with nt *rz- 0' It follows iftat Sp(S) - {*' t1*nz: u}i t equation r*U:o' T. sp(s) : {* in R2 : x : krb * n," .,^:::"i"fi , ,"1"1];.l"ii;:; ofr.::il'] *:ilJdifiJ#*di: :'ol "n.''' sp(s) : {x : Srt * 2r2 : 0\; so SP(S) is the [ " ] the equatio ttkla+kzb+ ksd: x has augmented matrix B. If x: Lr, I I-l -',1 :t\ ":.:::::::H,;j,1,-? :A;'ll; h toUo*"that the.system is consrsterrL tur -" - \ \rr:'' - i "t l 9.Sp(S):t*inR2:7:k1b+kzc+ksdforscalarskl'k2'k3\'Withx:L"l' the equation k1b *.k" + ksd: x has augmented matrix | -3 -ri '*'r\ The red'uction reveals that the system is consistent for every x ' so iP(s) : R2' 10....
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This note was uploaded on 05/30/2011 for the course MATH 308 taught by Professor Milakis during the Spring '08 term at University of Washington.

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308solutionhw4 - EXAMPLtrS OF SUBSPACtrS 65 30. Since d is...

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