2025pr6 - WMWge/la Math 2025, Problems (Fall 2003) 1)...

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Unformatted text preview: WMWge/la Math 2025, Problems (Fall 2003) 1) Determine) if the following set of vectors are linearly independent or not. If the vectors are linearly" dependent, write one of them as a combination 0f the others: a) (2,0,1),(1,1;O),(l,1,1) h) (1,1)] (1,““1) \ c) (1,—2,1),(1,2,1),(1;1,w~1),(1,0}1). d) (1,2),(m2,w4). e) (1.,1,2),(1,M1]0), (M2,O,1). f)(171,m1,1),(1,273,2),(152120). 2) Determine if the following sets of functions are linearly dependent on the interval [0, l.) or not. '51-) X§O,1/2):X[1/2,1)- b) X{0,1)=X[0.1/2)- e) (:ee(27rt), $119711). cl) m :c,g($) m 51:27 x 1. fl X[O,l/2)-, tel; t2,t+ 1. 3) Show that the vectors [1, and (—2, l) are a basis for R2. Then determine the cenetant (1,3) such that (5, m9) : a{l, 2) + fab-2,1) . 4) Show that the vectors (l7 el, 1), (l, 170) (l, 71:. 7‘2) form an orthogonal basis for R3 7 and then determine the cozzstents a, b, c E R such that (2, —4:? 3) x 03(1, —1, 1) + M}, 1., G) + C(l, -—l, —2). 5) Show that the vectors [1, 2, U)J (l? 1: l), (1, O, 1) form a basis for R3. Then determine the constants ct, Z35 C E R such that (4: 2) x mm) + 5(1,1,i:)~+~c(13071)- 6) Which of (he following Sets; is, generating for R2? a) (1,1) b)(~1;1),<1§217 ) (1,1), (0,1), (741,0). (:1) (5,1),(——1,5). 7) Which of the following sets is generatizzg for 1113? O a) (1, —-1, 1), (1, 1,8), 1)) (1,1,M3),(1,2,1),(1,~1,1) (3) (1,1,0), (1, -—-1,0),(1,0,—1),(2,1,0). 8) )erte the following vectors as :3 1111881" combination of (1, 1) and (1, W 1): J (2, 4), b) (333), ) (5,~w3). 9) Express the following; vectors as a linear combination of ('2, 1, G), (1, (l, l) and O, 1, {1): a) (1,0,8), 1)) (3,1,0), 0) (5, -1, 1). {3) (2,3,2). 10) Which of the, fol1owing sets is a basis for R2? 21 O 6) (3,6),(w7,14). 11) \Nhioh of the, following sets is a basis for R3? a) (1, 2, 1), 7'1, 1). b) (1,0,1),(1,1_,~~1),(2,0,1). c) (1, 2, 3), (1, 91, ‘ ), (2, G, G), (2, w1, 2) Math 2025, Quiz #4 (Fall 2003) Name: 1) Determine: if the following set of vectors are Eineai'ly independent or Hot. Give a reason for your answer. Functions are elements of the space of piecewise continuous functions on {9,2}. a} {1,0 1),{l,1,0)}{3,1,1} L I b)(171),(1,m12. OyWofiomccQ)‘r\-Qmu Li I c) {1,7Li.f2?1),(2,2). 3 MM as {a} m cowaxjs LB d) Xial)? XQOJ/gj e) .23, 2:2 and x3 2}_W§riich of the foélowing sets of vectors is generating for R3? 3,.) (1,0,1), (1,0,-—1),(0;1,0). Emma b)(1,1,0);(1,2,:>,<o,151>. 1-3 i W 304 WW3 (w 3 H mm“ R 412mm WWCGW) 3) Show that the vectors (1., —1, 1), {1, 1‘ O); (1, —1,—2} form an orthogonal basis for R3 hand then determine the constants c, iv, o E R such that (2, W4, 3) m a(1,—1,1)«E~5(1;1,G)+ {3(1, ~1, ~2). (2,-~1[3>‘c{,-W g+v+3 m 3 CL? .3: 3‘" 3 {V (xi—ms). mo} 3 2 —-‘-/ r m, .-— 9“ a C: (fly-'L'figjp (Cm/1M2) in 4 Math 2025, Quiz #5 Name: 1) Which of the following sets of vectors is linearly independent? (1) (1,0, 1), (0, 1,0), (1,0, W1). They are past wise orthogonal and hence linearly indepen- dent. (2) (1,0,0),(1,w1, 1),(1,2,2), (1,2,5). Four vectors in R3 are eiways linearly dependent. Another solution is to notice that 9 (1,2,5) : when) + 2) Which of the following sets is a basis for R2? lQiDJ (1,m1,1)+£(1,2,4) (1) (1,0), ( 1, Basis. The two vectors are perpenticnlar and hence linearly independent. In R2 any set of two linearly independent vectors is a. basis. (2) (1, 1), (1, 0), (1, —1). Not a basis. The set is generating but not lineariy independent. 3) Which of the ioiiowing sets is a basis for R3"? (1) (1,0,0), (1,1,0), (1,1,1). The set is linearly independent. The equation 01(17090) + 62(11 Wi“ (33(1: 1: 2 (Cl Mi“ C2 + (Bic? Mi“ 63363) Z: give: c1 + Cg + (:3 m 0 C2 + C3 = 0 c3 = U . The last equation gives c3 2 0. Then the second implies that c; : O, and then finaliy the first gives c, 2 0. Hence all the constants have to be zero. (2) (1,2, 0), (1, (l, 1). This is not a basis. The vectors are linearly independent, but we need three vectors for a, basis for R3. 4) Write the foliowiiig vectors as a linear combination of the vectors (1, 1) and (1, W1). (3(ZQe3flfltdteU (1, —3) : —(1, 1) + 2(1, -—1) Solution: Notice that 1i(1,1)|§2 2:: ((1, M1)H2 = 2 and that <(1,1),(1,~+~1)>=1m~ 1 = 0. Hence they are orthogonal. Thus, if fi : (3:, is a vector in R2 we have y) : c1(i, 1) + 62(1, —1) with 9+ em<ctexto>/2=L,y and c2 m< (3:,y),(1,w1) > /2 z t" g 1" if "3 = (2,4) this gives (:1: Egg 2 3 and c2 = = wl. if T)? = (1, m3) we get c1: : ii and m 13 Math 2025, Problems / linear independent, generating Recall that a set of vectors 111} ugfi . . . Inn is called linearly independent if C1111 "i" €21.12 “i” .1. + onun : 0 implies that all the numbers cl: . . . ,cn are zero. if the vectors are not linearly independent then they are called linearly dependent. Remember that a set of vectors is linearly independent if and only if one of the vectors is a linear combinations of the others. 1) Which of the following set of vectors is linearly independent? If the set is linearly dependent express one of the vectors as a linear combination of the other. (1) (1,2), (W13). Linearly independent If 01(1,2) + C2(W1,2) 2 (cl w 202}2C1+ 2C2) m (050) then 01 M 2(22 2 0 2c} + 202 : 0 Adding we get 301 = 0 or Cl = 0. The second equation gives c2 2 Mel. Whence 02 m 0. (2) (1, 0), (1,1), (2,1). Linearly dependent. Every set of three or more vectors in R2 is linearly dependent. (3) (1,1,0), (22,0), (0,0,1). Lineariy dependent: (2,2,0) 2 2(111,0). (4) (1,—1,1)) (170,—1), (0,170). Linearly independent: If 01(1, “*1, Jr Cg(1,0, “1) t (33(0D = (01+ ngwc1~l~ 83,61m Cg) =5- then c}_ + CE W 0 —c1 + (:3 : 0 c1 — (:2 m 0 Adding the first and the last equation gives 201 m 0 or c1 = 0 The second and the equations says that 02 _ c1 cg — (:1 Hence CQ 2 0 and (:3 = 0 (5) (170,1), (11,0; wl), (0,170). Linearly independent: if cl(1?0) + 620-? D: —1) 1: 3 (Ci CZ>CBa Cl M C2) fl (0: Then 61 Jr“ Cg = O 63 = 0 (1 W Cg -—— 0 The second equation gives C3 3 0 Adding the first and the third gives 201 2 U or (:1 r 0 The third equation then implies that (32 m D (6) (1,070): (1,1,0) (1,1,1): (1,2,0). Linearly dependent. Every set in R3 containing more .a then three vectors is linearly dependent. 2) Express the following vectors as a linear combination of (1,1) and (1, O): (1) (1,2) 2 2 X (131) — (1,0). How would we solve it? If (1,2) : (1(1, 1) + b(1,0):(a + he) Then a + b : 1 (L m 2-. The second equation gives the number a. Snbstraetion gives the second number h: bm~1. (2) (3) (1,m1) = w(1,1) + 2 x (1,0) 3) Express the following vectors as a. linear combination of (2, 1) and (1) W1): C l 1 (1,0):~§(2,1)+%(1,w1).1f (1,0) = a(2,1) + 5(1, m1) e (201 + he — b) (—1,2)22 >< (1,1)—3>< (1,0) Then The second equation gives that a 3.2"." Adding the second equation to the first gives 3a, : 1 07" a, : 1/3 (2) (0,1) 2: %(2,1) m gnu, W1). Just as above we get 201+?) = O e—b :1 Addng gives 3a rs 1 or a 2 1/3. Then the second equation gives 1 2 b: ,—1:——1m~m 0 3 3 (3) (2,2) 3 39,1) — §(15—1) We have 2a+b 2 2 aw?) 2-:2 2 Or 3&24 awe/3 Hence 4 -—2 a 3 3 (4) (2—2) 2 2 >< (Ii—1) 4) Express the following vectors as a iineai combination of (271,0)? (1:03 1), (0,1;0). The equations that we get if we set (any, 2) = (1".(2> 1, 0) + 5(1,071)+ C(U, 1,0) : (2a+b?a+c,b) are 2e + b x 3: a + c : y b : z From the third equation we get I) r: 2 Putting this information into the first equation gives 2ewmmb=mmz or 1 a m m z) 113 Now put this into the Second equation to get cry—amy—§($—z) (1) (1,0,0) 2 é(2,1,0) w §(O,1,0). Using the above we get 1 1 a = b 2 2:0 1 1 c = ye§(miz)=i~2~ (2) (0,1,0) 2 (0,1,0) Here (135:: (3.11:1 (3) (0,0,1) m —§(2,1,0)+(1,0,1)+%(0,1,0). Here 1 1 a m §(m—z)m—§ b : 2:1 1 1 c = ym§1wmzl=§ (4) (2,2,—1):g(2,1,(})—(1,0,1)+%(0,1,0). 1 3 2 “2+1 2“ a 21 l 2 b = i1 1 3 1 ‘ X 2—"2 1 C“ 2( +1 2 2 We can also solve this by noticing that the solution is 2 ><(the solution in 1)+2 X (the solution in 2)—(the solution from 3) (5) (0,1,1) : —-;~(2,1,0) +(1,0,1)+§(0,1,0) Recall that a set of vectors 111, . . . , un in a vector space is called generating if every vector u in the space can be written as a combination of those vectors, i.e., we can find scalars (:1, .1 . ,cflrL such that u r: clul + c2112 + . . . + emun 5) Which of the foilowing set of vectors generating for R2? (1) (1,0), (0,1). Generating (2) (1,1),(2,2). Not generating (3) (23111-2 (4) (2:0);(11 6) Which of the following set of vectors is generating for R3? (1) (1,0,2),(O,1,U),(0,0,2). Generating (2) (1: M1, 2), (13 0, 0). Not generating. We need at least three vectors to generate R3. 11). Generating 1), (3, 1). Generating, but we only need two of them (3,1) 2 (2,0) + (1,1) 17 (3) (1, 2,3), (1,~m1,{J), (0, 1,1). Not generating because (1,13% 3(8,1.1):(:.e1,0> So those three vectors are linearly dependent, but we need three lineariy independent vectors to generate R3. Recall that a set is called a basis if it is generating and linearly independent. 7) Which of the following set is a basis for R2? (1) (1,f}),(0,1). Basis (2) (1,2), (—1, .1). Basis (3) (1, 1), (2, 2). Not a basis, because linearly dependent. (4) (—1,1). ot a basis, because not generating. A basis contains exactly two elements! (5) (1, 2), (*1, 1), (1, 0). Not a basis. It is generating but not linearly independent. 8) Which of the following set is a basis for R3? 05 2)? (17 17 0)? 07 (2) (2,0,1),(0,1,(}) (3) (2,0,1), (mi, 1,0), (1,0, 1),({}, 1, 1) (4) (1,2, 71), (0, 71,2), (1,0,3) 9) Let V be a vector space with an inner product and let 111, . . . ,un be a set of non-zero vectors of length one which are orthogonal to each other. Thus < 11,-, u, >2 0 if i % j and < 11,-, u,- >2 1, Are then the following statement correct or not? (1) Then the vectors 11;, . . . ,u,, are linearly independent. (2) If the vector u is in the span of u1,...,u,,, is, u can be written as a combination of 111, . . . ,0.” then we much have u:<u,u1>u1+.,.+<u,un>un SOLUTION: Both statements are correct. Assume that c1u1+02+...+cnunmu for some vector 11. Take the inner product to get: 7': < E lelj,11é> 2:1 <u,u,> l l ' M ~37 /\ Lg: j; v FE _.o_ :53" l I g; 18 Where we have used in the third line that < 1133117; >: G if j 31$ :5. In the fourth line we used that < 117;,111; >: E. This Show that the second statement is correct. In the first statement we assume that u 2 0. Then < u, 111- >2 G for a1} 3' m 1, . . . ,n. Hence all the numbers C7; are zero. But that means that the vectors are linearly independent. ...
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This note was uploaded on 11/23/2011 for the course MATH 2025 taught by Professor Staff during the Spring '08 term at LSU.

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2025pr6 - WMWge/la Math 2025, Problems (Fall 2003) 1)...

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