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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) + fab2,1) . 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>. 13 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) (231112 (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 nonzero 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.
 Spring '08
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