SolutionToHW2

# SolutionToHW2 - 2.2 Consider the set of all polynomials in...

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Unformatted text preview: 2.2 Consider the set of all polynomials in s of degree less than or equal to four, with real coefﬁcients, with the conditions: P'(0)+P(U)= 0 Pa): 0 Show that this set is a linear vector space, and ﬁnd its dimension. Construct a basis f0r this Space- Solution: Since any polynomials with a deﬁned degree forms linear vector space, to show that above set of polynomial IS a linear space, we have to prove it is a subspace of the polynomial spacev So, we have to prove 6P1(x)+bg (x) e P(x) where a and b are scalar values and Plifx), Plot} c P(x}. i) Let L(x) = aﬁ(x)+ 2:50;) and am: am + a,",x my:2 + my:3 +al'dx4 for f = 1,2 3.1.10) + Lw) = new + bum + no” + bar” = 0mm + al‘l)+ Ham + a“) = a(*"1’(0}+i’1(0}J+ bU’iWH 5(0)) = 0 and, .'.L(l) = an”, + ham Hm” + b 02‘, + aol‘z + bani-1:10L3 + but3 +a a” + be;4 = atom + am + all + am 4» al14J+ Now + (111+ an + a” + a“) = aﬂ(l}+bP2(l] = 0 LU} E P(x) P(x) forms a subspace. ii) P (U)+P[0) = 0 and P0) = 0 implies 1‘1000 Tc- 11111P‘JHPQP3P‘]=0 The dimension of the null space of P{x) is 2. So, the dimensionality of the space: P(x) is S-2=3 because a degree 4 polynomial has 5 dimensions. 2.4 2.5 Are the following vectors in 91‘ linearly independent over the reals‘? 2 l l f: , j , and _; 3 0 2 Solution: Let 2 l l 0 —-3 i y = —1 4 —2 3 0 2 Since [email protected])=2, only two of the vectors are linearly independent. So, the set is not linearly independent. Which of the following sets of vectors are linearly independent? —i l 2 a) —9 , 3 F2 in the space ofrea] 3-tuples Over the ﬁeld of reals. 0 0 1 Solution: -1 I 2 —9 3 —2 = 6 LineaIly independent 0 0 l b) {[2 Vii], [1 + [ Ff in the space of complex pairs over the ﬁeld of reals, —l' —t 3 + 4: I Solution: 2—: H2: —:' o I _ _ “l H lelwlelolta—M“ Thus, they are linearly independent c) {252 + 25 — l, e 232 + 23 + s, 52 - s - S} in the space ofpoiynornials over the ﬁeld ofreals, Soiution: 0(252 +25—l) +b(—2:2 +23+5)+c(52 —s—5]= meta: b= c=0 Thus. they are linearly independent 2.10 For the following matrices, ﬁnd the ranks and degeneracies. l 2 -3 a) -2 2 2 4 1 -4 Let 1 2 —3 = -2 2 2 4 1 —4 AA) = 3 and degeneracy q(A) = 0 l 2 —3 Basis for the range space= —2 2 2 4‘ l —4 Basis for the null space: {E} 020 b)200 020 Let 0 D 0 MON 0 A=2 0 M) = 2 and degeneracy 4104) =1 D 2 Basis for the range space: 2 0 0’2 0 Basis for the null space= 0 l 1 6 3 2 -—2 c) 2 12 4 6 —10 3 l8 0 15 —15 ‘ Let I 6 3 2 —2 A=2 12 4 6 —10 m4; M33 3 18 0 15 —15 '0" r(A):3 [email protected] 901): 5—3 = 2 I‘M-n. Basis forthe range space new” ’35 MMT‘]. BNCC: ﬂ 74;“ [6/5444 (E) .1938 -.5869 .736! r ,4 2: ,l1 =011h(A)= .5101 —.6242 ~5918 { ) m m" 3)) )d My .8330‘ .5157 ‘ .1734 Dbﬁwm/frs IKE.) I?" —.9398 0 “(J/’3 2020 7’r5iMIUC .1225 .5071 a / ’ Basis for the null space: null(A)= .0510 —0.6092 .0510 —0.6092 0 G 2J2 If it is known that a valid basis for the space of polynomials of degree less than or equal to three is {Lm‘2 , t3 } , show that every such polynomial has a unique representation as p(!‘) : a3r3 + a2t2(1— r)+a,:(1 — :)2 +ao(l —l)3 Solution: We can write any polynomial into pa) = 133:3 + 232:2 + b1: + be p(t) = (23:3 +a2:2(1—:)+ altﬂ -r)2 +a0(l —t)3 2 = a3:3 + a2! — {23:3 + alt— a1t2+ 01:3 +00 “3a0t+3a0t2 —a0t3 =(a3 ~ar2 +a‘I —aro)t3 +(a2 —al +3a9)r2 +(al —3a0):+ao Thus, [a3 a2 01 a0][r3 :2(1—:):(1—:)2 (1—03]T =[(a3—a2+arao) (az’aii‘3ao) (oi—3cm) ao][r3r2r1]T Therefore, b3 1 —1 l —1 a3 b2 0 1 —1 3 a2 b, = 0 0 1 "a a] 150 0 0 0 i an 1 —l I —l 0 1 ~1 3 _ I _ ‘ 0 0 l #3 \$0 So, there 15 an umque representation for every polynomial. 0 0 0 1 ...
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SolutionToHW2 - 2.2 Consider the set of all polynomials in...

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