Solutions:Shankar Chapter1, Some Exercises
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1 Solutions:Shankar Chapter1, Some Exercises February 6, Exercise Vanishing Functions F{The functions having property f0 fl 0} We can check whether the set of these functions form a vector space or not be checking all the properties one by one. 1. If fx and gx belongs to F dene h0 f0 + g0 0 fl + gl hl So fx+gx belongs to F. 2. IdentityThere exist a functin Ox 0,in F such that fx + Ox fx for all fx in F. So F has an additive identity. 3.Inverse For every fx in F there is fx such that and f0 fl 0 fx + fx fx + fx 0 Ox So there every element of F has an inverse. For scalars a and b, otther properties of a vector space hold trivially. 1.2 Periodic Functionsf0fL Let A be the set of all such periodic functions. i.e A{All fx such that f0fl} Now again check all the properties one by one. 1.Dene h0 f0 + g0 fl + gl hl 1
2 hence fx + gx is also periodic so belongs to A 2. IdentityThere exist a functin Ox 0, which is periodic and hence is in A. This Ox satises fx + Ox fx + 0 fx + for all fx in A. So A has an additive identity. 3.Inverse For every fx in A there is fx such that f0 fl and fx + fx fx + fx 0 Ox So there every element of A has an inverse. For scalars a and b, other requirements for A being a vector space are fullled trivially. You should notice that the 1.1 is a specice case of 1.2, i.e periodic functions which vanish at boundaries. 1.3 Functions f04 Let B be the set of such functions i.e B{All fx such that f04} 1.Dene h0 f0 + g So fx + gx is not in B. First thing goes wrong! Lets see what else. 2. Identity Clearly There does not exist a functin Ox in B which can serve as identity because being identity requires fx + Ox fx f0 + O0 f0 O0 0 and by the denition of B, such Ox are not in B. 3.Inverse Again no inverse for fx in B. Infact if there is no identity in B so it does not even make sense to talk about inverse! For scalars a and b and fx and gx in B So rest of the properties hold nicely for B but it is not a vector space because it is not closed under addition and does not have an identity. 2
3 2 Exercise A nite number of vectors 1 >, 2 >, 3 >,... n >are linearly independent i a i i > 0 a i 0 i for all i. Important thing to realize is that the zero on the right side is also a vector, i.e it is of the [ same nature as the vectors i >, if vectors are matrix the the zero is infact the additive identity matrix For given vectors: [ 0 1 a a 1 > +b 2 > +c 3 > 0 1 [ b 0 1 [ c 0 2 [ [ b 2c a + b c 0 b 2c [ b 2c 0; a + b c 0 So we have two equations for three unknowns and the relations c a; b 2a So the equation 1 is satised for any a, so we can have non zero solutions to a, b and c showin that the given set of vectors are not linearly independent. It would be worth noting that the just by a little bit inspection we can show that the 1 > as linear combination of 2 >and 3 >i.e 1 > 2 2 > + 3 > So you can just identify that to show that vectors are linearly dependant and don't have to use the abstract denition. 3 Exercise ,1 0 1,3 2 1 are linearly dependant First lets show this using denition: a b c a + b + 3c a + 2c b + c 0 This gives three equations: a + b + 3c 0; a + 2c 0; b + c 0 Again this yeilds solution to only two unknowns and we have: a 2c; b c; c c So we can have non zero values of a, b and c satisfying 2 showing that these vectors are not linearly independent. In fact one can see linear dependance as:
4 , are linearly independant Again using denition: a b c a + b a + c b + c 0 a b; a c; b c First two equations also demand that b c which according to the third equation require that: c c c 0 a 0, b 0 And hence these three vectors are linearly independant. 4 Exercise A 3 i + 4 j ; B 2 i 6 j Let C and D be the new orthogonal basis, we can always take one of the given basis as it is and make the other orthogonal with respect to it so take \overrightarrow{c}\overrightarrow{a} and after normalization we have: C C 1 C /. C 5 3 i + 4 j From here on we can follow the technique of Gram-Shmidt process, i.e subtract out the part of B which is along C. D B B. C C D 2 i 6 j i + 4 j After rescaling we will get: D i j 25 D D D. D So D and C are two new orthonormal basis i 103 j 5 Exercise 1.3.4:The Triangle Inquality We know that V + W 2 < V + W V + W > V + W 2 < V V > + < W W > + < V W > + < W V > V 2 + W 2 + < V W > +< V W > Since in general for a complex number c, c + c 2Rec i,etwice the real part of c using this fact: V + W 2 V 2 + W 2 + 2Re< V W > 4
5 Using the relation given in the problem: Re< V W > < V W > ; we will have V + W 2 V 2 + W < V W > Now using the Shwarz Inequality, < V W > V W : V + W 2 V 2 + W V W V + W 2 V + W 2 V + W V + W 3 Where the nal relation is the well known Triangle Inequality. For V>a W>,a>0 In this case, The Left side of Triangle Inequality is: V + W aw + W a + 1 W Since a > 0 a a and also a + 1 > 0 a + 1 a + 1; so V + W a + 1 W and the other side is: V + W aw + W a W + W a + 1 W So V + W a + 1 W V + W Hence the inequality in 3 becomes equality for V > a W >; a > 0. 5
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