n α j x j = 0 j=1 has a nontrivial solution. Here A is the n k matrix whose jth column is the vector for all t j=0

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1 MODULE 2 Topcs: Lnear ndependence, bass and dmenson We have seen that f n a set of vectors one vector s a lnear combnaton of the remanng vectors n the set then the span of the set s unchanged f that vector s deleted from the set. If no one vector can be expressed as a combnaton of the remanng ones then the vectors are sad to be lnearly ndependent. We make ths concept formal wth: Defnton: The vectors {x 1, x 2,..., x n } V are lnearly ndependent f α j x j = 0 has only the trval soluton α 1 = α 2 = α 3 = = α n = 0. We note that ths condton says precsely that no one vector can be expressed as a combnaton of the remanng vectors. For were there a non-zero coeffcent α k then x k can be expressed as a lnear combnaton of the remanng vectors. In ths case the vectors are sad to be lnearly dependent. 1) Gven k vectors {x } k =1 wth x R n then they are lnearly dependent f the lnear system α 1 x 1 + α 2 x α k x k = 0, whch can be wrtten as A α = 0, has a nontrval soluton. Here A s the n k matrx whose jth column s the vector x j and α = (α 1,..., α k ). 2) The k + 1 vectors {t j } k j=0 are lnearly ndependent n Cn (, ) for any n because f k α j t 0 j=0 for all t then evaluatng the polynomal and ts dervatves at t = 0 shows that all coeffcents vansh. 6

2 3) The functons sn t, cos t and cos(3 t) are lnearly dependent n C k [a, b] for all k because they are k tmes contnuously dfferentable and cos(3 t) = cos 3 cos t + sn 3 sn t. The frst example shows that a check for lnear ndependence n R n or C n reduces to solvng a lnear system of equatons A α = 0 whch ether has or does not have a nontrval soluton. The test for lnear dependence n a functon space seems more ad-hoc. Two consstent approaches to obtan a partal answer n ths case are as follows. Let {f 1,..., f n } be n gven functons n the (functon) vector space V = C n 1 (a, b). We consder the arbtrary lnear combnaton H(t) α 1 f 1 (t) + α 2 f 2 (t) + + α n f n (t). If H(t) 0 then the dervatves H (j) (t) 0 for j = 0, 1,..., n 1. We can wrte these equatons n matrx form W (t) α = f 1 (t) f 2 (t) f n (t) f 1(t) f 2(t) f n(t) f (n 1) 1 (t) f (n 1) 2 (t) f n (n 1) (t) α 1. α n = 0. If the matrx W s non-sngular at one pont t n the nterval then necessarly α = (α 1,..., α n ) = (0,..., 0) and the functons are lnearly ndependent. However, a sngular W at a pont (even zero everywhere on (a, b)) does not n general mply lnear dependence. We note that n the context of ordnary dfferental equatons the determnant of W s known as the Wronskan of the functons {f }. Hence f the Wronskan s not zero at a pont then the functons are lnearly ndependent. The second approach s to evaluate H(t) at n dstnct ponts {t }. If H(t ) = 0 for all mples α = 0 then the functons {f j } are necessarly lnearly ndependent. Wrtten n matrx form we obtan A α = 0 7

3 where A j = f j (t ). Hence f A s not sngular then we have lnear ndependence. As n the other test, a sngular A does not guarantee lnear dependence. Defnton: Let {x 1,..., x n } be a set of lnearly ndependent vectors n the vector space V such that span{x 1,..., x n } = V Then {x 1,..., x n } s a bass of V. Defnton: The number of elements n a bass of V s the dmenson of V. If there are nfntely many lnearly ndependent elements n V then V s nfnte-dmensonal. Theorem: Let {x 1,..., x m } and {y 1,..., y n } be bases of the vector space V then m = n,.e., the dmenson of the vector space s unquely defned. Proof: Suppose that m < n. Snce {x j } s a bass we have m y j = α j x for j = 1, 2,, n. =1 Snce the matrx A = (α j ) has fewer rows than columns, Gaussan elmnaton shows that there s a non-zero soluton β = (β 1,, β n ) of the non-square system Aβ = 0. But then m β j y j = α j β j x = 0 =1 whch contradcts the lnear ndependence of {y j }. ) If {x 1,..., x k } are lnearly ndependent then these vectors are a bass of span{x 1,..., x k } whch has dmenson k. In partcular, the unt vectors {ê }, 1 n, where ê = (0, 0,..., 1, 0,..., 0) wth a 1 n the th coordnate, s a (partcularly convenent) bass of R n or C n. ) The vectors x = (1, 2, 3) and x = (2, 1, 4) are lnearly ndependent because one s not a scalar multple of the other; hence they form a bass for the plane 11x + 2y 5z = 0,.e. for the subspace of all vectors n R 3 whose components satsfy the equaton of the plane. 8

4 ) The vectors x = t, = 0,..., N form a bass for the subspace of all polynomals of degree N n C k (, ) for arbtrary k. Snce N can be any nteger, the space C k (, ) contans countably many lnearly ndependent elements and hence has nfnte dmenson. v) Any set of n lnearly ndependent vectors n R n s a bass of R n. The norm of a vector: The norm of a vector x, denoted by x, s a real valued functon whch descrbes the sze of the vector. To be admssble as a norm we requre the followng propertes: ) x 0 and x = 0 f and only f x = 0. ) αx = α x, α F. ) x + y x + y (the trangle nequalty). Certan norms are more useful than others. Below are examples of some commonly used norms: 1) Settng: V = R n and F = R or V = C n and F = C ) x = max x 1 n ( ) 1/2 ) x 2 = x j 2 ) x 1 = n x j v) Let C be any non-sngular n n matrx then x C = Cx 2) Settng: V = C 0 [a, b], F = R v) f = max f(t) a t b ( b v) f 2 = a f(t) 2 dt v) f 1 = b a f(t) dt ) 1/2 To show that 1-) satsfes the condtons for a norm consder: x > 0 for x 0 and 0 = 0 by nspecton 9

5 αx = max x + y = max max αx = α max x = α x x + y max{ x + y } x + max y = x + y. The verfcaton of the norm propertes for 1-) and 1-v) as well as for 2-v) and 2-v) s also straghtforward. However, the trangle nequaltes for 1-) and 2-v) are not obvous and wll only be consdered after we have ntroduced nner products. ) x 2 for x R 3 s just the Eucldean length of x. ) x 1 s ncknamed the taxcab (or Manhattan norm) of a vector n R 2. ) f 2 s related to the root mean square of f defned by rms(f) = f 2 / b a whch s used to descrbe, e.g., alternatng electrc current. To gve an llustraton: The voltage of a 110V household current s modeled by E(t) = E 0 cos(ωt α). Let us consder ths functon as an element of C 0 [0, T ] where T = 2π/ω s one perod. Then E 2 = ( E 2 / T ) T = 110 T where the term n parentheses s the root mean square of the voltage.e., 110V. Snce also, by drect computaton, E 2 = E 0 T/2 we see that E 0 = and hence that the peak voltage s gven by E = E 0 =

6 Module 2 - Homework 1) Let x 1 = (, 3, 1, ), x 2 = (1, 1, 2, 2), x 3 = ( 1,, A, 2). Prove or dsprove: There s a (complex) number A such that the vectors {x 1, x 2, x 3 } are lnearly dependent. 2) Plot the set of vectors n R 2 for whch ) x 1 = 1 ) x 2 = 1 ) x = 1. 3) For a gven vector x = (x 1,..., x n ) C n wth x 1 show that f(p) x p = x j p s a decreasng functon of p for p [1, ). Show that ths result s consstent wth your plots of homework problem 2. 4) Fnd an element x n span{1, t} C 0 (0, 1) such that x 1 = 1 x 2 = 1. Is ths element unquely determned or are there many such elements? 5) Show that the functons e αt and e βt are lnearly ndependent n C 0 ( 5, 5) for α β. 6) Prove or dsprove: The functons f 1 = 1 + t, f 2 = 1 + t + t 2 and f 3 = 1 t 2 are lnearly dependent n C 0 (, ). 7) Let f(t) = max{0, t 3 } and g(t) = mn{0, t 3 }. ) Show that these functons are lnearly ndependent n C 0 (, ). ) Show that the Wronskan of these two functons s always zero. 8) Let f(t) = t(1 t) and g(t) = t 2 (1 t). Let H(t) denote a lnear combnaton of f and g. Show that H(0) = H(1) = 0 but that the functons are lnearly ndependent. 1/p 11

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