Cosider the differetial equatio y '' k y 0 has particular solutios y1 si( kx) ad y cos( kx) I geeral, ay liear combiatio of y1 ad y, cy 1 1 cy where c1, c is also a solutio to the equatio above The reaso for this lies i Liear Algebra Through our study of Liear Algebra, we will m lear that the set of differetiable fuctios from is a vector space, which we will defie below Vector Space: A vector space is a set,v, alog with two operatios called vector additio, +, ad scalar multiplicatio with the followig properties (11)[Zero Vector] There exists 0V such that for all vv, v0 v (1)[Additive Iverse Vector] For all vv, there exist s vv such that vv 0 (13) [Associativity] If vwu,, V, the ( v w) u v( w u) (14) [Commutativity] For all vw, V, vw w v (15) [Vector Additio Closure] For all vw, V, vw V (15) [Scalar Closure] If c, the cvv (16) [Idetity Scalar] 1 v v for all v V (17) [Scalar Associativity] ( cc 1 ) v c1( cv) for all c1, c, vv (18) [Distributivity] cv ( w) cvcwad ( vwc ) vc wc Problems: First you will eed to defie these spaces 1) Show the set of all cotiuous fuctios, ( C ) 0, from m is a vector space ) Are the set of polyomials with real coefficiets, [ x], a vector space? Justify your aswer 3) Are the set of coverget sequeces,, a vector space? Justify your aswer
Subspaces ad Liear Combiatios Subspace: A subspace, W, of a vector space, V, is a subset of V which is also a vector space uder the same vector additio ad scalar multiplicatio Subspace criteria: To show W properties 1) The additive idetity 0 W V is a subspace, it suffices to show that W satisfies the followig ) If c1 ad c, w1 ad ww, the cw 1 1 cw W We call this property "closure uder additio ad scalar multiplicatio" Example: Show that the set of all polyomials of degree less tha or equal to, [ x ], forms a subspace of [ x] Solutio: The Zero polyomial is the polyomial whose costatly 0, so therefore, with all coefficiets zero To show closure, suppose there exists two polyomials p( x), q( x) [ x] The ap( x) bq( x) has real coefficiets, ad caot have degree greater tha Therefore, ap( x) bq( x) [ x] Problems: 1 Show that the set of all differetiable fuctios, ( C ) 1, is a subspace of the vector space of cotiuous fuctios, C ( ) 0 Determie which of the followig subsets are subspaces of the idicated vector space: a) {(1, y) y } i x b) {( xy, ) xy, {0} ad } (0,0)i y c) {( xyz,, ) xyz,, ad x yz0} i 3 d) {( xy,, z) x, y, z ad x y z 1} i 3
Liear Depedece ad Idepedece: We say a subspace, W, is geerated by a set of vectors { v1, v v } if for all ww, there exists real k 1, umbers a1, a,, a such that w akvk I other words, if every elemet i w is a liear combiatio of the v i 's We also say { v1, v, v } spas W or Sp( v1, v, v ) W However, we also would like the smallest possible set that spas W For example, cosider the vectors 1, x, x, x 1, x i the vector space [ x] It is clear that Sp(1, x, x, x 1, x ) [ x], but do we really eed all of them? We kow we ca write ay vector i [ x] as ax bx c where a, b,ad c are real umbers Does't this mea, by defiitio, that they ca be writte as a liear combiatio of x, x,ad 1? The we really do't eed the other two vectors What we are lookig for is a spaig set of liearly idepedet vectors It is easier, however, to defie liear depedece tha to defie liear idepedece Liear Depedece: We say that a set of vectors { v1, v, v } is liearly depedet if there exists a1, a,, a where ot all ai 0, such that 0 av 1 1 av av If this does ot occur, the we call the set of vectors liearly idepedet Why is this the desired property that tells us that we have o extra vectors? Well, from our example, we could see that the other two vectors could be writte as liear combiatios of x, x, ad 1 I particular, x 1( x ) 1(1) x1( x) (1) It looks a little silly, but ow we see that this meas x 1( x ) 1(1) 0 x 1( x) (1) 0 I geeral we ca do the same thig From this poit o, whe we metio zero, we mea the zero vector If a liear combiatio of a set of vectors equals zero, ad at least oe coefficiet is ozero, the we ca divide by that ozero coefficiet, ad isolate the correspodig vector, v This will yield a liear combiatio of the other vectors equal to v, makig v uecessary To show a set is liearly idepedet, we eed to show o such combiatio exists Usually, oe does this by cotradictio or by cotrapositive
Example: Show that the set of vectors {(1,0,0), (0,1,0) ad (0,0,1) is liearly idepedet Solutio: Suppose there exists a1, a, a3 such that a1(1,0,0) a(0,1,0) a3(0,0,1) (0,0,0) ( a, a, a ) (0,0,0) 1 3 This ca oly happe if a1 a a3 0, which meas that the set is liearly idepedet, by defiitio Basis: A basis of a vector space is a liearly idepedet spaig set of the vector space There may be may possible bases to a give vector space, but they will always have the same umber of elemets The proof of this is ivolved ad heavy o the otatio, so we will defer it to your liear algebra class The umber of vectors i a basis of a vector is called the dimesio of the vector space Problems: 1 Show that the set {(,1,0), (0,,-1), (1,1,)} is liearly idepedet Is the set of polyomials { x1, x, x 3x5} liearly depedet? Justify your aswer 3 Ca you fid a liear combiatio of (4,-,1) ad (-3,1,) that equals the vector, (6,-4, 7)? If so, give the coefficiets If ot, prove it
Liear Trasformatios A liear trasformatio is a fuctio betwee vector spaces, T : V W, such that Tav ( av) atv ( ) atv ( ) 1 1 1 1 This meas that liear trasformatios preserve the liear combiatios of vectors I particular, if the liear trasformatio is oe-to-oe, the the liear trasformatio seds a basis of V to a basis of W You will study this more i depth i Liear Algebra, so we will oly cover the more importat theorems for Differetial Equatios, ad leave most of the proofs to your Liear Algebra course All liear trasformatios ca be writte as matrices We have bee exposed to matrices with real m coefficiets i the past, ad they have usually bee liear trasformatios from to I geeral, if we have a liear trasformatio, T : V W, betwee two vector spaces, V ad W with respective bases { v1, v, v} ad { w1, w,, wm}, we ca create a matrix v1 v v w1 a 11 a 1 a 1 w a 1 a a am 1 am am3 w where Theorem: A liear trasformatio, T : V m W a) Tv ( ) 0 implies v 0 b) det( T ) 0 c) The colum vectors ad row vectors are liearly idepedet d) The matrix row reduces to the idetity matrix The proof of this theorem will be doe i Liear Algebra,is oe-to-oe if ad oly if the followig are true:
Problem: 1) Show that the compositio of liear trasformatios is a liear trasformatio ) Let T : [ x] [ x] be defied by T( p( x)) p' ( x) Show that it's a liear trasformatio 3) Let L: [ x] [ x] be defied by L( p( x) ) xp( x) Let T be defied as above Compute L T( p( x)) ad T L( p( x))