Abstract Vector Spaces. Abstract Vector Spaces
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1 Astract Vector Spaces The process of astractio is critical i egieerig! Physical Device Data Storage Vector Space MRI machie Optical receiver Icreasig astractio 6.1 Astract Vector Spaces Dimesio of a vector space Defiitio: A vector space is -dimesioal if it possess a set of idepedet vectors, ut every set of + 1 vectors is a depedet set Defiitio: If for every +, we ca fid idepedet vectors i, the is -dimesioal. Defiitio: A fiite set of vectors e 1,, e is a asis for a vector space if: (a) The vectors e 1,, e are idepedet, ad () every vector x ca e writte as a liear comiatio of the asis vectors, i.e., x = 1 e e, where the j if is a real vector space ad j if is a real vector space Theorem: The represetatio of x i terms of a give set of asis vectors is uique. Theorem: If is -dimesioal, the ay set of idepedet vectors forms a asis 6. 1
2 Examples of vector spaces Astract Vector Spaces Example 1: The vector spaces ad. We write 0 = (0,, 0) ad the vectors e 1 = (1, 0, 0), e = (0, 1, 0), e = (0, 0,, 0) form a asis. These vector spaces are used as astract represetatio of the row vectors i matrix algera. Example : Let f 1 (t),, f (t) e ay real fuctios. All fuctios of the form 1 f 1 (t) + + f (t), costitute a -dimesioal vector space. I particular, we see that all polyomials 1 + t + + t of degree < costitute a -dimesioal vector space Example 3 (Ifiite dimesioal spaces): (a) All real- or complex-valued fuctios x(t) defied i a t. () The space of all real- or complexvalued fuctios for which x dt exists. This space is -dimesioal. a 6.3 Metric Spaces ad Norms Metric spaces: Defiitio: A collectio of poits (ot ecessarily a vector space) is called a metric space if to each pair of elemets x ad y, there correspods a umer d(x, y) that satisfies the properties: (a) d(x, y) = d(y, x); () d(x, y) 0, d(x, y) = 0 if ad oly if x = y; (c) d(x, z) d(x, y) + d(y, z) [triagle iequality] Defiitio: Cosider a sequece of elemets {x }. We write x x or x coverges to x if for each > 0, there exists a idex N such that d(x, x ) wheever > N. Ofte, we do ot ow if the elemets coverge to a fixed elemet i the space. So, we defie a Cauchy sequece. Defiitio: A sequece {x } is a Cauchy sequece if to each > 0, there exists a idex N such that d(x m, x p ) < wheever m, p > N. If {x } is a Cauchy sequece, we will write lim d( x, x ) 0 mp, m p 6.4
3 Metric spaces: Metric Spaces ad Norms Theorem: If a sequece coverges, it is a Cauchy sequece Proof: Let x e the limit of a sequece. By the triagle iequality, we have d(x m, x p ) d(x m, x) + d(x, x p ). Sice x x, there is some N, such that d(x, x) wheever > N. Hece, if m ad p are oth larger tha N, the d(x m, x p ) NOTE: The coverse does ot ecessarily hold! Defiitio: A metric space is complete if every Cauchy sequece is a coverget sequece. Remar: If the metric space is ot complete, we may add the limits of the Cauchy sequeces to complete the space. 6.5 Metric spaces: Metric Spaces ad Norms Example of icompleteess: The space of ratioal umers with the metric d(x, y) = x y is a metric space, ut it is ot complete Cosider the sequece x1 1, x 1,, x 1 1! 1!! ( 1)! Its limit is e, which is irratioal. Hece, the Cauchy series does ot coverge to a elemet i the space. A similar coclusio ca e reached for the series x1, x,, x ( 1)( 3) which coverges to /8. Remar: By cotrast, the space of real umers is complete with this metric
4 Examples of metric spaces: Metric Spaces ad Norms Example 1: The space of real or the space of complex umers with the metric d(x, y) = x y are oth complete metric spaces. Example : The spaces ad with the metric d( x, y) 1/ 1 1 where x = ( 1,,, ) ad y = ( 1,,, ) are complete metric spaces Example 3: The space C(a, ) of all cotiuous complex-valued fuctios. With the metric d1 ( x, y) max x( t) y( t), the space is complete. This metric a t assures us that if ay Cauchy sequece will coverge uiformly. Ay uiformly coverget sequece of cotiuous fuctios is cotiuous. With the metric d ( x, y) x() t y() t dt, the space is ot complete. 1/ a 6.7 Metric Spaces ad Norms Examples of metric spaces: Example 3 (cotiued): The space C(a, ) of all cotiuous complex-valued 1/ a fuctios is ot complete with the metric d ( ) ( ( x, y ) x() t y () t dt. Examples of icompleteess: (a) Cosider the sequece (with a = 1 ad = 1) 1 1 x ( t) arcta t, 1 t 1 it coverges to: y(t) = 1, 0 < t 1; y(t) = 0, 1 t < 0, which is ot cotiuous. The itegral equals 0 for all ad hece the Cauchy sequece coverges
5 Examples of metric spaces: Metric Spaces ad Norms Example 3 (cotiued): The space C(a, ) of all cotiuous complex-valued 1/ a fuctios is ot complete with the metric d ( x, y) x() t y() t dt. Examples of icompleteess: () Cosider the sequece (with a = ad = 1) 1; t r/ s, 1 s1, 0 r s x () t 0; otherwise It coverges to a fuctio that equals 0 whe t is irratioal ad equals 1 whe t is ratioal. The stadard Riema itegral of elemetary calculus exists for all ad yields zero. Hece, the Cauchy sequece coverges to zero. However, the Riema itegral does ot exist! We must use the Leesgue itegral!! 6.9 Metric Spaces ad Norms Properties of the Leesgue itegral: (1) The space of all square-leesgue-itegrale fuctio (a, ) completes the space C(a, ) of all cotiuous complex-valued fuctios with the metric 1/ d ( x, y) x() t y() t dt. a Hece, the space C(a, ) is dese i (a, ) ad ay elemet i (a, ) ca e approximated y a Cauchy sequece of elemets i C(a, ). () The Riema itegral of all Riema-itegrale fuctios equals the Leesgue itegral. I practice, itegratio always meas Leesgue itegratio (3) Two fuctios are equal almost everywhere if d (x, y) = 0. A set of fuctios that are equal almost everywhere are treated as a sigle poit i (a, )
6 Normed vector spaces: Metric Spaces ad Norms Defiitio: A ormed vector space is a vector space with a real-valued fuctio x with the properties: (1) x 0, x = 0 if ad oly if x = 0; () x = x ; (3) x + y x + y Remars: A ormed vector space is a metric space with the atural metric d(x, y) = x y. The coverse is ot true; d(x, y) = 0 whe x = y ad d(x, y) = 1, otherwise is a metric, ut it is ot a orm (Why?) Defiitio: A ormed vector space that is complete i its atural orm is a Baach space Ier Product Spaces Defiitio: A ier product x, y o a real vector space (r) is a real-valued fuctio of a order pair of vectors x, y with the properties: (1) x, y = y, x; () x, y = x, y; (3) x 1 + x, y = x 1, y + x, y; (4) x, x 0, with equality holdig oly for x = 0 Remar: A ier product ca also e defied o complex vector spaces. The oly chage is that x, y y, x. Note that x, y = x, y Theorem (Schwarz Iequality): For ay two vectors x ad y, we have Proof: We write x, y x, x y, y 0,,,,, x y x y y y y x x y x x We ow let = r x, y/ x, y, where r is a aritrary real umer, ad we fid 0,,, r y y r x y x x which is oly possile for all r if the theorem holds
7 Ier Product Spaces Theorem (Schwarz Iequality): For ay two vectors x ad y, we have x, y x, x y, y Proof (cotiued): The quadratic o the right is a miimum whe r x, y y, y The theorem follows Remar 1: I three dimesios, this theorem ecomes a a cos Remar : We ow fid 1/ 1/ 1/ 1/ x y, x y x, x Re x, y y, y x, x y, y It follows that x x, x 1/ defies a atural orm for the ier product space. The Schwarz iequality may ow e writte x, y x y 6.13 Hilert Space: Ier Product Spaces Defiitio: A ier product space that is complete i its atural orm is a Hilert space. Defiitio: A Hilert space is separale if there are a coutale umer of elemets f 1, f,, f, such that, give f i ad > 0, there exists a N that satisfies N f f 1 Remar:,, ad (a, ) are separale Hilert spaces. The ier products for or ad (a, ) are x, y x y, x, y x( t) y( t) dt 1 a Separaility is ovious for or, ut must e proved for (a, )
8 Ier Product Spaces Hilert Space: Remar: The atural orm for these spaces is the -orm 1/ 1/ x x, x x( t) dt a 1 Other importat orms are the p-orm (particularly p = 1) 1/ p p p x, ( ) p x x x t dt p a 1 ad the -orm, x max x, x max x( t) 1/ p 6.15 Orthogoality: Ier Product Spaces Defiitio: Two vectors are orthogoal if x, y = 0. Theorem 1: Two orthogoal vectors satisfy the Pythagorea theorem, x + y = x + y Theorem : A orthogoal set of o-zero vectors is idepedet. Hece, orthogoal vectors are a asis for a -dimesioal vector space. Theorem 3. Cosider the lie geerated y y, which is all vectors y, the ay vector x may e uiquely decomposed ito the sum of a compoet x p that proportioal to y ad a compoet z that is orthogoal to y. The compoet x p is called the projectio of x o y. We have specifically xy, xy, x xp z, xp y, z x y, z, y 0 y y
9 Gram-Schmidt procedure: Orthoormal ases Defiitio: A liear maifold is a suset of a vector space that is itself a vector space. Example: Cosider the vector space geerated y idepedet vectors x 1,, x m, where m <. The, is a liear maifold ad so is, the space that cosists of all vectors that are orthogoal to. Developig a orthoormal asis allows us to separate matrix operatios ad more geeral trasformatios ito their rage ad ull spaces! 6.17 Gram-Schmidt procedure: Orthoormal ases Give idepedet vectors {e 1, e,, e }, we wish to costruct a orthoormal set { 1,,, } Step 1: We let 1 = e 1 / e 1 Step : We let g = e e, 1 1, = g / g. We fid that 1 1, 1 g, 1 e, 1 e, 1 0 g g Cotiuig iductively, we fid for all j Step j: We let g j = e j e j, 1 1 e j, e j, j j j = g j / g j, l, m = lm (l, m j )* *The fuctio lm is called the Kroeecer delta-fuctio. It equals 1 whe l = m ad is zero otherwise
10 Orthoormal ases Gram-Schmidt procedure: Example: We let e 1 = (1,, 3), e = (1, 1, ), e 3 = (,, ), 1 3 1,, (0.67, 0.535, 0.80) g 3 11 ( 1) (1, 1, ),,,, ,, (0.313, 0.835, 0.45) g 3 (,, ) 3.1 (0.67, 0.535, 0.80) ( 0.139) )( (0.313, 0.835, 0.45) 3 (1.19, 0.170, 0.509) (0.911, 0.130, 0.391) 6.19 Gram-Schmidt procedure: Example: We let e 1 = (1,, 3), e = (1, 1, ), e 3 = (,, ) I MATLAB: Orthoormal ases >> E = [1 1 ; -1 ; 3 ], F = E >> F(1:3,1)=F(1:3,1)/orm(F(1:3,1)) >> F(1:3,) = F(1:3,) - (F(1:3,)'*F(1:3,1))*F(1:3,1) >> F(1:3,) = F(1:3,)/orm(F(1:3,)) >> F(1:3,3) = F(1:3,3) - (F(1:3,3)'*F(1:3,1))*F(1:3,1) - (F(1:3,3)'*F(1:3,))*F(1:3,) >> F(1:3,3) = F(1:3,3)/orm(F(1:3,3)) Remar: The classic Gram-Schmidt procedure ca e used to produce a QR decompositio (ut is a computatioally poor way to do it) Compare i MATLAB: >> G = F *E >> [Q, R] = qr(e)
11 Gram-Schmidt procedure: Orthoormal ases Similar ideas ca e used to costruct a orthoormal asis for (a, ). I fuctioal otatio, we have g () t e () t () t e () t () t dt () t e () t () t dt, a a 1/ () t g() t g() t dt a Example: From the Weierstrass approximatio theorem, we ow that the polyomials are dese i C(a, ) ad hece i (a, ). Startig with the polyomial set, e = = = 0 1, e 1 t,, e t, o the iterval 11 t 1 ad ormalizig so that all elemets equal 1 at t = 1, we otai the Legedre polyomials 1 0() t 1, 1() t t, () t 3t 1, 6.1 Gram-Schmidt procedure: Orthoormal ases I may applicatios, we wat to use a complete orthoormal asis i (a, ) to study how a trasformatio affects a iput sigal. A complete asis is oe for which we may write x j j, where j = x, j. Calculatig j0 j j j j1 j1 x x, x x,, we otai Bessel s iequality x x, j, j11 The goal is to show that as, the iequality ecomes a equality. We will do that later for Fourier series
12 Fuctioals Defiitio: Cosider, where is a Hilert space. If to each x, there is a complex umer T [x], the T is a fuctioal o. A fuctioal is ouded o if there is a costat c such that for all x, T [x] c x. The smallest c for which this iequality holds is called the orm of T ad is writte T. A fuctioal is cotiuous if for every x ad for every sequece {x } x, we have T[x ] T[x]. Defiitio: A fuctioal is liear if T [x 1 + x ] = T [x 1 ] + T [x ]; T [x] = T[x] Remar: T[0] = 0; T jxj jt x j (for fiite) j1 j1 Remar: T[x] ] = x, y is a liear fuctioal. Whe represeted dy matrices, we would have Tx = y T x. Note that the row vector is represetig a fuctioal, ot a astract vector. Note that y, x is ot liear; it is sometimes called ati-liear 6.3 Fuctioals Theorem 1: If a liear fuctioal is cotiuous at x = 0, the it is cotiuous o the etire domai of defiitio D T Theorem : A fuctioal is ouded if ad oly if it is cotiuous Theorem 3 (Riesz represetatio ti theorem): Each cotiuous liear fuctioal T[x] o a Hilert space ca e expressed i the form x, f, where f is a fixed elemet i. Proof: Cosider the set of vectors i the ull space of T[x]. Either =, i which case we set f = 0, or dim 1, ad we may fid a elemet f 0 such that f 0 = 1 i. We ow set f T[ f. Let y = T[x] f 0 xt[ f 0 ]. The 0] f0 T[ y] = 0 for all x, ad y. It follows that Txf [ ] 0 xt [ f0 ], f0 Tx [ ] T [ f0 ] xf, 0 so that Tx [ ] T[ f0] xf, 0 xt, [ f0] f0 To demostrate uiqueess, we ote that if x, f = x, g for all x, the x, f g = 0 ad usig x = f g, we coclude f g. Remar: We ifer dim = 0 or
13 Fuctioals Theorem 4: Every liear fuctioal i is ouded ( ) ( ) Proof: It suffices to show that limt x 0 for every { x } 0 0 We tae a orthoormal asis { 1,,, }, ad we write ( ) ( ) ( ) ( ) x j j, where j x, j j1 ( ) j 0 ( ) ( ) limt x lim j T j 0 j 1 From cotiuity ad {x () } 0, we ifer. Hece, Remar: This theorem does ot hold for liear fuctioals i. The classic couter-example example is the -fuctio: T[x] = x(0). Taig, 1/ t 1/ x () t 0, otherwise we see that x (0) / x as 6.5 Fuctioals Dual Bases Let {e 1,, e } e a fixed aritrary asis i. The effect of T o ay vector x is completely descried y complex umers T[e 1 ],, T[e ] sice TxT jej jt e j j1 j1 Coversely, give T[e 1 ] = 1,, T[e ] =, there is exactly oe liear fuctioal that it defies. I particular, if we let T [e j ] = j, the the Riesz theorem tells us that there is a uique vector e* that satisfies the relatio ej, e* j The set of vectors { e1*, e*,..., e*} is called the dual or reciprocal asis. (How do we ow that this set of vectors is a asis?) Remar: Dual ases are very importat i lossy (o-hermitia) prolems, where it is ofte coveiet to use a o-orthoormal asis sice we may write
14 Dual Bases Fuctioals Remar: Dual ases are very importat i lossy (o-hermitia) prolems, where it is ofte coveiet to use a o-orthoormal asis sice we may write x e, where x, e* j1 j j j j Developig ad usig dual ases i large systems remais a active area of research! 6.7 Trasformatios Defiitio: A trasformatio A: is a fuctio that maps x ito y, where,. The domai, the rage, ad the ull space of the trasformatio A are deoted A (= ), A, ad A respectively Examples: (1) x = ( 1,,, m ) y = m : I this case, we may choose = m, ad we fid =, A =, A = (x : x 0), A = {0} () x( 1,,, : I this case, we may choose = m m) y x ( 1,,, m), ad we fid = A = A = m, ad A = {0} (3) The Fourier trasform: y( f ) x( t)exp( ift) dt I this case, we may choose = (, ), ad we fid = A = A = (, ), ad A = {0}
15 Trasformatios Defiitio: A trasformatio A: is liear if A (x 1 + x ) = Ax 1 + Ax ; A (x) = A (x) Remar: I this case, A, A, ad A are all liear maifolds Remar: All liear trasformatios from m ca e represeted as matrices ad vice versa. Liear trasformatios from/to ifiite-dimesioal systems are also represeted as matrices whe discretized for computatio. Defiitio: A trasformatio is ouded o its domai if for all x A, there exists a costat c such that Ax c x. We the defie the orm of A, A, as the lowest upper oud of Ax / x. Theorems: As i the case of ffuctioals, we fid: (1) If a liear trasformatio ti is cotiuous at x = 0, it is cotiuous o all of A. () A liear trasformatio is cotiuous if ad oly if it is ouded. (3) All liear trasformatios m are ouded. 6.9 Trasformatios Adjoit Trasformatios i ( from ) Defiitio: Let A e a liear trasformatio ad f a fixed vector i. The Ax, f is a liear fuctioal that must e cotiuous. From the Riesz theorem, we ow there is some g such that Ax, f = x, g. We may write g = A*f, where we fid that A* is a liear trasformatio. The trasformatio A* is the adjoit of A. For every x ad y, we have Ax, y = x, A*y. The correspodig matrices A = [a j ] ad A* = [ a* j ] are Hermitia cojugates with respect to a orthoormal asis. Alterative Theorem: The equatio Ax = f has solutios if ad oly if f, z = 0 for every solutio that is a solutio of A*z = 0, i.e., f A if ad oly if f is orthogoal to the ull space of A*; that is A = ( A* )
16 Adjoit Trasformatios i Trasformatios Alterative Theorem: The equatio Ax = f has solutios if ad oly if f, z = 0 for every solutio that is a solutio of A*z = 0, i.e., f A if ad oly if f is orthogoal to the ull space of A*; * that is A = ( A* ) Proof: We will prove A* = ( A ), which is equivalet sice () = for ay liear maifold. (a) We first show ( A ) A* : Let z e i ( A ) ; the f, z = 0 for every f A. Sice the domai equals the whole space, we have for every x, Ax, z = 0. Hece, x, A*z = 0 for all x ad A*z = 0. () We ow show A* ( A ) : Let z A* ad f A. Hece, A*z = 0 ad there exists a x such that Ax = f. The, f, z = Ax, z = x, A*z = x, 0 = 0 Corollary: A = ( A* ). We ote (A*)* = A Theorem: ra A = ra A*; ullity A = ullity A*. Proof: We have ra A + ullity A* = ra A* + ullity A =. We earlier showed usig matrices that ra A + ullity A =. The result follows
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