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1 Lear Algebra ( 線性代數 ) 絕大部分的自然科學現象和工程問題是非線性的, 但要了解這些可能超過個 100 個變數的非線性問題, 一個主要方式即先將此問題線性化 (Learzato) 因此, 線性代數 ( 以下簡稱線代 ) 和矩陣理論是 : () 很多數學分枝, 如方程微分幾何機率等的基本工具 ; () 也是一個 actve ad alve 的研究領域, 如頂級應數期刊 SIAM 系列, 就有一期刊是專門討論矩陣理論和其應用 (SIMAX); () 在其他領域有極大應用價值 : 如工程領域物理領域和經濟領域正因為線性代數的重要性多樣性和應用性, 如何選擇一本線代的書是要費心思的, 以下先介紹三本好書 : 1 G. Strag, Lear Algebra ad Its Applcatos, 4th edto, Brooks/ Cole, K. Hoffma ad R. Kuze, Lear Algebra, 2d edto, Pretce-Hall, S.H. Fredberg, A.J. Isel, L.E Spece, 4th edto, Lear Algebra, Pretce-Hall, 第一本書作者,Glbert Strag 是在 MIT 任教, 又此書的 Lectures 被錄製為開放式課程書的寫法非常能傳達線代的美麗和應用價值作者嘗試解釋觀念而不是只導出觀念同時也提供許多直觀且能 vsualzed 的想法書的寫法另一特色是非常對話式, 不像定義定理證明的傳統式的寫法幾年前, 我曾經以這本書教本系大二學生, 不少學生非常不習慣這種對話式的寫法, 因為他們覺得不容易找到重點這本書也非常適合唸第二次線代的學生或老師來讀, 當更可 1

2 apprecate 書中欲傳達的訊息相較於第一本書, 第二本和第三本的寫法是較理論抽象第二本書是本 classcal 的理論線代書可惜內容未能反應線代是門 " 會成長 " 的學科, 例如在現今各種應用領域常需具備 LU QR SVD 和 least squares 的知識我後來選擇第 3 本為現代的教科書, 基於以下理由 : 1. 個人雖然非常欣賞第一本書, 但其原作者已有 Webpage 和 Vdeo Lectures 放在網路讓大家開放式的學習 2. 本校尚缺較理論線性書的開放式課程, 第三本書寫法雖抽象, 但也重視應用, 同時也反應了線代是門會成長的學科 2

3 Ch1 Vector Spaces ( 向量空間 ) H.W. 1,9,11-19 初淺來說, 線代是學習作用在向量空間 (Vector Spaces, 以下簡稱 VS) 的線性函數的一些共同性質 VS 是一個代數系統且有某些運算性質課本第一章第一節空間中向量所成的集合即是 VS 的原型 (prototype) Deftos: A vector space V over a feld F cossts of a set o whch two operatos (called addto ad scalar multplcato) are defed, so that the followg 10 propertes hold. (VS-1) x yv wheever x, yv. ( 加法封閉 ) (VS 0) xv wheever F ad xv. ( 乘法封閉 ) (VS 1) x y yx for all x, yv. ( 可交換 ) (VS 2) ( x y) z x( yz) for all x, y, zv. ( 可結合 ) (VS 3) 0V so that x0 x for all xv. ( 有加法單位元素 ) (VS 4) For each xv, yv s.t. x y 0. ( 有加法反元素 ) (VS 5) For each elemet x V, 1 x x. Here 1 F. ad s the detty elemet for multplcato. 3

4 (VS 6) ( ab) x a( bx) for all a, bf ad x V. ( 分配律 ) (VS 7) a ( x y) axay for all af ad x, yv. ( 分配律 ) (VS 8) ( ab) xaxbx for all a, bf ad xv. ( 分配律 ) Remark: () The elemets of the feld F are called scalars. () The elemets of the VS are called vectors. () 一個體 (feld) 即是一個集合上是有某些性質的 4 種運算 ( 加減乘除 ) 複數實數和有理數皆是體, 但整數不是體 ( 因為除法無封閉性 ) 見附錄 C Ex1. a1 () F : a F, 1, 2,...,, F s a feld, R, 皆為 VS. a a1 () V : a R, 1, 2,..., F. a Is V a VS over the complex umbers? No, (VS0) s volated. ( 課本 1.2- Ex15). 4

5 Ex2. () M ( F) A a : a F, 1 m, 1 j s a VS. m j m j () V A a, a R, F Q. V s a VS. j m j () V A a, a Q, F R. V s ot a VS. j j Ex3. FSF (, ) f: S F, where Fs a feld s a VS. Ex4. P f( x) : f( x) a x a x... a, a F, 1, N s a VS Ex5. S ( a ) : a F s a VS. 1 以下例子不是 VS, 哪些條件不滿足? Ex6. S ( a, a ): a, a R. F R 加法 : ( a, a ) ( b, b ) ( a b, a - b ) Scalar multplcato (SM): ca (, a) ( ca, ca) (VS1), (VS2) ad (VS8) fal to hold. 5

6 Ex7. S 同上, F R. 加法 : ( a, a ) ( b, b ) ( a b, 0) SM: ca (, a) ca (,0) (VS3), (VS5) (X) Ex8. V ( a, a ): a, a R 加法 : ( a, a ) ( b, b ) ( a 2 b, a +3 b ) SM: ca (, a) ( ca, ca) (VS1)(VS2)(VS8) (X) Ex9. V 同上. 加法 : 正常 SM: (0,0) f c 0, ca ( 1, a2) a2 ca1, f c 0. c (VS8) (X) 6

7 Thm1.1 If xyz,, are V, a VS, Vsuch that x z y z, the x y. ( 消去法成立 ) Pf: 由 (4) 知 vv wth zv 0. (3) (4) (2) 已知 (2) x x0 x( zv) ( x z) v ( y z) v y( zv) y. Cor1. 0 (VS3) s uque. Pf: Let 0 ad 0 be such that for all x V. x0 x. x 0= x. Thm 1.1 x0 = x0 00. Cor2. The vector y (VS4) s uque. Pf: Let y ad y V be such that x y 0 x y. The y y. Remark: Sce y s uque, we shall deote such y by - x. Thm1.2 V : a vector space. The (a) 0 x0 for xv (0 F, 0V). (b)( - ax ) -( ax) a(- x) for a Fad xv. (c) a00 for a F. 7

8 Pf: (a) 0x0 x(0 0) x0x0 x0. Thm1.1 0 x 0. (b) - ( ax) s the uque elemet of V such that ax -( ax) 0. Wated: ax (- a) x 0. (*) (The - ( ax) =(- ax ).) To see(*), we have that: (8) (a) ax (- a) x =( a (- a)) x 0 x = 0. -( ax) = (- a) x. -(1 x)= (-1) x - x(-1) x. Hece, a(- x)= a (-1) x = a(-1) x= (- a) x. (c) a 0+a 0= a( 0+ 0)= a 0 = a 0+ 0 a 0= 0. 8

9 H.W. 1,8,17-21,23,26,30, Subspaces ( 子空間 ) 子空間是具有原向量空間結構的部份集合這節我們將學以下事項 1 什麼是 subspace 2 如何驗證 3 subspace 的一些性質 Def: Let V be a VS over a feld F. The W V s called a subspace of V f W s a VS over the F uder the operatos of addto ad SM defed o V. Remarks: 0 () V ad are subspaces of V. () (VS 1-2, 5-8) 對 W來說會自動滿足, 因此檢查 W是否 subspaces只需檢查 (VS -1,0,3,4). () 事實上, 只需檢查 (VS -1,0,3) 即可若 V, 則只需檢查 (VS -1,0). Thm1.3 W V, where V s a VS. The W s a subspace of V (a) 0 W. (VS3) (b) x yw wheever, x, yw. (VS-1) (c) cx W wheever, c F, x W. (VS0) Pf: ( ) 證明 V和 W的零向量是一樣的即可 Let 0 ad 0 be zero vectors for W ad V, respectvely. Let xw. The (3) (3) x x0 ad x x0. Hece, x0 x0. Thm

10 ( ) To complete the proof, we eed to show that ( a),( b),( c) For each xw, yw s.t. x y 0. Thm1.2-b To see ths, let xw V. The - ( x) = (-1) x. By (c) (-1) xw. Hece, -( x) W. Remark: 由 Thm1.2-a 得若 W, 則只需檢查 ( b),( c) 即可.(Ex.17) t t Def: Let A( a ). Defe the tramspose Aof A by A ( a ). j m j m T A s symmetrc A A. Ex1. The set W of all symmetrc matrces M ( F) s a subspace of M ( F). Ex2. P ( F) P( F), a subspace. Ex3. CR ( ) FRR (, ), a subspace. Ex4. The set of dagoal matrces s a subspace of M ( F). Ex5. W AM ( F) : Trace A0. Ex6. W A( a ) M ( F) : a 0 s ot a subspace of M ( F). j j 2 Ex7. W ( x, mx) : xr, m s fxed s a subspace of R. 2 W ( x, mxb) : xr, m ad b 0 are fxed s ot a subspace of R. Thm 1.4 Ay tersecto of subspace of a vector space V s a subspace of V. Is W W a subspace of V?

11 (Ex19. W W a subspace of V W W or W W.) Def. S S { x y: xs ad y S } Def. A vector space V s called the drect sum of W ad W f () W, 1, 2, 1 2 are subspace, () W W {0} ad () W W V. We deote that V s the drect sum of W ad W by wrtg V W W

12 H.W. 1, 3(a),4(a),5(g),9-12,15, Lear Combatos ad Systems of Lear Equatos 給一非空集合 S, 其中 S V且 V為一向量空間, 這節將介紹如何找出一個包含 S的最小子空間技巧上常引進 lear combato ( 線性組合 ) 的觀念, 並進而訂出 spa S Spa S 即為包含 S 的最小子空間本節也透過問 " 給一向量是否能用其他向量線性組合起來 " 作為討論線性聯立方程式的一個起點 Def. V: a vector space S V, S. A vector vv s called a lear combato ( 線性組合 ) of vectors of S f u V, a F, 1,2,...,. s.t. v au. 1 I ths case, we also say that v s a lear combato ( 簡稱 l.c.) of u, u,..., u ad 1 2 call a, a,..., a the coeffcets of the lear combato. 1 2 Def: S V, S, V : a vector space. The spa of S spa ( S) = The set of all l.c. of the vectors of S au : us, af, 1,2,...,, N. 1 For coveece, we defe spa( ) 0. 12

13 Def: A subset S of a vector V geerates (or spas) V f Spa ( S) V. I ths case, we also say that the vectors of S geerate (or spa) V. Ex1. Let V R, S {(0,1),(1,0)}. Let x ( x, x ) R. The x x (1,0) x (0,1) Hece, ay vector s a l.c. of vectors of S. Thm1.5 () Spa ( S) : s a subspace of V. () W : a subspace of V ad W S. The W Spa ( S) ( Spa ( S) 是最小包含 S 的子集空間 ). Proof: () S Spa( ) 0 s a subspace. S 0z 0 spa ( S), where zs. If x ad yspa ( S), the a F, u S, 1, 2,.... ad b F, v S, j 1, 2,... m. s.t. j j m m x au, y b u, ad so x y au b u Spa ( S). j j j j 1 j1 1 j1 Moreover, for c F. cx ( c a ) u Spa ( S). 1 Spa ( S) s a subspace of V. () If x Spa ( S), the x au, where a F, u S W au W. 1 13

14 au W xw Spa ( S) W. 1 Ex2. Ca a vector be expressed as a l.c. of other vectors? Such questo ofte reduces to the problem of solvg a system of lear equato. (2,6,8) au au au au au u 1 u2 u3 u4 u5 (1, 2,1), (-2,-4,-2), (0, 2,3), (2,0,-3) ad (-3,8,16). a - 2 a 2a - 3a 2, a - 4a 2 a 8a 6, a - 2a 3a - 3a 16a 利用一連串的列運算使其變成如右上圖 ( 高斯消去法 ) a - 2 a a -4, a 3a 7, 3 5 a -2a

15 a 2a -a -4, a -3a 7, 3 5 a 2a a, a 2 5 可任意給. Ex3. P( R) Is 2x - 2x 12x- 6 a l.c. of x - 2x - 5x- 3, ad 3x - 5x -4x-9? 3 2 How about ? x x x Sol: (Case1) 2x - 2x 12x- 6 a( x - 2x - 5x- 3) b(3x - 5x - 4x- 9). a3b 2, - 2a- 5b -2, - 5a- 4b 12, -3a-9b (Case2)

16 3 xy R Ex4. spa (1,0,0)(0,1,0) = the -plae. Re mark.1 0 spa ( S). S 3 Ex5. (1,1,0),(1,0,1) ad (0,1,1) spas. R a a a a - a a a a a a -a a - a a a3 a2 -a ( a2- a3+ a1) ( a3- a2+ a1) ( a3+ a2- a1) 2 S x x x x x x P R Ex6. 3 2, 2 5-3, spa(s)= 2 ( ). Ex () S,,, spa(s) M2 2( R) () S, ad The Spa S M ( R)

17 1 1 1 a a a a2 a a a3 a a a a a2 2a a2 a a2 a a1 a a1 a a a a a a1 a2 a3 a2 a1. a1 a3 a3 a2 a4 a1 a3 取 a3 a2 a4 a1 a3, 則 A 無法由 S 生成出來 a2 a Ex8. Spa, ad R. ay lear combato of the vectors of the above set has equal dagoal etres. 17

18 H.W. 1, 3, 8,12,16,17, Lear Depedece ad Lear Idepedece Goal: Let W be a subspace of a vector space V. Fdg a smallest" fte subset S that geerates W. Fdg such smallest" fte subset leads to the cocept of lear depedece ( l.d.) ad lear depedece (l..). Def s () A subset S of a vector space V s called l.d. f dstct vectors u, 1,..., S ad scalars a, 1, 2,...,, ot all zero, s.t. 1 au 0. I ths case, we also say that the vectors of S are l.d. (Ay set S cotag the zero vector of l.d.). 一集合是 l.d., 即是零向量可由集合中之向量非 " tral" 的線性組合起來. 或是說此集合有一向量可用集合中其他向量線性組合起來. 此處 " tral" 是指組合係數皆為 0. () If S s ot l.d., the S s called l.. (For ay fte umber of dstct vectors u, u,.. u S ad au 0, where a F, the a a... a 0)

19 Ex1. S (1,3, 4,2),(2,2, 4,0),(1, 3,2, 4),( 1,0,1,0) u, u, u, u The 4u 3u 2u 0u S s l.d.. Ex S,, u -2, u 3, u 5. S s l.d

20 Thm1.6 Let V be a vector space, ad let S S V. If S s l.d., the S s l.d Cor. If S s l.., the S s l Thm1.7 S : l..ad vv - S. The S v s l.d. vspa(s), or equvaletly vspa ( S) S v s l... ( 一個線性獨立集 S, 多加一個向量 v S, 則新集合是否 l.d. 是看 v是否在 Spa( S)). Pf: ( ) S v s l.d. dstct vectors u,..., u S v ad a F, 1, 2,...,, ot all zero; such that au 0. Moreover, v u, u..., u. Otherwse, a= for all 1, WLOG, let u v. The u S, 1,2,..., au av 1 0. Hece a 0. Otherwse, a 0, for all 1, 2,..., -1, -1 a v - u vspa ( S). 1 a ( ) vspa(s) u S, a F, 1,2,...,, s.t. v au. Here we may assume that u, are dstct. 1 Hece, vu,,..., u are dstct. 1 20

21 0 (-1) v au. 1 1 vu,,..., u s l.d.. Thm1.6 S v s l.d.. Re mark: 1. If o proper subset of S geerates the spa of S, the S must be learly depedet. Otherwse, a proper subset S of S wth Spa( S) = Spa( S), a cotradto. 2. If Spa( S) W, where W s a subspace of a vector space V, ad o proper subset of S s a geeratg set for W, the S s a l.. Such S s called a learly depedet geeratg set for W. 21

22 H.W. 1,2(a),3(a),5,10(a),13,14,21,22,30, Bases ad Dmeso ( 基底和維度 ) 一個向量空間的 buldg block 是一組 learly depedet geeratg (l..g.) set, 就如基本粒子之於物理世界本章節是要找一組不多也不少的集合能生成一個向量空間此恰恰好的集合及稱為此向量空間的基底 (bass), 此基底元素的個數即稱為此向量空間的維度 (dmeso) Def: A bass for a vector space V s a l.. subset of V that geerates V. Ex1. Spa( ) 0 = 1 Ex2. e (0,0,,...,0).The e, e,..., e for F. 1 2 j Ex3. E = that matrx whose oly ozero etry s a 1 the th row ad the jth colum. j The E,1 m,1 j for M ( F). Ex4. = 1, x,..., x for P ( F). 2 Ex5. = 1, xx,... for PF ( ). 1 2 m Thm1.8 V: a vector space. u, u,..., u. The s a bass for V each vv.! a F, 1, s.t. v au. 1 22

23 Pf: ( ) Let a, b F, 1, so that v au bu. The ( a b ) u 0. Hece, a b for all. ( ) Let au 0. Sce 0u 0, 1 1 we have, va the uqueess of a, that a =0 for all. 1 u s l... Remark: 1. 一個向量空間若其基底個數是有限, 則上述定理顯示 V F. 2. 本書主要討論的是有限基底的向量空間. 一個向量空間基底的元素個數不一定是有限的, 但 Thm1.9 告訴我們若此向量空間可由一有限集合所生成, 則此向量空間必有一個有限基底 Thm1.9 Let V Spa( S), where #( S). The some subset of S s a bass for V. Hece V has a fte bass. Pf. Case () S or S 0 The V 0 ad s a bass for V. Case () S cotas a ozero vector u. 1 1 The u s a l.. set. 由已知 #( S), 我們可得一線性獨立集合 u, u,..., u, 其中 u, u... u 在 S中, 且若 v S -, 則 v 1 2 k 1 2 k 是線性相依 ( 即可找到一個包含於 S中之 " 最大 " 的線性獨立集合 ). 23

24 Clam: s a bass for V ( 只需再證明 S Spa( ), 因為由 Thm1.5-(,) 可得 Spa( ) Spa( S) V). Let vs. If v,the vspa( ). If v, v the 是線性相依由 Thm1.7 得 vspa( ). Thus S Spa( ). Remark: 此定理的證明是一個 Costructve proof, 告訴我們如何找到有限的向量空間的一組基底 Step1: Fd a fte geeratg set S. That s, spa( S) V. Step2: Select a ozero vector u S. 1 Step3: Fd a "largest" possble l.. set, where u ad S. 1 Such s a fte bass. Ex: S (2,-3,5), (8,-12,20),(1,0-2), (0,2,-1), (7,2,0). 3 R Sol: (2,-3,5),(1,0,-2),(0,2,-1) s a bass for Spa( S). 接下來要回答一個基本且重要的問題 : 如果一個向量空間有一有限基底, 則其他基底的元素個數是否相同呢? Thm1.10 ( Re placemet Theorem) Let V Spa{ G}, where #( G). Let L s a l.. subset of V wth #( L) m. () m. ( 由 G spa 出來的 vector space中的線性獨立向量的個數不可能大於 G 集合的個數 ). 24

25 () H G, #( H) - m, s.t. spa( LH) V. 註 : 由定理可得, 若一向量空間是由有限多個向量所張出的空間, 則此向量空間不可能含有一個無限的線性獨立集 Proof: The proof s by mathmatcal ducto o m. () Let m0 L. The m 且令 H= G. 因此當 m0, () ad () 皆成立. () 當 m0 假設 () 和 () 成立, 欲證 () 和 () 對 m1 也成立. Let L v,..., v be a l.. subset of V, 則 L = v,..., v 也是 l.. 由 ducto 假設, 1 m1 1 1 m - H u,..., u G, s.t. #( H) - m ad spa( LH) V. 若 m, 則 H, m Spa( L) V. v Spa( L) L s l.d. ( ). m1 Hece, m, 因此 m1. 接下來證 () 成立 : v V Spa ( L H'), 因此 m1 m -m v av bu 其中 b 必不全為 0 ( 否則 v 可寫成 v, 1, 2,..., m, 的線性組合 ) m1 m1 1 1 m -m 1 WLOG, we may assume that b 0, 因此 u v - av bu. 1 1 m1 b1 1 2 令 H u,..., u, 則 #( H) - m-1, u spa( LH). 由定理 1-5, 2 m - 1 spa( LH) spa( LH) V. LH G. 所以 spa( LH) V. 25

26 以下說明 ( 有限 ) 基底的個數皆相同. Corollarly: () Let V spa ( G) ad # ( G). If L s a l.. subset of G, the #( G). () Let V be a vector space havg a fte bass. The every bass for V cotas the same umber of vectors. Pf of (): Let L be a l.. subset of G. If #( L), the a subset L of L cotag 1 vectors. Moreover, by Thm 1.6, L s a l.. set. By Thm1. 10-() #( L) 1, a cotracto. Thus #( L). Aga, by Thm1.10-() #( L). Pf of (): 由 Thm1.10, 加上已知, 可得任何 V的基底都是有限. Let ad be ay two fte bases for V that cotag ad m vectors, respertvely. 由定理 1.10分別可得 m 和 m. 因此 m. Deftos: () A vector space, 因為有之前的 Corollary, 以下維度定義是 well-defed, s called fte-dmesoal f t has a fte bass. () The umber of vectors a bass s called the dmeso of V ad s deoted by dm( V). () A vector space that s ot fte-dmesoal s called fte-dmetsoal. Ex7. V 0. dm( V ) 0. ( ) Ex8. V F. dm( V). Ex9. V M ( F). dm( V) m. m Ex10. V P ( F). dm( V) 1. Ex11. F R, V Rdm( V) 1 1. Ex12. F R, V C dm( V) 2, 1,. 26

27 Corollary2. Let dm( V). (a) Let G V ad Spa( G) V. The #( G). Moreover, f #( G), the G s a bass for V. ( 基底為 " 最小 " 的 geetatg set). (b) Let L V ad L s l.. The #( L). Moreover, f #( L), the L s a bass for V. ( 基底為 " 最大 " 的 l.. set). (c) Every l.. subset of V ca be exteded to a bass for V. Pf: (a) Let be a fte bass of V. 由定理 1.10-() #( ) #( G). 由定理 1.9, G的部份集合 H 使得 H 為 V的基底, 但 #( H). 若 #( G), 則 G H. (b) 第一個敘述是 Thm1.10-(). 第二個敘述也由 Thm1.10-() 可得. ( c) 由 Thm1.10 得知. Ex: If uvw,, s a bass for a vector space V, the u v wv, ww, s also a bass for V. Pf: By Cor 2(b), t suffes to show that uvw, vw, w s l... 因為 Let au ( vw) bv ( w) cw0 au( ab) u( abc) w0. u, v, w s a bass a 0, ab 0, abc 0 a b c 0. The dmeso of subspaces. Thm1.11 W: a subspace of a vector space V, where dm( V). The () dm( W) dm( V). () If dm( W) dm( V), the W V. 27

28 Pr oof: () 由 Thm1.10-() 可得. Let be a fte bass for V. The spa( ) V. Let be a bass for W. The V ad s l... Hece, #( ) #( ). That s, dm( W) dm( V). ( ) 由 Thm1.10-() 得 H, where #( H) dm( V) - dm( W), s.t. spa( H) V. 由已知得 #( H) 0 spa( ) V W V. Ex17. W ( a, a, a, a, a ) F : a a a 0, a a dm( W) Ex18. W= The set of dagoal matces dm( W). 1 Ex19. W= The set of symmetrc matces dm( W) ( 1). 2 Corollarly: If W s a subspace of a fte-dmesoal vector space V, the ay bass for W ca be exteded to a bass for V. Ex. Descrbe subspaces of R 3 that have dmetso 3,2,1 ad 0, represetvely. The Lagrage Iterpolato Formula. Lagrage polyomals assocated woth c, c,... c. 0 1 x- ck f ( x). c - c k 0 k k 0 j f( cj). 1 j 0 1 Clam: f, f,... f s a l.. subset of P ( F). Let a f 0 ( 零函數 ) for some scalars a F, 0,1...,. 0 af( c) 0 ( c) 0. 0 j j 28

29 But af( c) a for ay j 0,1,...,. 0 j j Hece, a 0. Sce dme( P ( F)) 1. j s a bass for P ( F). Every polyomal g P ( F) ca be uquely ecpress as a l.c. of. Let g = b f. 0 The gc ( ) bf( c) b. j j j 0 So g g( c ) f. 0 Ths represetato s called Lagrage terpolato formula. 利用上述公式可以算出過 ( 1) 點的次多項式同時也可得知若次多項式有 1個零根, 則此多項式必為零函數 29

Microsoft Word - Friedberg doc

Microsoft Word - Friedberg doc Chapter Lear Trasformatos ad Matrces - Lear Trasformatos, Null Spaces, ad Rages A fucto T wth doma V ad codoma W s deoted by T: V W 定義域為 V, 對應域為 W 的函數註記為 T: V W DEFINTION. Lear trasformato Let V ad W