Ch2 Linear Transformations and Matrices

Transcription

1 Ch Lea Tasfoatos ad Matces 上一章介紹抽象的向量空間, 這一章我們將進入線代的主題, 也即了解函數 能 保持 向量空間結構的一些共同性質 這一章討論的向量空間皆具有相同的 体 F 1 Lea Tasfoatos, Null spaces, ad ages HW 1, 9, 1, 14, 1, 3 Defto: Let V ad W be vecto spaces We call a fucto T : V W a lea tasfoato (o lea) fo V to W f fo all x, y V ad c F, we have (a) T( x y) T( x) T( y) ad (b) Tcx ( ) ctx ( ) Reaks: 1 若 F = Q, 則 (a) (b) (Ex37) 一般來說 (a) ad (b) 是彼此獨立的條件 (Ex38, 39) Let T be lea The the followg hold : 3 T(0) T(0) 4 T( x- y) T( x) - T( y) fo all x, yv 5 T( a x ) at( x ) fo all x V ad a F, 1,, 1 1 1

2 6 T s lea T( cx y) ct( x) T( y) fo all x, yv ad cf 7 I : V V I( x) x, I s called the detty tasfoato 8 T : V W T ( x) 0, T s called the zeo tasfoato Ta (, a) ( a a, a) Ex1 T : R R, Ex T : R R P( a1, a) T ( a, a ) 1 T 將點 p逆時鐘轉 角 T s lea Ex3 T : R R T( a1, a) ( a1, - a) T s called the eflecto about the x-axs Ex4 T : R R T( a1, a) ( a1, 0) t Ex5 T : M ( F) M ( F) T( A) A, Ex6 T : P ( R) P ( R) T( f( x)) f ( x) 1 Ex7 V C( R) Let a, br, a b T : V R T( f) f( t) dt b a Ex8 T : P ( R) M ( R), whee f(1) f() 0 T( f( x)) 0 f (0)

3 cf (1) g(1) cf () g() 0 Tcf ( ( x) gx ( )) 0 cf (0) g(0) f(1) f() 0 g(1) g() 0 c 0 f(0) 0 g(0) =c T( f( x)) T( g( x)) T s lea Deftos: T : V W s lea () NT ( ) the ull space (o keel) of T xv: T( x) 0 () RT ( )= the age (o age) of T T x x V y W x V T x y ( ) : = : wth ( ) Th1 T : V W, whee V, W ae vecto spaces ad T s lea RT ( ) ad NT ( ) ae subsoaces of Wad V, espectvely Poof : NT ( ) xv, Tx ( ) 0 () T(0) 0 0 N( T) () If xy, NT ( ), the Tx ( ) 0 ad Ty ( ) 0 T( x y) T( x) T( y) x yn( T) () Let cf ad xn( T) The T( cx) ct( x) c0 0 cx N( T ) NT ( ) s a subspace of V 3

4 RT ( ) yw: xvwth T( x) y T( x) : xv () T(0) 0 0 R( T) () If y, y N( T), the x, x V st 1 1 y T( x ), y T( x ) y y T( x ) T( x ) T( x x ) y y N( T) 1 () If y NT ( ), the xv st y T( x) the cy ct ( x) T ( cx) cy N( T ) RT ( ) s a subspace of W Th T : V W T: lea, V ad W: vecto spaces 1 If v, v, v s a bass fo V, the 1 RT ( ) spa ( T( )) = spa T( v), T( v),, T( v) Pf : Needed: RT ( ) spa ( T(( )) ad spa ( T(( )) R( T) Sce R( T) s a subspace of W ad T( v ), 1,,, ae R( T), we have RT ( ) spa ( T(( )) 4

5 If y R( T), the xv st T( x) y Sce s a bass fo V, c F, 1,,,, st x cv Now 1 y T( x) T( cv ) ct( v ), 1 1 yspa T( )) RT ( ) spa ( T ( )) 3 Ex: T : R R Ta (, a, a) ( a,- a, a) NT ( ) ( aa,,0) : ar RT ( ) R d( NT ( )) 1, d( RT ( )) = d ( R ) 3 Th3 (Deso Th) T : V W : lea If d( V ) the ullty ( T) ak( T) d V, 5

6 whee ullty( T) d ( N( T)) ad ak( T)= d ( R( T)) Poof: (1) Clealy, NT ( ) ad RT ( ) ae subspaces of Vad W, espectvely () Let d V ad v,, v be a bass fo N( T) 1 k Needed: (3) By Co to Th111 (p51), we ay assue that v,, v V st 1 k k+1 v,, v, v,, v s a bass fo V k1 (4) Cla : S T( v ),, T( v ) fos a bass fo R( T) If so, the we ae doe (a) Spa( S) R( T) 由定理 R( T) spa T( v ),, T( v ), T( v ), T( v ) k+1 1 k k1 = spa 0,,0, Tv ( ), Tv ( ) k 1 = spa Tv ( ), Tv ( ) spa S (b) S s a l set To pove ths, let 0 bt( v ) The 0 T( bv ) k1 k1 k1 bv = bv fo soe b,, b F 1 k k1 1 1 k1 1 bv- bv =0 Sce, v,, v s l, we have that b 0 fo all 1,, Hece, S s a l, set as claed Re ak: 1 直觀上來說, ullty of T 愈大, ak of T 就愈小, 反之也然 6

7 lea Th4 T : V W The T s 1-1 N( T) 0 由已知 Pf: ( ) Sce T(0) 0, N( T) 0 ( ) Suppose T( x) T( y) The T( x- y) 0 Hece, x- y 0 x y T s 1-1 Th4 的 V和 W, 不需要 V ad W是有 Th5 Let d( V) d( W) Let T : V W be lea THAE (The followg ae equvalet) () T s 1-1 () T s oto () ak( T) d( V) Pf: () () 由 Th4 得 1-1 NT ( ) 0 由 Deso Th 得 NT ( ) 0 ak( T) d( V) ak( T) d( W) Th 111 RT ( ) W Re ak: I Th 5, f V s ot fte desoal, the the assetos of Th5 ay fal (see ex 15, 16, ad 1) 線性函數另一常用且重要的性質是它的行為可由 bass 的行為來全部決定 7

8 Th6 Let v,, v be base fo V 1 Let Tv ( ) w, 1,, The! lea tasfoato T: V W st T( v ) w fo 1,, Pf: (1) Let xv The! a F, 1,,, st x 1 av Defe T( x) a w 1 The () T s lea () T( v ) w () T s uque Coollay ST, : V Wae lea ad v,, v s a (fte) bass fo V 1 Suppose, S( v ) T( v ) fo 1,, The S T 8

9 HW 1, (e), 3, 5, 10,13, 16 The Matx Repesetato of a Lea Tasfoato 有限維上的線性函數其實也用矩陣來表現出來, 但選擇不同的基底, 其相對的矩陣則不同, 但都是 類似 Def: A odeed bass fo V s a fte sequece of lealy depedet vectos V that geeates V 1 Ex1 e, e,, e s the stadad odeed bass fo F Ex 1, xx,,, x the stadad ode ed bass fo P ( F) Def: Let u, u,, u be a odeed bass fo V Fo xv, let a,, a be the 1 1 uque scalas st x au - 1 We defe the coodate vecto of x elatve to, deoted x, by x a1 a a = Re aks: () u e () T : V F by T( x)= x s a lea tasfoato Ex: V P ( R), 1, x, x Let f( x) 4 6x- 7x The 9

10 f 4 = 6-7 矩陣表示法 : Let u, u,, u ad w, w,, w be odeed bases fo V ad W, espectvely 1 1 Let a1 j a j Tv ( j) aw j Tv ( j) 1 j 1 1 a j ad set A a j Def: The atx A defed above s called the atx epesetato of T the odeed bases ad ad wte A T If V W ad, the we wte A T 3 Ex: T : R R Ta (, a) ( a3 a,0,a - 4 a) ad stadad odeed bases fo R ad R T(1,0) (1,0,) a e a e a e 1e 0e e T(0,1) (3, 0,-4) a e a e a e 3e 0e -4 e T

11 3 1 Let e, e, e -4 T Ex:,,, , xx, ad 1 t (a) Defe T : M ( F) M ( F) by T( A) A T T T T T

12 f (0) f(1) ( b) Defe T : P( R) M ( F) by T( f( x)) 0 f (3) T (1) T( x) ( ) 0 0 T x T ( c) T : M ( F) F by T( A) t( A) T 1, T 0, T 0, T T 1,1,1,1 14 (d) T : P ( R) R by T( f( x)) f() T (1) 1 T( x) 1 T x ( ) T 1,, (e) A A

13 3 1 (f) f( x) 3-6 x x f -6 (g) af, a ( a) Def: TU, : VW, whee Vad Wae vecto spaces ove F, ad let af Defe: T U : V W by ( T U)( x) T( x) U( x) ad at: VWby ( at)( x) at( x) Theoe 7 Let TU, : V Wbe lea () Fo all af, at U s lea () L T : T : V W, T : lea, V ad W: vecto spaces The L s a vecto space ove F Deftos: () LV (, W) T: T: VW a lea tasfoato () LV ( ) LV (, V) Theoe 8 V (vecto space) a odeed base W (vecto space) a odeed base T U T U (a) (b) at = a T fo all scalas a Pf: Let v, v,, v 1 1 w, w, w 13

14 The Tv ( ) aw 1 j j j 1 ad Uv ( ) bw 1 j j j 1 ( T U)( v ) T( v ) U( v ) a w b w ( a b ) w j j j j j j j T U aj bj T U = = j j T U T U = Ex Ta (, a) ( a3 a,0, a - 4 a) Ua (, a) ( a- a, a,3a a) Let ad be the stadad odeed bass of R ad R, espetvely The T U ( T U)( a, a ) (a a, a,5a - a ) ( T U) =

15 HW 1,3,11, 1, 16,19,1,3 3 Coposto of Lea Tasfoato ad Matx Multplcato () UT: lea f U ad T ae lea UT U T () Tu T u () ( ) Th 9 The coposte of lea tasfoatos s lea Pf: UT ( ax y) U ( at ( x) T ( y)) au ( ( T( x)) U( T( y)) aut ( ( x)) UT ( y) Th 10 () TU ( U) TU TU, ( U U) TUTUT () TUU ( ) ( TU) U 1 1 () TI IT T (v) auu ( ) ( au) U U( au ) fo all scalas a Def: A ( a ) j B ( b ) p j Defe AB ( c ), whee j p c a b, 1 p, 1 j j k kj k 1 (1) AB有定義的先決條件是 A的行的個數要等於 B的列的個數 () AB不一定等於 BA t t t (3) AB BA 15

16 T U Th11 Let V W Z be leat, let B T A U The UT AB Reak: Lea tasfoato 的合成函數之對應矩陣等於其相對應矩陣的相乘 v1 v w1 w z1 z zp Pf: Let,,,,,,,, The UT ( v ) U ( T ( v )) U ( b w ) b U ( w ) j j kj k kj k k1 k1 p b ( a z ) b a z kj k kj k k1 1 k1 1 p p p p bazab zcz kj k k kj j 1 k1 1 k1 1 UT C AB p Note that: d d, whee p, k k1 1 1 k1 (I) (II) p k d d d p d d d 1 p d d d 1 p I = 先將列相加再將列和相加 II = 先將行相加再將行和相加 I = II 16

17 Def: (1) Koeke delta j by 1 f j j = 0 othewse () I ( ) j Th1 Read youself k 次 Motato: k A AA A A A 0 1 A A 0 0 Note that 0 Th 13 A: atx B: p atx 1 p 1 p Wte B v,, v ad AB u,, u () u Av () Be v j j j j Pf: AB A v, v,, v ( Av, Av,, Av ) 1 p 1 p () A AI Be v j j T V W V W lea Th14 :, d( ), d( ) ad ae odeed bases fo V ad W, espectoly fo each uv, we have Tu T u ( ) 17

18 Pf: Fx uv defe f : F V by f( a) au ad g : F W by g( a) at( u) The g ft Let = 1 be a bass fo F Let ad be bases fo W ad V, espetvely The Tu g g Tf T f ( ) (1) 1 1 Th11 T f(1) T u Ex: 利用以下例子檢查 Th14 T : P( R) P ( R) 3 3 x x x x x 1,,, 1,, T( f( x)) f ( x) The AT px x x x 3 Let ( ) T p x x x T p x 9 The ( ( )) -4 9 ( ( )) 18

19 -4 ad px ( ) Now = T( p( x)) = T p( x) = = Def: A: a atx Defe L : F F by L ( x) Ax A A L A = a left-ultplcato tasfoato Th15 L : F F, a left-ultplcato tasfoato A ad ae stadad odeed bases fo F ad F (a) L A A b L L A B A B (c) L = L + L ad L al fo all af A+B A B aa A (d) If T : F F s lea, the! atx C st T L T I fact, C C (e) E: a p atx, the L L L AE A E (f) If, the L I I F 19

20 Pf of (e): L ( e ) AE( e ) A( Ee ) L ( Ee ) L ( L ( e )) AE j j j A j A E j Th 6 L L L AE A E Th 16 Let ABC,, atces such that ABC ( ) s defed The () AB( C) s also defed ad () ABC ( )= ABC ( ) Pf: () Let the desos of A, B ad C ae, espectvely,, ad Sce ABC ( )s defed, we ust have = ad = Hece, ABs defed ad so s AB( C) () L LL L( LL) ( LL) L L L L ABC ( ) A BC A B C A B C AB C ( ABC ) ABC ( )=( ABC ) Reak: 1 此定理可直接證明成立 此處是利用 Lea tasfoatos 和矩陣之間的關係來證明 Applcatos: I copute scece, the clque poble efes to ay of the pobles elated to fdg patcula coplete subgaphs ( clques ) a gaph, e, sets of eleets whee each pa of eleets s coected Clque pobles clude but ot lted to 0

21 fdg the axu clque (a clque wth the lagest ube of vetces) lstg all axu clques solvg the decso poble of testg whethe a gaph cotas a clque lage the a gve sze These pobles ae all dffcult 以下我們介紹一些較簡單的相關問題 : a squae atx A s called a cdece atx, f a 0 ad a j 0 o 1 fo all 1, j aj 1 s elated to j, ( 例如 : 認識 j, o 可傳送信息到 j o 有路到 j ) a s ot elated to j, ( 例如 : 不認識 j, o 不能傳送信息到 j o 沒 j 0 路到 j ) 註 : 若 a1 1, a1 0 ( 可解釋為 到 1 是單行道 ) 例如 A1 a j 不可送訊息給 3 ( a3 0) 可解釋為 3 可傳送訊息給 ( a3 1) 但 令 A a j 則 a 4 a a 可解釋 在 步可送訊息給 j 的次數 j k kj k 1 同理 ( A A A ) j 在 步內可送視訊給 j的次數 1

22 A clque = a axu collecto of thee o oe people wth the popety that ay two ca sed to each othe 例如 A ,, 4, 1, 4, 3 不是一個 clque 則 1,, 3,, 3, 4 皆是一個 clque, 但 註 : 1,, 3,4 當然不是一個 clque 之前的例子 A 1沒有一個 clque Fact: (Execse 19) Gve ay cdece atx, A aj B b j by, we defe a assocated atx b j 1, f aj aj 1, 0, othewse Hee, 3 a B 3 The clque 0 1 Pf: ( ) Let k,, k belogs to a clque Hece, The a a a a a a 1 Hece, k k k1 k1 k1k kk1 b b b b b b 1 Ad so, k k k1 k1 k1k kk1 B 3 bbb k kl l bk b 1 k1k b k kl, 1 10

23 3 ( ) If B 0, the k, l, 1 k, l, k, l, k l, st j bbb k kl l 1 kl,, as a goup ca sed essages to each othe I fte steps, oe ca fd a clque cotag kl,, Re ak: () I s lage, the fdg a clque s a dffcult task () Fo A, oe see that B 0 fo all Hece, A has o clque () ( B ) 0 fo all 1,,3,4 皆屬於某一個 clque 3 Doace elato f the assocated cdece atx A ( a j ) has the popety that a 1 a 0 ( 任給兩個人,j, 只有一方可傳送訊息給另一 j j 方 ) Fact (Execse 1) Let A be a cdece atx that s assocated wth a doace elato Pove that A A has a ow [colu] whch each ety s postve except fo the dagoal ety Pf: () Let A A ( aj ) The aj aj ak akj k 1 Clealy, a 0 fo all 1 () Ideas of the poof: st a >0 fo all j j 3

24 A 1 1 A 3 A A = 和 1 有 1 step 聯結的數字所成集合 1 (WLOG, 假設 1 和有些數字是聯集的 ) A = 和 1 恰有 steps 聯結的數字所成集合 ( 也即 A恰一步連結 A )* 1 * A 和 A 有一步聯結即 fo aa, ba st a和 b有聯結 1 1 A 1,,, - 1 A A 若 A The we ae doe! Let A A 和 1, A 有一步聯結 且和 A有兩步聯集 3 1 WLOG, we ay assue A A, 也即在 A中再重覆上述過程 在重覆作有限次 3 3 後新的 A 3 必為 (v) A A 的行也有相同性質之證明同上 Re ak: () 有一 ow 的每一 ety皆大於 0, 表示這個編號在 步內可送訊息給其他人 () 有一 colu的每一 ety皆大於 0, 表示這個編號在 步內可接受任何其 他人傳送的訊息 4

25 HW 1, 3, 6, 14, 15, 4 Ivetblty ad Isoophss 本節主要說明任何兩有限維的向量空間 V ad W 若是維度一樣, 可 看成 是一樣的, 此處 看成 數學上來說是指存在一從 V 到 W 之可逆的線性轉換 若 d( V), 則可把 V 看成是 Def 可逆函數 : F lea T : V W A fucto U : W V s sad to be a vese of T f TU I w ad UT I If T has a vese, the T s sad to be vetble Note that f T s vetble, v -1 the the vese of T s uque ad s deoted by T Facts: (a) Ideed, the vese of T s uque To see ths, let U, U be veses of T The 1 U ( w) U T( U ( w)) U ( TU ( w)) U ( w) fo all ww (b) T, U: vetble () ( TU ) U T () ( T ) T; patcula, T s vetble (c) T s vetble T s 1-1 ad oto (d) If d( V) d( W) ad T s lea, the T s vetble ak( T) d( V) Th 17 T : V W: lea ad vetble T -1 : lea ad vetble Def: A: atx A s vetble f a atx B st AB BA I -1 (Such B s uque ad s deoted by A ) 5

26 Lea T : V W: lea ad vetble The V s fte-desoal W s fte-deesoal I ths case, d( V) d( W) Pf: ( ) Let x, x,, x be a bass fo V 1 Th Th5 T( ) spas R( T) = W W s fte desoal by Th19 ( ) 利用 Th17, 其餘同上 1-1 oto 若 d( V) ad d( W) ae fte 則 d( V) Rak( T) d( W) d( V), d( W) < lea Th18 T : V W (odeed bases) T The T s vetble s vetble T -1-1 Futheoe, T = Pf ( ) T s vetble d( V) d( W) Let d( V) So, T s a atx Now, T : W V satsfes TT Iw ad T T Iv Ths -1-1 w I I TT T T v -1-1 T I I T T T 6

27 -1-1 So T s vetble ad T = T ( ) T s vetble d( V) d( W) v v w w Let,, ad, 1 1 Let A T By assupto, a atx B st AB BA I Th6! U L( W, V) st U( w ) B v j 1,,,, j j 1 whee w, w,, w ad v, v,, v U B 1 1 Wated: U T -1 Th 11 UT U T BA I I = v So UT I, ad slaly, TU I v w lea Co1 T : V V The -1-1 s vetble s vetble Futheoe, = T T T T Co Let A be a atx The A s vetble L s vetble A -1 L L -1 Futheoe, = A A M a b ( F) c d F 4 ( a, b, c, d) 7

28 4 vecto space F M ( F) soophc 從之觀點和可被視為一樣的空間這類的相同稱為 Deftos Let V ad W be vecto spaces We say that V s soophc to W f lea T : V W that s vetble Such tasfoato s called a soophs fo V oto W Reak: 1 V s sooplc to W W s sooplc to V So we eed oly say that V ad W ae sooplc o, syhollocally, V ~ W Ex: F ~ P1 ( F) Ex: P( R) ~ M ( R) 3 Pf: Defe T : P( R) M ( R) by 3 f(1) f() T( f) f(3) f(4) () T s lea 0 0 () If T( f), the f( x) 0 fo all xr (3次方程式, 不可能有 4 個零根 ) 0 0 T s 1-1 T s vetble Th 19 d( V), d( W) The V ~ W d( V) = d( W) Pf: ( ) Lea to Th18 ( ) Let v, v,, v, ad w, w,, w be bases fo V ad W, espectvely 1 1 8

29 Th6! T: lea st T( v ) w fo 1,, Th RT ( ) spa T( ) =spa W Th5 T s oto T s 1-1 T s a soophs Co V ~ F d( V) Th 0 LV (, W) ~ M ( F), whee d( V) ad d( W) Pf: Let v,, v ad w,, w be odeed bases fo V ad W Defe 1 1 ( T) T, whee T L( V, W) () s lea () s oto(hece, s 1-1 ad vetble) To see ths, let A a M ( F) Defe j Tv ( ) aw, j1,, j j 1 Th6 T s uque ad T = A o ( T) A Co: d LV (, W) = Def: Let d( V) Stadad epesetato of V wt s the fucto : V F by ( x) x fo each xv Ex: (1,0),(0,1) ad (1, ),(3, 4) 9

30 ad ae odeed bases fo R 1-5 Fo (1,-), - x x x Th 1 d( V ) s a soophs Fact: d( V) V ~ F d( W) W ~ F V F T L A W F T, A That s, V ad W ae detfy wth F ad F va ad, espectvely We ay "detfy" T wth L A Clealy, T L A Note that Tv () ( Tv ()) Tv () T v Av L () v A fo vv Hece, T L A 30

31 HW 1, 4, 6(a), 7, 10, 1 5 The Chage of Coodate Matx 若 T : V W 是 lea 且 和 為 V的兩個 odeed bases, 則 T 和 T 的關係為何? 這問題的答案是和座標轉換有關 x -4xy5y 1 代表如下的橢圓 y y x x 作適當的座標變換可得 ( x) 6( y) 1 如何變成上述的簡潔式子在 6節, R 為例, let e1, e 會提出但這可看成由不同的基底所得的座標顯示, 以 ( 標準基底 ), x x, 若 P = 以 為基底的 P點座標 = P =, 則 51 5 y y x x 其中 ad 的關係為何? y y Defe I : R R I( P) P, P R, Cosde the odeed bases ad fo Do( I)= R ad the age R( I) R 31

32 The I I I x 5 5 x y 1 y Th Let ad be two odeed bases fo a fte-desoal vecto space V, ad Q I v The (a) Q s vetble (b) Fo ay v V, v Q v ( 矩陣 Q乘向量 ) v ) Th18 P oof: (a) Sce I s vetble, Q s vetble v Th14 v I I v Qv (b) = (v) = v v Re ak: Q s called a chage of coodate atx Moe pecsely, Q chages -coodate to -coodate 3

33 Ex1: (1,1),(1,-1) ad (, 4),(3,1) (, 4) 3(1,1) -1(1,1) ad (3,1) (1,1)+1(1,-1) 3-1 Q 1 x 3-1x = y 1y 3-1 o P = P 1 lea Re ak: T : V V s called a lea opeato Th3 T : V V s a lea opeato -1 T Q T Q whee Q I v I T I v v I T I P oof: V V V V T I T I T I T I v 33

34 a 3a-b Ex T b a 3 b 1 1 3,,, Q Q , T 3 = T Q T Q - -1 = = Ex3 y (-,1) ( ab, ) (1,) Tab (, ) x Reflecto about the le y x s lea (why?) I geeal eflecto about the le y xb s also lea Gve ( ab, ) R, how ca we fd T( ab, )? T (1, ) (1, ) 1(1, ) 0(-,1) T (-,1) (, -1) 0(1, ) -1(-,1) 1 0 = (1,),(-,1) = (1,0),(0,1) 0-1 T Q I = Q T a 1-3a 4b T b 5 4 a 3 b 34

35 Co: AM ( F),,,, a oded bass fo F 1-1 L Q AQ Q A, whee,,, 1 Pf: I LA I F F F F -1 L A Q AQ, whee Q I ad = the stadad odeed bass of F F 1 j Note that I ( j) e j, whee j 1 j Def: AB, M ( F) We say that Bs sla to Af a vetble -1 atx Q st B Q AQ Re aks: () B s sla to A A s sla to B So we eed oly say that A ad B ae sla () Th3 ca be stated as follows: If T s a lea opeato o a fte desoal vecto space V, the T s sla to T 35