LECTURES O N LINEA R GROUP S 5 dim R + dim 7 = n, oR = R, oP = P, res o = 0 == a = \ v . Obviously o an d o~ l hav e th e sam e R, P y res an d P = { x G K | a x = jc}. I f / ? H P = 0 , then V= R ®P an d / ? is , so to speak , wher e a' s nontrivia l actio n reall y occurs . How - ever we will see in examples tha t i ? n ? ^ 0 , indee d R CP, ar e possible. I f R i s a line, plane, hyperplane, we also refer t o i t a s the residua l line, etc., of o. Similarl y wit h th e fixed line, etc. Convention. Wheneve r a a i n GL n (V) i s under discussion , th e letter R wil l auto- matically refe r t o th e residua l spac e o f a , th e letter P t o it s fixed space . I n the sam e way Rt an d P t wil l be associate d wit h a o t i n GL n (V). 1.3.1. Let o x and o 2 be elements of GL n (V) and put o=o l o2. Then R CR X +R 2 , PDP X H/ 2 , res o l o2 re s ox + res o 2 . 1.3.2. EXAMPLES . Th e residua l spac e o f a radiation ¥= \v i s all of V. I f w e take a direct su m V = U 0 W an d put o x = (1^) 0 ( a lw) wit h a ¥= 0 , 1 , the n i t i s easily see n that R x = W an d P 1 = U.' Putting o 2 = (al^) 0 (1^ ) give s an example o f equalit y throughout 1.3.1 . O n th e othe r hand, i f o i s any elemen t o f GL n (V) wit h o ^ \ v an d we put o x = a, a 2 = a" 1 , w e get inequalitie s throughou t 1.3.1 . 1.3.3. Let o x and o 2 be elements of GL n {V) and put o~o x o2. Then (1) V = PX + P 2 =R = RX +R 2 , (2) / ^ n / ? 2 = 0=/ = P1 OP 2 . PROOF. Firs t let u s prove (1) . Her e Rl=(ol-lv)V=(al-\v)(Pl +P 2 ) C(o-lv)V=R. Looking a t o~ l = o2lo1~l give s R 2 CR. Henc e R = Rx + R2 b y 1.3.1 . No w we must prove (2). Fo r an y x i n P w e have o2x - x = - (aj(a 2 x) - o 2 x) £ R x ^ R 2 = 0, so PCP 2 . S o P C P j . S o P = PxnP2. Q.E.D . 1.3.4. Le f a i 2 be elements of GL n {V). Then the residual and fixed spaces of 2 a 2 - 1 ar e 2/ ? i 2 P respectively. In particular re s 2 a 2 - 1 = res a and a 2 = 2a imp/ie s tfwf ?1R=R and 2 P = P. 1.3.5. Let o x and a 2 be elements of GL n {V). Then R l ^LP2 and R 2 CPX makes o x o2 = o 2 ox.
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