Аналитическая геометрия и линейная алгебра. Умнов А.Е. - 45 стр.

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cÈÏ Ëã
49
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¯ÈÏ ¹¯Èm ¯º®} Ò mÈ mË}º¯È
a
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11 2 2 33
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º°mº®°mÈämË}º¯Óºº¹¯ºÒÏmËËÓÒ«
[,] [ , ]
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ab g g g g g g
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→→ → → → → → →
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11 22 3311 22 33
11 1 1 12 1 2 13 1 3
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ηηη
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c È Ï  Ë ã                                                      49
¯ºÒÏmËËÓÒ«mË}ˆº¯ºm



{©¯ÈÎËÓÒËmË}ˆº¯Óºº¹¯ºÒÏmËËÓÒ«m}ºº¯ÒÓȈȲ
                  
                  
                  
                                                                                                        →       →       →                                                     →         →    →
                  ‚°ˆ ÏÈÈÓÈ°Ò°ˆËäÈ}ºº¯ÒÓȈ {O, g1 , g 2 , g 3 } ˆÈ}È«ˆºmË}ˆº¯© g1 , g 2 , g 3 º­
                                                                                                   →             →                                          →           →               →          →
¯Èς ˆ ¹¯Èm‚  ˆ¯º®}‚ Ò mÈ mË}ˆº¯È a  Ò b  ã« }ºˆº¯©² a = ξ1 g1 + ξ2 g 2 + ξ3 g 3  Ò
→             →              →             →
b = η1 g1 + η2 g 2 + η3 g 3 
                  
                  º°mº®°ˆmÈämË}ˆº¯Óºº¹¯ºÒÏmËËÓÒ«
                           →→                       →            →            →          →             →             →
                         [a , b ] = [ξ1 g1 + ξ2 g2 + ξ3 g3 ,η1 g1 + η2 g2 + η3 g3 ] =
                                                      →     →                    →     →                    →       →
                                        = ξ1η1[ g1 , g1 ] + ξ1η2 [ g1 , g2 ] + ξ1η3 [ g1 , g3 ] +
                                                        →    →                    →      →                    →      →
                                        + ξ2η1[ g 2 , g1 ] + ξ2η2 [ g 2 , g 2 ] + ξ2η3 [ g 2 , g 3 ] +                                                                                  
                                                      →      →                    →     →                    →       →
                                        + ξ3η1[ g3 , g1 ] + ξ3η2 [ g3 , g2 ] + ξ3η3 [ g3 , g3 ] =
                                               3               →      →                    →      →                      →    →              3     3              →     →
                                        = ∑ ( ξj η1[ g j , g1 ] + ξj η2 [ g j , g2 ] + ξj η3 [ g j , g 3 ]) = ∑ ∑ ξj ηi [ g j , gi ]                                                .
                                             j =1                                                                                           j =1 i =1

                                                                 →      →               →
                  |­ºÏÓÈÒä ˯ËÏ f1 , f 2  Ò f 3  ¹º¹È¯Ó©Ë mË}ˆº¯Ó©Ë ¹¯ºÒÏmËËÓÒ« ­ÈÏÒ°Ó©²
                         →       →
mË}ˆº¯ºm [ g i , g j ] °ãË‚ Ò亭¯ÈϺä
                                                               →            →      →               →            →       →             →            →      →
                                                               f1 = [ g2 , g3 ] ;                  f 2 = [ g 3 , g1 ] ;               f 3 = [ g 1 , g 2 ] 
                  
                                                                                                                  → →
º°ˆÈmã««ªˆÒº­ºÏÓÈËÓÒ«mm©¯ÈÎËÓÒËã« [a , b ] ÒÒ°¹ºã ς«‚ˆm˯ÎËÓÒˈ˺¯Ëä©
¹ºã‚Òä
         
                      →→                                     →                                 →                                  →
                  [a, b ]= (ξ 2η3 − ξ 3η 2 ) f1 − (ξ1η3 − ξ 3η1 ) f 2 + (ξ1η 2 − ξ 2η1 ) f 3 =
                                                                                                                                          →         →         →
                                                                                                                                   f1 f 2 f 3
                                 ξ 2 ξ3 →        ξ ξ3 →          ξ ξ2                                                   →
                      = det             f1 − det 1     f 2 + det 1                                                      f 3 = det ξ1 ξ 2 ξ 3 
                                 η 2 η3          η1 η3           η1 η 2
                                                                                                                                  η1 η 2 η3
                  
                  
                  
vã‚È®º¯ˆºÓº¯äÒ¯ºmÈÓÓºº­ÈÏÒ°È
                  
                                                                                → → →
                  ‚°ˆ  Ò°²ºÓ©® ­ÈÏÒ° {e1 , e2 , e3 }  º¯ˆºÓº¯äÒ¯ºmÈÓÓ©® º­¯Èς Ò® ¹¯Èm‚ 
                                                                                                             →          →         →         →           →         →
ˆ¯º®}‚mË}ˆº¯ºmˆºÈ¹ºº¹¯ËËãËÓÒ  f 1 = e1 ,                                                                         f 2 = e2 ,            f 3 = e3 
       
       

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