Straipsnio aptarimas iš Vikipedijos, laisvosios enciklopedijos.
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{\displaystyle (a\alpha \cdot b\beta )=ab(\alpha \cdot \beta )}
N-mačių eilučių tikrojoje erdvėje skaliarinę sandaugą galime įvesti panašiai kaip trimatėje erdvėje. Vektorių-eilučių
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{\displaystyle [\alpha ]=[a_{1},a_{2},...,a_{n}]}
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{\displaystyle [\beta ]=[b_{1},b_{2},...,b_{n}]}
skaliarinę sandaugą taip apibrėšime
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{\displaystyle [\alpha ]\cdot [\beta ]=[a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}]}
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{\displaystyle [\alpha ]\cdot [\beta ]=\sum _{s=1}^{n}a_{s}b_{s}.}
Euklidinę eilučių erdvę galima apibrėžti ir taip:
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{\displaystyle [\alpha ]\cdot [\beta ]=}
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{\displaystyle =s_{1,1}a_{1}b_{1}+s_{1,2}a_{1}b_{2}+...+s_{1,n}a_{1}b_{n}+}
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{\displaystyle +s_{2,1}a_{2}b_{1}+s_{2,2}a_{2}b_{2}+...+s_{2,n}a_{2}b_{n}+}
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{\displaystyle .................................}
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{\displaystyle +s_{n,1}a_{n}b_{1}+s_{n,2}a_{n}b_{2}+...+s_{n,n}a_{n}b_{n}.}
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{\displaystyle =\sum _{j,k=1}^{n}s_{j,k}a_{j}b_{k}.}
Tegu turime n-matę unitarinę (euklidinę) erdvę ir jos bazę
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{\displaystyle {\omega }={w_{1},w_{2},...,w_{n}}}
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{\displaystyle w_{1},w_{2},...,w_{n}}
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{\displaystyle [\alpha ]\cdot [\omega ]=[a_{1}w_{1}+a_{2}w_{2}+...+a_{n}w_{n}]}
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{\displaystyle [\beta ]\cdot [\omega ]=[a_{1}w_{1}+a_{2}w_{2}+...+a_{n}w_{n}]}
skaliarinę sandaugą.
Kadangi formulė (1) matematinės indukcijos metodu lengvai pakeičiama n dėmenų atvejui, tai
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{\displaystyle (\alpha \cdot \beta )=(a_{1}w_{1}+a_{2}w_{2}+...+a_{n}w_{n}\cdot b_{1}w_{1}+b_{2}w_{2}+...+b_{n}w_{n})=}
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{\displaystyle =\sum _{j,k=1}^{n}a_{j}b_{k}(w_{j}\cdot w_{k}).}
Vektorių skaliarinė sandauga yra skaliaras, todėl
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{\displaystyle (w_{j}\cdot w_{k})=s_{j,k}}
Matrica S (kuri tapati ermitinei matricai H ) turi būti simetrine, t.y. jos elementai turi tenkinti šią sąlygą:
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{\displaystyle s_{j,k}=s_{k,j}}
Pavyzdžiui turime vektorius a=(3, 5, 7) b=(4, 6, 8) ir w=(2, 3, 4).
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{\displaystyle \alpha =a\cdot \omega =[3\cdot 2+5\cdot 3+7\cdot 4]=6+15+28=49}
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{\displaystyle \beta =b\cdot \omega =[4\cdot 2+6\cdot 3+8\cdot 4]=8+18+32=58}
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{\displaystyle (\alpha \cdot \beta )=49\cdot 58=2842}
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{\displaystyle \sum _{j,k=1}^{n}a_{j}b_{k}(w_{j}\cdot w_{k})=}
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{\displaystyle =3\cdot 4\cdot (2\cdot 2)+3\cdot 6\cdot (2\cdot 3)+3\cdot 8\cdot (2\cdot 4)+}
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{\displaystyle +5\cdot 4\cdot (3\cdot 2)+5\cdot 6\cdot (3\cdot 3)+5\cdot 8\cdot (3\cdot 4)+}
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{\displaystyle +7\cdot 4\cdot (4\cdot 2)+7\cdot 6\cdot (4\cdot 3)+7\cdot 8\cdot (4\cdot 4)=}
=48+108+192+120+270+480+224+504+896=348+870+1624=2842
Kitas pavyzdys. Turime vektorius a=[3, 4, 5, 6], b=[2, 5, 3, 7], c=[6, 2, 4, 7], d=[5, 4, 6, 3].
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{\displaystyle \alpha =a\cdot c}
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{\displaystyle \beta =b\cdot d}
Skaliarų sandauga
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{\displaystyle \alpha \cdot \beta =88*69=6072}
Formulė atrodys taip:
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{\displaystyle \sum _{j,k=1}^{n}a_{j}b_{k}(c_{j}\cdot d_{k})=}
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{\displaystyle =a_{1}b_{1}(c_{1}\cdot d_{1})+a_{1}b_{2}(c_{1}\cdot d_{2})+a_{1}b_{3}(c_{1}\cdot d_{3})+a_{1}b_{4}(c_{1}\cdot d_{4})+}
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{\displaystyle +a_{2}b_{1}(c_{2}\cdot d_{1})+a_{2}b_{2}(c_{2}\cdot d_{2})+a_{2}b_{3}(c_{2}\cdot d_{3})+a_{2}b_{4}(c_{2}\cdot d_{4})+}
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{\displaystyle +a_{3}b_{1}(c_{3}\cdot d_{1})+a_{3}b_{2}(c_{3}\cdot d_{2})+a_{3}b_{3}(c_{3}\cdot d_{3})+a_{3}b_{4}(c_{3}\cdot d_{4})+}
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{\displaystyle +a_{4}b_{1}(c_{4}\cdot d_{1})+a_{4}b_{2}(c_{4}\cdot d_{2})+a_{4}b_{3}(c_{4}\cdot d_{3})+a_{4}b_{4}(c_{4}\cdot d_{4})=}
=3*2*6*5+3*5*6*4+3*3*6*6+3*7*6*3+
+4*2*2*5+4*5*2*4+4*3*2*6+4*7*2*3+
+5*2*4*5+5*5*4*4+5*3*4*6+5*7*4*3+
+6*2*7*5+6*5*7*4+6*3*7*6+6*7*7*3=
=180+360+324+378+
+80+160+144+168+
+200+400+360+420+
+420+840+756+882=
=1242+
+552+
+1380+
+2898=
=6072 . Panagrinėkime atvejį, kai a=(3, 4, 5), b=(6, 7, 8), w=(1, 1, 1).
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{\displaystyle \alpha =a\cdot w}
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{\displaystyle \beta =a\cdot w}
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{\displaystyle \alpha \cdot \beta }
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{\displaystyle \alpha \cdot \beta =\sum _{j,k=1}^{n}a_{j}b_{k}(w_{j}\cdot w_{k})=}
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{\displaystyle =a_{1}b_{1}(1\cdot 1)+a_{1}b_{2}(1\cdot 1)+a_{1}b_{3}(1\cdot 1)+}
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{\displaystyle +a_{2}b_{1}(1\cdot 1)+a_{2}b_{2}(1\cdot 1)+a_{2}b_{3}(1\cdot 1)+}
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{\displaystyle +a_{3}b_{1}(1\cdot 1)+a_{3}b_{2}(1\cdot 1)+a_{3}b_{3}(1\cdot 1)=}
=3*6*1+3*7*1+3*8*1+
+4*6*1+4*7*1+4*8*1+
+5*6*1+5*7*1+5*8*1=
=63+84+105=252 Pavyzdžiai iš pagrindinio straipsnio iškelti į diskusijas, nes tai bereikalingas straipsnelio apkrovimas. Orionus 12:35, 2007 Gegužės 4 (EEST)
Kairėje sumos distributyvumo lygybės pusėje vietoj sandaugos turėtų būti skliaustas.
Galite pataisyti. :) --Homo 12:30, 2008 rugsėjo 20 (EEST)
![{\displaystyle [\alpha ]=[a_{1},a_{2},...,a_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/649e681e916bf197df29ae77c2f4b8ee6408a486)
![{\displaystyle [\beta ]=[b_{1},b_{2},...,b_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2871585a8c6f1da772ba21ac8f417c5185cedb1a)
![{\displaystyle [\alpha ]\cdot [\beta ]=[a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f2632216a341bdb11a491e87e8bc50e38fa5347)
![{\displaystyle [\alpha ]\cdot [\beta ]=\sum _{s=1}^{n}a_{s}b_{s}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/785bf47e62a688f834623456ccbbe01a84e12eee)
![{\displaystyle [\alpha ]\cdot [\beta ]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fb702004f2c291f6b0c5b289e47763e904c225a)
![{\displaystyle [\alpha ]\cdot [\omega ]=[a_{1}w_{1}+a_{2}w_{2}+...+a_{n}w_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/433ca0dbff57011ef6c813a28c495db77d2a2e59)
![{\displaystyle [\beta ]\cdot [\omega ]=[a_{1}w_{1}+a_{2}w_{2}+...+a_{n}w_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/350748270da6d1637c885ea586db34d51dc4fe3e)
![{\displaystyle \alpha =a\cdot \omega =[3\cdot 2+5\cdot 3+7\cdot 4]=6+15+28=49}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7844131d465e152ab57244061c083b17d415ead)
![{\displaystyle \beta =b\cdot \omega =[4\cdot 2+6\cdot 3+8\cdot 4]=8+18+32=58}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef4858b5bf47e5fd1b32f70d134513e79a33ee46)
