"대칭곱 (symmetric power)과 대칭텐서"의 두 판 사이의 차이

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===$\dim V=2$인 경우===
 
===$\dim V=2$인 경우===
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Block[{columnheading, row, rowheading, maintable, F},
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$$
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  columnheading = {"\!\(\*SuperscriptBox[\(Sym\), \(n\)]\)A"};
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\begin{array}{c|c}
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  row = Table[x, {x, 0, 3}];
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n & \operatorname{Sym}^nA \\
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  rowheading = Prepend[row, ""];
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\hline
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  F = {SymPower[2, #][a]} &;
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0 & \left(
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  maintable = Table[F[n], {n, row}];
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\begin{array}{c}
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  Grid[MapThread[
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1 \\
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    Prepend, {Prepend[maintable, columnheading], rowheading}],
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\end{array}
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  Frame -> All]
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\right) \\
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  ] // TraditionalForm
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1 & \left(
−
 
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\begin{array}{cc}
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a(1,1) & a(1,2) \\
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a(2,1) & a(2,2) \\
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\end{array}
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\right) \\
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2 & \left(
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\begin{array}{ccc}
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a(1,1)^2 & a(1,1) a(1,2) & a(1,2)^2 \\
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2 a(1,1) a(2,1) & a(1,2) a(2,1)+a(1,1) a(2,2) & 2 a(1,2) a(2,2) \\
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a(2,1)^2 & a(2,1) a(2,2) & a(2,2)^2 \\
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\end{array}
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\right) \\
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3 & \left(
 +
\begin{array}{cccc}
 +
a(1,1)^3 & a(1,1)^2 a(1,2) & a(1,1) a(1,2)^2 & a(1,2)^3 \\
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3 a(1,1)^2 a(2,1) & a(2,2) a(1,1)^2+2 a(1,2) a(2,1) a(1,1) & a(2,1) a(1,2)^2+2 a(1,1) a(2,2) a(1,2) & 3 a(1,2)^2 a(2,2) \\
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3 a(1,1) a(2,1)^2 & a(1,2) a(2,1)^2+2 a(1,1) a(2,2) a(2,1) & a(1,1) a(2,2)^2+2 a(1,2) a(2,1) a(2,2) & 3 a(1,2) a(2,2)^2 \\
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a(2,1)^3 & a(2,1)^2 a(2,2) & a(2,1) a(2,2)^2 & a(2,2)^3 \\
 +
\end{array}
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\right) \\
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4 & \left(
 +
\begin{array}{ccccc}
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a(1,1)^4 & a(1,1)^3 a(1,2) & a(1,1)^2 a(1,2)^2 & a(1,1) a(1,2)^3 & a(1,2)^4 \\
 +
4 a(1,1)^3 a(2,1) & a(2,2) a(1,1)^3+3 a(1,2) a(2,1) a(1,1)^2 & 2 a(1,2) a(2,2) a(1,1)^2+2 a(1,2)^2 a(2,1) a(1,1) & a(2,1) a(1,2)^3+3 a(1,1) a(2,2) a(1,2)^2 & 4 a(1,2)^3 a(2,2) \\
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6 a(1,1)^2 a(2,1)^2 & 3 a(2,1) a(2,2) a(1,1)^2+3 a(1,2) a(2,1)^2 a(1,1) & a(1,2)^2 a(2,1)^2+4 a(1,1) a(1,2) a(2,2) a(2,1)+a(1,1)^2 a(2,2)^2 & 3 a(2,1) a(2,2) a(1,2)^2+3 a(1,1) a(2,2)^2 a(1,2) & 6 a(1,2)^2 a(2,2)^2 \\
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4 a(1,1) a(2,1)^3 & a(1,2) a(2,1)^3+3 a(1,1) a(2,2) a(2,1)^2 & 2 a(1,2) a(2,2) a(2,1)^2+2 a(1,1) a(2,2)^2 a(2,1) & a(1,1) a(2,2)^3+3 a(1,2) a(2,1) a(2,2)^2 & 4 a(1,2) a(2,2)^3 \\
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a(2,1)^4 & a(2,1)^3 a(2,2) & a(2,1)^2 a(2,2)^2 & a(2,1) a(2,2)^3 & a(2,2)^4 \\
 +
\end{array}
 +
\right) \\
 +
\end{array}
 +
$$
  
 
==관련된 항목들==
 
==관련된 항목들==

2015년 5월 6일 (수) 20:10 판

개요

  • 벡터공간 $V$에 대하여 대칭곱 $\operatorname{Sym}^n V$를 정의할 수 있다
  • $V$에 작용하는 선형변환 $A$에 대하여 $\operatorname{Sym}^n A$를 정의할 수 있다


행렬의 대칭곱

  • $V$에 작용하는 선형변환 $A$를 생각하자


$\dim V=2$인 경우

$$ \begin{array}{c|c} n & \operatorname{Sym}^nA \\ \hline 0 & \left( \begin{array}{c} 1 \\ \end{array} \right) \\ 1 & \left( \begin{array}{cc} a(1,1) & a(1,2) \\ a(2,1) & a(2,2) \\ \end{array} \right) \\ 2 & \left( \begin{array}{ccc} a(1,1)^2 & a(1,1) a(1,2) & a(1,2)^2 \\ 2 a(1,1) a(2,1) & a(1,2) a(2,1)+a(1,1) a(2,2) & 2 a(1,2) a(2,2) \\ a(2,1)^2 & a(2,1) a(2,2) & a(2,2)^2 \\ \end{array} \right) \\ 3 & \left( \begin{array}{cccc} a(1,1)^3 & a(1,1)^2 a(1,2) & a(1,1) a(1,2)^2 & a(1,2)^3 \\ 3 a(1,1)^2 a(2,1) & a(2,2) a(1,1)^2+2 a(1,2) a(2,1) a(1,1) & a(2,1) a(1,2)^2+2 a(1,1) a(2,2) a(1,2) & 3 a(1,2)^2 a(2,2) \\ 3 a(1,1) a(2,1)^2 & a(1,2) a(2,1)^2+2 a(1,1) a(2,2) a(2,1) & a(1,1) a(2,2)^2+2 a(1,2) a(2,1) a(2,2) & 3 a(1,2) a(2,2)^2 \\ a(2,1)^3 & a(2,1)^2 a(2,2) & a(2,1) a(2,2)^2 & a(2,2)^3 \\ \end{array} \right) \\ 4 & \left( \begin{array}{ccccc} a(1,1)^4 & a(1,1)^3 a(1,2) & a(1,1)^2 a(1,2)^2 & a(1,1) a(1,2)^3 & a(1,2)^4 \\ 4 a(1,1)^3 a(2,1) & a(2,2) a(1,1)^3+3 a(1,2) a(2,1) a(1,1)^2 & 2 a(1,2) a(2,2) a(1,1)^2+2 a(1,2)^2 a(2,1) a(1,1) & a(2,1) a(1,2)^3+3 a(1,1) a(2,2) a(1,2)^2 & 4 a(1,2)^3 a(2,2) \\ 6 a(1,1)^2 a(2,1)^2 & 3 a(2,1) a(2,2) a(1,1)^2+3 a(1,2) a(2,1)^2 a(1,1) & a(1,2)^2 a(2,1)^2+4 a(1,1) a(1,2) a(2,2) a(2,1)+a(1,1)^2 a(2,2)^2 & 3 a(2,1) a(2,2) a(1,2)^2+3 a(1,1) a(2,2)^2 a(1,2) & 6 a(1,2)^2 a(2,2)^2 \\ 4 a(1,1) a(2,1)^3 & a(1,2) a(2,1)^3+3 a(1,1) a(2,2) a(2,1)^2 & 2 a(1,2) a(2,2) a(2,1)^2+2 a(1,1) a(2,2)^2 a(2,1) & a(1,1) a(2,2)^3+3 a(1,2) a(2,1) a(2,2)^2 & 4 a(1,2) a(2,2)^3 \\ a(2,1)^4 & a(2,1)^3 a(2,2) & a(2,1)^2 a(2,2)^2 & a(2,1) a(2,2)^3 & a(2,2)^4 \\ \end{array} \right) \\ \end{array} $$

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