"양의 정부호 행렬(positive definite matrix)"의 두 판 사이의 차이

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21번째 줄: 21번째 줄:
 
==2×2 행렬의 경우==
 
==2×2 행렬의 경우==
  
−
*  행렬<br><math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math><br>
+
*  행렬:<math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math><br>
−
*  principal submatrix<br><math>\left( \begin{array}{c}  a_{1,1} \end{array} \right)</math>, <math>\left( \begin{array}{c}  a_{2,2} \end{array} \right)</math>, <math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math><br>
+
*  principal submatrix:<math>\left( \begin{array}{c}  a_{1,1} \end{array} \right)</math>, <math>\left( \begin{array}{c}  a_{2,2} \end{array} \right)</math>, <math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math><br>
−
*  leading principal submatrix<br><math>\left( \begin{array}{c}  a_{1,1} \end{array} \right)</math>, <math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math><br>
+
*  leading principal submatrix:<math>\left( \begin{array}{c}  a_{1,1} \end{array} \right)</math>, <math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math><br>
  
 
 
 
 
31번째 줄: 31번째 줄:
 
==3×3 행렬의 경우==
 
==3×3 행렬의 경우==
  
−
*  행렬<br><math>\left( \begin{array}{ccc}  a_{1,1} & a_{1,2} & a_{1,3} \\  a_{2,1} & a_{2,2} & a_{2,3} \\  a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)</math><br>
+
*  행렬:<math>\left( \begin{array}{ccc}  a_{1,1} & a_{1,2} & a_{1,3} \\  a_{2,1} & a_{2,2} & a_{2,3} \\  a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)</math><br>
−
*  principal submatrix<br><math>\left( \begin{array}{c}  a_{1,1} \end{array} \right)</math>,<math>\left( \begin{array}{c}  a_{2,2} \end{array} \right)</math>,<math>\left( \begin{array}{c}  a_{3,3} \end{array} \right)</math><br><math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math>, <math>\left( \begin{array}{cc}  a_{1,1} & a_{1,3} \\  a_{3,1} & a_{3,3} \end{array} \right)</math>, <math>\left( \begin{array}{cc}  a_{2,2} & a_{2,3} \\  a_{3,2} & a_{3,3} \end{array} \right)</math><br><math>\left( \begin{array}{ccc}  a_{1,1} & a_{1,2} & a_{1,3} \\  a_{2,1} & a_{2,2} & a_{2,3} \\  a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)</math><br>
+
*  principal submatrix:<math>\left( \begin{array}{c}  a_{1,1} \end{array} \right)</math>,<math>\left( \begin{array}{c}  a_{2,2} \end{array} \right)</math>,<math>\left( \begin{array}{c}  a_{3,3} \end{array} \right)</math>:<math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math>, <math>\left( \begin{array}{cc}  a_{1,1} & a_{1,3} \\  a_{3,1} & a_{3,3} \end{array} \right)</math>, <math>\left( \begin{array}{cc}  a_{2,2} & a_{2,3} \\  a_{3,2} & a_{3,3} \end{array} \right)</math>:<math>\left( \begin{array}{ccc}  a_{1,1} & a_{1,2} & a_{1,3} \\  a_{2,1} & a_{2,2} & a_{2,3} \\  a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)</math><br>
−
*  leading principal submatrix<br><math>\left( \begin{array}{c}  a_{1,1} \end{array} \right)</math><math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math>, <math>\left( \begin{array}{ccc}  a_{1,1} & a_{1,2} & a_{1,3} \\  a_{2,1} & a_{2,2} & a_{2,3} \\  a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)</math><br>
+
*  leading principal submatrix:<math>\left( \begin{array}{c}  a_{1,1} \end{array} \right)</math><math>\left( \begin{array}{cc}  a_{1,1} & a_{1,2} \\  a_{2,1} & a_{2,2} \end{array} \right)</math>, <math>\left( \begin{array}{ccc}  a_{1,1} & a_{1,2} & a_{1,3} \\  a_{2,1} & a_{2,2} & a_{2,3} \\  a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)</math><br>
  
 
 
 
 
43번째 줄: 43번째 줄:
 
==예==
 
==예==
  
−
*  다음과 같은 5x5 행렬을 생각하자<br><math>\left( \begin{array}{ccccc}  2 & -1 & 0 & 0 & 0 \\  -1 & 2 & -1 & 0 & 0 \\  0 & -1 & 2 & -1 & 0 \\  0 & 0 & -1 & 2 & -1 \\  0 & 0 & 0 & -1 & 1 \end{array} \right)</math><br>
+
*  다음과 같은 5x5 행렬을 생각하자:<math>\left( \begin{array}{ccccc}  2 & -1 & 0 & 0 & 0 \\  -1 & 2 & -1 & 0 & 0 \\  0 & -1 & 2 & -1 & 0 \\  0 & 0 & -1 & 2 & -1 \\  0 & 0 & 0 & -1 & 1 \end{array} \right)</math><br>
−
*  leading principal submatrix와 그 행렬식을 구하면 다음과 같다<br><math>\begin{array}{ll}  \left( \begin{array}{c}  2 \end{array} \right) & 2 \\  \left( \begin{array}{cc}  2 & -1 \\  -1 & 2 \end{array} \right) & 3 \\  \left( \begin{array}{ccc}  2 & -1 & 0 \\  -1 & 2 & -1 \\  0 & -1 & 2 \end{array} \right) & 4 \\  \left( \begin{array}{cccc}  2 & -1 & 0 & 0 \\  -1 & 2 & -1 & 0 \\  0 & -1 & 2 & -1 \\  0 & 0 & -1 & 2 \end{array} \right) & 5 \\  \left( \begin{array}{ccccc}  2 & -1 & 0 & 0 & 0 \\  -1 & 2 & -1 & 0 & 0 \\  0 & -1 & 2 & -1 & 0 \\  0 & 0 & -1 & 2 & -1 \\  0 & 0 & 0 & -1 & 1 \end{array} \right) & 1 \end{array}</math><br>
+
*  leading principal submatrix와 그 행렬식을 구하면 다음과 같다:<math>\begin{array}{ll}  \left( \begin{array}{c}  2 \end{array} \right) & 2 \\  \left( \begin{array}{cc}  2 & -1 \\  -1 & 2 \end{array} \right) & 3 \\  \left( \begin{array}{ccc}  2 & -1 & 0 \\  -1 & 2 & -1 \\  0 & -1 & 2 \end{array} \right) & 4 \\  \left( \begin{array}{cccc}  2 & -1 & 0 & 0 \\  -1 & 2 & -1 & 0 \\  0 & -1 & 2 & -1 \\  0 & 0 & -1 & 2 \end{array} \right) & 5 \\  \left( \begin{array}{ccccc}  2 & -1 & 0 & 0 & 0 \\  -1 & 2 & -1 & 0 & 0 \\  0 & -1 & 2 & -1 & 0 \\  0 & 0 & -1 & 2 & -1 \\  0 & 0 & 0 & -1 & 1 \end{array} \right) & 1 \end{array}</math><br>
  
 
 
 
 

2013년 1월 12일 (토) 10:57 판

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개요

  • 실계수 n×n 행렬 M이 모든 0이 아닌 벡터 v 에 대하여, \(v^{T}M v > 0 \) 를 만족시킬 때, 양의 정부호 행렬이라 한다
  • 실베스터 판정법 - leading principal minor 가 모두 양수이면 양의 정부호 행렬이다
  • 다변수함수의 극점을 분류하는 헤세 판정법 에 응용할 수 있다

 

 

 

2×2 행렬의 경우

  • 행렬\[\left( \begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array} \right)\]
  • principal submatrix\[\left( \begin{array}{c} a_{1,1} \end{array} \right)\], \(\left( \begin{array}{c} a_{2,2} \end{array} \right)\), \(\left( \begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array} \right)\)
  • leading principal submatrix\[\left( \begin{array}{c} a_{1,1} \end{array} \right)\], \(\left( \begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array} \right)\)

 

 

3×3 행렬의 경우

  • 행렬\[\left( \begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)\]
  • principal submatrix\[\left( \begin{array}{c} a_{1,1} \end{array} \right)\],\(\left( \begin{array}{c} a_{2,2} \end{array} \right)\),\(\left( \begin{array}{c} a_{3,3} \end{array} \right)\)\[\left( \begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array} \right)\], \(\left( \begin{array}{cc} a_{1,1} & a_{1,3} \\ a_{3,1} & a_{3,3} \end{array} \right)\), \(\left( \begin{array}{cc} a_{2,2} & a_{2,3} \\ a_{3,2} & a_{3,3} \end{array} \right)\)\[\left( \begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)\]
  • leading principal submatrix\[\left( \begin{array}{c} a_{1,1} \end{array} \right)\]\(\left( \begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array} \right)\), \(\left( \begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \end{array} \right)\)

 

 

 

예

  • 다음과 같은 5x5 행렬을 생각하자\[\left( \begin{array}{ccccc} 2 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & -1 & 1 \end{array} \right)\]
  • leading principal submatrix와 그 행렬식을 구하면 다음과 같다\[\begin{array}{ll} \left( \begin{array}{c} 2 \end{array} \right) & 2 \\ \left( \begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array} \right) & 3 \\ \left( \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right) & 4 \\ \left( \begin{array}{cccc} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array} \right) & 5 \\ \left( \begin{array}{ccccc} 2 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & -1 & 1 \end{array} \right) & 1 \end{array}\]

 

 

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