"양의 정부호 행렬(positive definite matrix)"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) |
Pythagoras0 (토론 | 기여) |
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13번째 줄: | 13번째 줄: | ||
==2×2 행렬의 경우== | ==2×2 행렬의 경우== | ||
− | * 행렬:<math>\left( \begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array} \right)</math | + | * 행렬:<math>\left( \begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array} \right)</math> |
* principal submatrix | * 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 | + | <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> |
* leading principal submatrix | * 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}{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> |
25번째 줄: | 25번째 줄: | ||
==3×3 행렬의 경우== | ==3×3 행렬의 경우== | ||
− | * 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 | + | * 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> |
* principal submatrix | * 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}{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}{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 | + | <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> |
* leading principal submatrix | * 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 | + | <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> |
41번째 줄: | 41번째 줄: | ||
==예== | ==예== | ||
− | * 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 | + | * 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> |
− | * 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 | + | * 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> |
87번째 줄: | 87번째 줄: | ||
==수학용어번역== | ==수학용어번역== | ||
− | * [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집] | + | * [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집] |
** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=definite | ** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=definite | ||
** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=minor | ** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=minor |
2020년 11월 13일 (금) 06:25 판
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\[\begin{array}{ll} \left( \begin{array}{c} 2 \end{array} \right) & 2 \\ \left( \begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array} \right) & 3 \\ \left( \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right) & 4 \\ \left( \begin{array}{cccc} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array} \right) & 5 \\ \left( \begin{array}{ccccc} 2 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & -1 & 1 \end{array} \right) & 1 \end{array}\]
역사
메모
- Math Overflow http://mathoverflow.net/search?q=
관련된 항목들
매스매티카 파일 및 계산 리소스
수학용어번역
사전 형태의 자료
- http://ko.wikipedia.org/wiki/
- http://en.wikipedia.org/wiki/Positive-definite_matrix
- http://en.wikipedia.org/wiki/Sylvester's_criterion
리뷰논문, 에세이, 강의노트
관련논문
- Gilbert, George T. 1991. “Positive Definite Matrices and Sylvester’s Criterion”. The American Mathematical Monthly 98 (1) (1월 1): 44-46. doi:10.2307/2324036.