"코쉬 행렬과 행렬식"의 두 판 사이의 차이

2021년 2월 17일 (수) 06:01 기준 최신판

개요

행렬 \(A=({\frac{1}{x_i-y_j}})_{1\le i,j\le n}\)를 크기 n인 코쉬 행렬이라 함
행렬식

\[ \det \left(\frac{1}{x _i-y _j}\right) _{1\le i,j \le n}=(-1)^{\binom{n}{2}}\frac{\prod _{1\le i < j\le n} (x_j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i-y _j)} \] \[ \det \left(\frac{1}{x _i+y _j}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x_j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)} \]

예

n=1인 경우

\(\left( \begin{array}{c} \frac{1}{x_1-y_1} \end{array} \right)\)

n=2인 경우

코쉬 행렬

\[\left( \begin{array}{cc} \frac{1}{x_1-y_1} & \frac{1}{x_1-y_2} \\ \frac{1}{x_2-y_1} & \frac{1}{x_2-y_2} \end{array} \right)\]

행렬식

\[ \frac{\left(x_1-x_2\right) \left(y_1-y_2\right)}{\left(x_1-y_1\right) \left(y_1-x_2\right) \left(x_1-y_2\right) \left(x_2-y_2\right)} \]

n=3인 경우

코쉬 행렬은

\[\left( \begin{array}{ccc} \frac{1}{x_1-y_1} & \frac{1}{x_1-y_2} & \frac{1}{x_1-y_3} \\ \frac{1}{x_2-y_1} & \frac{1}{x_2-y_2} & \frac{1}{x_2-y_3} \\ \frac{1}{x_3-y_1} & \frac{1}{x_3-y_2} & \frac{1}{x_3-y_3} \end{array} \right)\]

행렬식은

\[-\frac{\left(-x_1+x_2\right) \left(-x_1+x_3\right) \left(-x_2+x_3\right) \left(y_1-y_2\right) \left(y_1-y_3\right) \left(y_2-y_3\right)}{\left(x_3-y_1\right) \left(-x_1+y_1\right) \left(-x_2+y_1\right) \left(x_2-y_2\right) \left(x_3-y_2\right) \left(-x_1+y_2\right) \left(x_1-y_3\right) \left(x_2-y_3\right) \left(x_3-y_3\right)}\]

n=4인 경우

코쉬 행렬은

\[\left( \begin{array}{cccc} \frac{1}{x_1-y_1} & \frac{1}{x_1-y_2} & \frac{1}{x_1-y_3} & \frac{1}{x_1-y_4} \\ \frac{1}{x_2-y_1} & \frac{1}{x_2-y_2} & \frac{1}{x_2-y_3} & \frac{1}{x_2-y_4} \\ \frac{1}{x_3-y_1} & \frac{1}{x_3-y_2} & \frac{1}{x_3-y_3} & \frac{1}{x_3-y_4} \\ \frac{1}{x_4-y_1} & \frac{1}{x_4-y_2} & \frac{1}{x_4-y_3} & \frac{1}{x_4-y_4} \end{array} \right)\]

역사

수학사 연표

메모

http://mathoverflow.net/questions/20609/what-role-does-cauchys-determinant-identity-play-in-combinatorics

매스매티카 파일 및 계산 리소스

https://docs.google.com/leaf?id=0B8XXo8Tve1cxM2E1ODYzMGUtYTJhMi00MmYxLWEzZDMtZDI2NmZmMWZmMDdm&sort=name&layout=list&num=50

사전 형태의 자료

http://en.wikipedia.org/wiki/Cauchy_matrix

메타데이터

위키데이터

ID : Q2915997

Spacy 패턴 목록

[{'LOWER': 'cauchy'}, {'LEMMA': 'matrix'}]

@@ 1번째 줄: / 1번째 줄: @@
 ==개요==
-* 행렬 $A=(a_{i,j})_{1\ge i,j\le n}$를 크기 n인 코쉬 행렬이라 함. 여기서
+* 행렬 <math>A=({\frac{1}{x_i-y_j}})_{1\le i,j\le n}</math>를 크기 n인 코쉬 행렬이라 함
-:<math>a_{ij}={\frac{1}{x_i-y_j}}</math>
+* 행렬식
-* 행렬식의 계산
+:<math>
+\det \left(\frac{1}{x _i-y _j}\right) _{1\le i,j \le n}=(-1)^{\binom{n}{2}}\frac{\prod _{1\le i < j\le n} (x_j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i-y _j)}
+</math>
+:<math>
+\det \left(\frac{1}{x _i+y _j}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x_j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)}
+</math>
+==예==
-==n=1인 경우==
+===n=1인 경우===
 * <math>\left( \begin{array}{c}  \frac{1}{x_1-y_1} \end{array} \right)</math>
-==n=2인 경우==
+===n=2인 경우===
-* <math>\left( \begin{array}{cc}  \frac{1}{x_1-y_1} & \frac{1}{x_1-y_2} \\  \frac{1}{x_2-y_1} & \frac{1}{x_2-y_2} \end{array} \right)</math>
+* 코쉬 행렬
+:<math>\left( \begin{array}{cc}  \frac{1}{x_1-y_1} & \frac{1}{x_1-y_2} \\  \frac{1}{x_2-y_1} & \frac{1}{x_2-y_2} \end{array} \right)</math>
+* 행렬식
+:<math>
+\frac{\left(x_1-x_2\right) \left(y_1-y_2\right)}{\left(x_1-y_1\right) \left(y_1-x_2\right) \left(x_1-y_2\right) \left(x_2-y_2\right)}
+</math>
+===n=3인 경우===
-==n=3인 경우==
 * 코쉬 행렬은
 :<math>\left( \begin{array}{ccc}  \frac{1}{x_1-y_1} & \frac{1}{x_1-y_2} & \frac{1}{x_1-y_3} \\  \frac{1}{x_2-y_1} & \frac{1}{x_2-y_2} & \frac{1}{x_2-y_3} \\  \frac{1}{x_3-y_1} & \frac{1}{x_3-y_2} & \frac{1}{x_3-y_3} \end{array} \right)</math>
@@ 21번째 줄: / 30번째 줄: @@
 :<math>-\frac{\left(-x_1+x_2\right) \left(-x_1+x_3\right) \left(-x_2+x_3\right) \left(y_1-y_2\right) \left(y_1-y_3\right) \left(y_2-y_3\right)}{\left(x_3-y_1\right) \left(-x_1+y_1\right) \left(-x_2+y_1\right) \left(x_2-y_2\right) \left(x_3-y_2\right) \left(-x_1+y_2\right) \left(x_1-y_3\right) \left(x_2-y_3\right) \left(x_3-y_3\right)}</math>
-==n=4인 경우==
+===n=4인 경우===
 * 코쉬 행렬은
 :<math>\left( \begin{array}{cccc}  \frac{1}{x_1-y_1} & \frac{1}{x_1-y_2} & \frac{1}{x_1-y_3} & \frac{1}{x_1-y_4} \\  \frac{1}{x_2-y_1} & \frac{1}{x_2-y_2} & \frac{1}{x_2-y_3} & \frac{1}{x_2-y_4} \\  \frac{1}{x_3-y_1} & \frac{1}{x_3-y_2} & \frac{1}{x_3-y_3} & \frac{1}{x_3-y_4} \\  \frac{1}{x_4-y_1} & \frac{1}{x_4-y_2} & \frac{1}{x_4-y_3} & \frac{1}{x_4-y_4} \end{array} \right)</math>
 ==역사==
 * [[수학사 연표]]
 ==메모==
+* http://mathoverflow.net/questions/20609/what-role-does-cauchys-determinant-identity-play-in-combinatorics
 ==관련된 항목들==
+* [[반데몬드 행렬과 행렬식 (Vandermonde matrix)]]
 * [[힐버트 행렬]]
 ==매스매티카 파일 및 계산 리소스==
 * https://docs.google.com/leaf?id=0B8XXo8Tve1cxM2E1ODYzMGUtYTJhMi00MmYxLWEzZDMtZDI2NmZmMWZmMDdm&sort=name&layout=list&num=50
-* [[매스매티카 파일 목록]]
+==사전 형태의 자료==
+* http://en.wikipedia.org/wiki/Cauchy_matrix
-==사전 형태의 자료==
+==관련논문==
-* http://en.wikipedia.org/wiki/Cauchy_matrix
+* Ishikawa, Masao, Soichi Okada, Hiroyuki Tagawa, and Jiang Zeng. “Generalizations of Cauchy’s Determinant and Schur’s Pfaffian.” Advances in Applied Mathematics 36, no. 3 (2006): 251–87. doi:10.1016/j.aam.2005.07.001.
+* Chen, William Y. C., Christian Krattenthaler, and Arthur L. B. Yang. “The Flagged Cauchy Determinant.” Graphs and Combinatorics 21, no. 1 (2005): 51–62. doi:10.1007/s00373-004-0593-9.
 [[분류:선형대수학]]
+[[분류:대칭다항식]]
+[[분류:행렬식]]
+==메타데이터==
+===위키데이터===
+* ID :  [https://www.wikidata.org/wiki/Q2915997 Q2915997]
+===Spacy 패턴 목록===
+* [{'LOWER': 'cauchy'}, {'LEMMA': 'matrix'}]