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Linear Algebra Lecture #3 [齋藤xx]は「齋藤正彦著『線型代数入門』東京大学出版会のxxページ参照」の意味。 [語句] augmented matrix ここでは行列 A と単位行列を並べたもの[齋藤 53 例 2:この方法で逆行列を求 めることを講義では Gauss-Jordan idea と呼んでいる] block multiplication 区分けを使った掛け算[齋藤 38] determinant 行列式[齋藤 78] dot product 内積(a・b で表す。“a dot b”と読む;= inner product) entry(行列の)成分[齋藤 31] Gauss ガウス linearly dependent 線型従属[齋藤 99] linearly independent 線型独立[齋藤 99] Jordan ジョルダン(フランスの数学者;Jordan canonical form ジョルダンの標準形) left inverse 左逆元(e を単位元とすると、a の左逆元とは、xa=e を満たす x のこと) Lo and behold, the inverse will show up here. rectangular matrix 長方形の行列(正方行列でない行列のこと) right inverse 右逆元(e を単位元とすると、a の右逆元とは、ax=e を満たす x のこと;正方行列で は右逆元と左逆元は一致する。[齋藤 88 (5)]) square matrix 正方行列[齋藤 40] summation formula ここではシグマ記号を使って表すことを指す。 transpose 転置行列(この講義では A の転置行列を表すのに AT という記号(A transpose と読む)を使 っている) [齋藤 37] Multiplication and Inverse Matrices Overview of the lecture matrix multiplication (four ways) Inverse of A, AB and AT Gauss-Jordan way of finding A-1 How to multiply matrices First Way: regular way [ m×n ] [ n×p ] = [ m×p ] A B= C n the (i, j)-entry of C = cij = (row i of A)・(column j of B) = ∑ aik bkj k=1 “the sum from k equals 1 to n of a sub ik times b sub kj” Second Way: column way (matrix times column vector) [ m×n ] [ n×p ] = [ m×p ] A B= C A×(column 1 of B) = column 1 of C A×(column 2 of B) = column 2 of C A×(column p of B) = column p of C from lecture 2 ??? 3 ? ? ? × 4 = 3×(column 1) + 4×(column 2) + 5×(column 3) ??? 5 3×3 3×1 3×1 (matrix×column vector= linear combination of columns of the matrix) column k of C = A×(column k of B) = linear combination of columns of A Third Way: row way (row vector times matrix) (row 1 of A)×B = row 1 of C (row 2 of A)×B = row 2 of C (row m of A)×B = row m of C from lecture 2 ??? [1 2 7]×? ? ? = 1×(row 1) + 2×(row 2) + 7×(row 3) ??? 1×3 3×3 1×3 row×matrix = linear combination of rows of the matrix row k of C = (row k of A)×B = linear combination of rows of B Fourth Way: column times row (column of A)×(row of B) = m×1 1×p 2 2 12 3 ×[1 6] = 3 18 4 4 24 m×p the row of the product is a multiple of the row vector the column of the product is a multiple of the column vector n A B = ∑ (column k of A)×(row k of B) k=1 block multiplication (multiplication by blocks) Inverse Matrix A: square matrix Is it invertible (non-singular) or not? If the inverse matrix (A-1) exists, A-1A = I (left inverse) AA-1 = I (right inverse) For square matrices, the left inverses are also right inverses. the case where there is no inverse (singular case) A = 1 3 Why is A singular? (The determinant is zero.) 26 Suppose 1 3 ? ? =1 0 Column 1 of the product 1 is linear combination of 1 and 3 26 ? ? 01 0 2 6 This is impossible. The square matrix A is not invertible iff there is a non-zero x with Ax = 0. 1 3 3= 0 2 6 -1 0 Why is A not invertible? Because A-1Ax = 0, hence x = 0. (contradiction!) How to find the inverse? 13 a c =10 27 bd 01 A A-1 I 13 a =1 27 b 0 13 c =0 27 d 1 Finding the inverse is like solving two systems. In general, A×(column j of A-1) = column j of I Gauss-Jordan way of finding A-1 (solve n systems of equations at once) Do eliminations and get A to I. Use elimination upwards. 1 3 1 0 1 3 1 0 1 0 7 -3 2701 0 1 -2 1 0 1 -2 1 A I I A-1 Why did we get A-1? E [A I] = [I X] E = the product of all the elimination matrices E A = I. So E = A-1 X = E I = E = A-1