行列式det(A) 其实表示的只是一个值 ∣ a b c d ∣ = a d − b c \begin{vmatrix} a & b\\ c & d\end{vmatrix} = ad -bc acbd =ad−bc,其基本变化是基于这个值是不变。而矩阵表示的是一个数表。
矩阵与线性变换的关系
即得
( a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . . a m 1 a m 2 . . . a m n ) ( x 1 x 2 . . . x n ) = ( y 1 y 2 . . . y n ) \begin{pmatrix} a_{11} & a_{12} & ...& a_{1n}\\ a_{21} & a_{22} & ...& a_{2n}\\ ... & ... & ...& ....\\ a_{m1} & a_{m2} & ...& a_{mn}\end{pmatrix} \begin{pmatrix} x_1\\x_2\\...\\x_n\end{pmatrix} = \begin{pmatrix} y_1\\y_2\\...\\y_n\end{pmatrix} a11a21...am1a12a22...am2............a1na2n....amn x1x2...xn = y1y2...yn
可以推矩阵乘法
即得中的 y 1 = c 11 = a 11 x 1 + a 12 x 2 + . . . + a 1 n x m y_1=c_{11}=a_{11}x_1+a_{12}x_2+...+a_{1n}x_m y1=c11=a11x1+a12x2+...+a1nxm
矩阵乘法的提前: 第一个矩阵的列数和第二个矩阵的行数相同
( a 11 a 12 . . . a 1 n y 1 a 21 a 22 . . . a 2 n y 2 . . . . . . . . . . . . . . . . . a m 1 a m 2 . . . a m n y n ) ( x 1 x 2 . . . x n − 1 ) = 0 \begin{pmatrix} a_{11} & a_{12} & ...& a_{1n} & y_1\\ a_{21} & a_{22} & ...& a_{2n}& y_2\\ ... & ... & ...& .... & ....\\ a_{m1} & a_{m2} & ...& a_{mn} & y_n\end{pmatrix} \begin{pmatrix} x_1\\x_2\\...\\x_n\\-1\end{pmatrix} = 0 a11a21...am1a12a22...am2............a1na2n....amny1y2....yn x1x2...xn−1 =0
行和列的关系
( x 1 x 2 . . . x n ) ( a 11 a 21 . . . a m 1 a 12 a 22 . . . a m 2 . . . . . . . . . . . . . a 1 n a 2 n . . . a m n ) = ( y 1 y 2 . . y n ) \begin{pmatrix} x_1&x_2&...&x_n\end{pmatrix} \begin{pmatrix} a_{11} & a_{21} & ...& a_{m1}\\ a_{12} & a_{22} & ...& a_{m2}\\ ... & ... & ...& ....\\ a_{1n} & a_{2n} & ...& a_{mn}\end{pmatrix} = \begin{pmatrix} y_1&y_2&..&y_n\end{pmatrix} (x1x2...xn) a11a12...a1na21a22...a2n............am1am2....amn =(y1y2..yn)
E m ( i , j ) = ( 1 0 . . . 0 0 0 1 i 行 . . . 0 0 . . . . . . . . . . . . . . . . . 0 0 . . . 1 j 行 0 0 0 . . . 0 1 ) m 的 i 行与 j 行对调 ( 1 0 . . . 0 0 0 0 . . . 1 i 行 0 . . . . . . . . . . . . . . . . . 0 1 j 行 . . . 0 0 0 0 . . . 0 1 ) m E_m(i,j)=\begin{pmatrix} 1 & 0 & ...& 0& 0\\ 0 & 1_{i行} & ...& 0& 0\\ ... & ... & ...& ....& ....\\ 0 & 0 & ...& 1_{j行}& 0\\ 0 & 0 & ... & 0& 1\end{pmatrix}_m 的 i行与j行对调 \begin{pmatrix} 1 & 0 & ...& 0& 0\\ 0 & 0 & ...& 1_{i行}& 0\\ ... & ... & ...& ....& ....\\ 0 & 1_{j行} & ...& 0& 0\\ 0 & 0 & ... & 0& 1\end{pmatrix}_m Em(i,j)= 10...0001i行...00...............00....1j行000....01 m的i行与j行对调 10...0000...1j行0...............01i行....0000....01 m
E m ( i ( k ) ) = ( 1 0 . . . 0 0 0 1 i 行 . . . 0 0 . . . . . . . . . . . . . . . . . 0 0 . . . 1 0 0 0 . . . 0 1 ) m 的 i 行乘于常数 k ( 1 0 . . . 0 0 0 k i 行 . . . 0 0 . . . . . . . . . . . . . . . . . 0 0 . . . 1 0 0 0 . . . 0 1 ) m E_m(i(k))=\begin{pmatrix} 1 & 0 & ...& 0& 0\\ 0 & 1_{i行} & ...& 0& 0\\ ... & ... & ...& ....& ....\\ 0 & 0 & ...& 1& 0\\ 0 & 0 & ... & 0& 1\end{pmatrix}_m 的 i行乘于常数k \begin{pmatrix} 1 & 0 & ...& 0& 0\\ 0 & k_{i行} & ...& 0& 0\\ ... & ... & ...& ....& ....\\ 0 & 0 & ...& 1& 0\\ 0 & 0 & ... & 0& 1\end{pmatrix}_m Em(i(k))= 10...0001i行...00...............00....1000....01 m的i行乘于常数k 10...000ki行...00...............00....1000....01 m
E m ( i j ( k ) ) = ( 1 0 . . . 0 0 0 1 i 行 . . . 0 0 . . . . . . . . . . . . . . . . . 0 0 . . . 1 j 行 0 0 0 . . . 0 1 ) m i 行的 k 倍加到 j 上 ( 1 0 . . . 0 0 0 1 i 行 . . . 0 0 . . . . . . . . . . . . . . . . . 0 k j 行 . . . 1 j 行 0 0 0 . . . 0 1 ) m E_m(ij(k))=\begin{pmatrix} 1 & 0 & ...& 0& 0\\ 0 & 1_{i行} & ...& 0& 0\\ ... & ... & ...& ....& ....\\ 0 & 0 & ...& 1_{j行}& 0\\ 0 & 0 & ... & 0& 1\end{pmatrix}_m i行的k倍加到j上 \begin{pmatrix} 1 & 0 & ...& 0& 0\\ 0 & 1_{i行} & ...& 0& 0\\ ... & ... & ...& ....& ....\\ 0 & k_{j行} & ...& 1_{j行}& 0\\ 0 & 0 & ... & 0& 1\end{pmatrix}_m Em(ij(k))= 10...0001i行...00...............00....1j行000....01 mi行的k倍加到j上 10...0001i行...kj行0...............00....1j行000....01 m
A n ∗ m ( a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . . a m 1 a m 2 . . . a m n ) 转置为 A n ∗ m T ( a 11 a 21 . . . a m 1 a 12 a 22 . . . a m 2 . . . . . . . . . . . . . a 1 n a 2 n . . . a m n ) A_{n*m} \begin{pmatrix} a_{11} & a_{12} & ...& a_{1n}\\ a_{21} & a_{22} & ...& a_{2n}\\ ... & ... & ...& ....\\ a_{m1} & a_{m2} & ...& a_{mn}\end{pmatrix} 转置为 A_{n*m}^T \begin{pmatrix} a_{11} & a_{21} & ...& a_{m1}\\ a_{12} & a_{22} & ...& a_{m2}\\ ... & ... & ...& ....\\ a_{1n} & a_{2n} & ...& a_{mn}\end{pmatrix} An∗m a11a21...am1a12a22...am2............a1na2n....amn 转置为An∗mT a11a12...a1na21a22...a2n............am1am2....amn
例如:矩阵 B = ( 1 2 3 4 5 6 ) B = \begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6\end{pmatrix} B=(142536)的转置矩阵就是 B T = ( 1 4 2 5 3 6 ) B^T = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6\end{pmatrix} BT= 123456
根据行列式和矩阵乘法的公式刚好得出 A A ∗ = ∣ A ∣ E AA^*=|A|E AA∗=∣A∣E
利用初等变化转为对角矩阵,方便计算
我们通过验证分块矩阵乘法得到的元素与通用乘法得到元素是否一致,来证明分块乘法的可靠性,以 c 32 c_{32} c32为例:
c 32 = ( a 31 a 32 a 33 ) ( b 12 b 22 b 32 ) c_{32}= \begin{pmatrix} a_{31} & a_{32} &a_{33} \end{pmatrix}\begin{pmatrix} b_{12} \\b_{22} \\b_{32} \end{pmatrix} c32=(a31a32a33) b12b22b32
与他对应是 C 11 = A 11 B 11 + A 12 B 21 C_{11}=A_{11}B_{11}+A_{12}B_{21} C11=A11B11+A12B21中的 c 32 c_{32} c32
c 32 = ( a 31 a 32 ) ( b 12 b 22 ) + ( a 33 ) ( b 32 ) c_{32}= \begin{pmatrix} a_{31} & a_{32} \end{pmatrix}\begin{pmatrix} b_{12} \\b_{22} \end{pmatrix} + \begin{pmatrix} a_{33} \end{pmatrix} \begin{pmatrix} b_{32} \end{pmatrix} c32=(a31a32)(b12b22)+(a33)(b32)
《矩阵的转置》
《克拉默法则》
《共轭矩阵》
《分块矩阵的初等变换(3)行列式不变吗?》
《矩阵分块乘法的原理是怎么样的?》