MIT线性代数笔记-第十九讲

Formula for detA

先看一下determinant最核心的三个性质:
1.detI=1$detI=1$
2.sign reverse with row exchange
3.det is linear in each row separately
现在我们要用这三个性质来推出detA的一般公式

先看一下2*2的例子:
∣∣∣acbd∣∣∣=∣∣∣ac0d∣∣∣+∣∣∣0cbd∣∣∣=∣∣∣ac00∣∣∣+∣∣∣a00d∣∣∣+∣∣∣0cb0∣∣∣+∣∣∣00bd∣∣∣=∣∣∣a00d∣∣∣+∣∣∣0cb0∣∣∣=ad−cb$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=\left|\begin{array}{cc}a& 0\\ c& d\end{array}\right|+\left|\begin{array}{cc}0& b\\ c& d\end{array}\right|=\left|\begin{array}{cc}a& 0\\ c& 0\end{array}\right|+\left|\begin{array}{cc}a& 0\\ 0& d\end{array}\right|+\left|\begin{array}{cc}0& b\\ c& 0\end{array}\right|+\left|\begin{array}{cc}0& b\\ 0& d\end{array}\right|=\left|\begin{array}{cc}a& 0\\ 0& d\end{array}\right|+\left|\begin{array}{cc}0& b\\ c& 0\end{array}\right|=ad-cb$
由这个简单的例子我们可以推广到n*n
我们想象一下，如果是3*3，我们会分解出3*3*3 = 27个矩阵，但是里面很多矩阵的det都为0，那么哪些不为0，答案是排列，即为3！ = 6个。
看一下3*3的例子

∣∣∣∣a11a21a31a12a22a32a13a23a33∣∣∣∣=∣∣∣∣a11000a22000a33∣∣∣∣+∣∣∣∣a110000a320a230∣∣∣∣+∣∣∣∣0a210a120000a33∣∣∣∣+∣∣∣∣00a31a12000a230∣∣∣∣+∣∣∣∣0a21000a32a1300∣∣∣∣+∣∣∣∣00a310a220a1300∣∣∣∣=$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=\left|\begin{array}{ccc}{a}_{11}& 0& 0\\ 0& a_{22}& 0\\ 0& 0& a_{33}\end{array}\right|+\left|\begin{array}{ccc}{a}_{11}& 0& 0\\ 0& 0& a_{23}\\ 0& a_{32}& 0\end{array}\right|+\left|\begin{array}{ccc}0& a_{12}& 0\\ a_{21}& 0& 0\\ 0& 0& a_{33}\end{array}\right|+\left|\begin{array}{ccc}0& a_{12}& 0\\ 0& 0& a_{23}\\ a_{31}& 0& 0\end{array}\right|+\left|\begin{array}{ccc}0& 0& a_{13}\\ a_{21}& 0& 0\\ 0& a_{32}& 0\end{array}\right|+\left|\begin{array}{ccc}0& 0& a_{13}\\ 0& a_{22}& 0\\ a_{31}& 0& 0\end{array}\right|=$
a11a22a33−a11a32a23−a21a12a33+a31a12a23+a21a32a13−a31a22a13$a_{11}{a}_{22}{a}_{33}-a_{11}{a}_{32}{a}_{23}-a_{21}{a}_{12}{a}_{33}+a_{31}{a}_{12}{a}_{23}+a_{21}{a}_{32}{a}_{13}-a_{31}{a}_{22}{a}_{13}$

OK，我们来看看n * n 的一般公式:

看一个例子:

通过big formula,我们很容易就得出这个矩阵的det为0，为奇异矩阵

Cofactor formula

我们回到这个例子
∣∣∣∣a11a21a31a12a22a32a13a23a33∣∣∣∣=∣∣∣∣a11000a22000a33∣∣∣∣+∣∣∣∣a110000a320a230∣∣∣∣+∣∣∣∣0a210a120000a33∣∣∣∣+∣∣∣∣00a31a12000a230∣∣∣∣+∣∣∣∣0a21000a32a1300∣∣∣∣+∣∣∣∣00a310a220a1300∣∣∣∣=$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=\left|\begin{array}{ccc}{a}_{11}& 0& 0\\ 0& a_{22}& 0\\ 0& 0& a_{33}\end{array}\right|+\left|\begin{array}{ccc}{a}_{11}& 0& 0\\ 0& 0& a_{23}\\ 0& a_{32}& 0\end{array}\right|+\left|\begin{array}{ccc}0& a_{12}& 0\\ a_{21}& 0& 0\\ 0& 0& a_{33}\end{array}\right|+\left|\begin{array}{ccc}0& a_{12}& 0\\ 0& 0& a_{23}\\ a_{31}& 0& 0\end{array}\right|+\left|\begin{array}{ccc}0& 0& a_{13}\\ a_{21}& 0& 0\\ 0& a_{32}& 0\end{array}\right|+\left|\begin{array}{ccc}0& 0& a_{13}\\ 0& a_{22}& 0\\ a_{31}& 0& 0\end{array}\right|=$
a11a22a33−a11a32a23−a21a12a33+a31a12a23+a21a32a13−a31a22a13$a_{11}{a}_{22}{a}_{33}-a_{11}{a}_{32}{a}_{23}-a_{21}{a}_{12}{a}_{33}+a_{31}{a}_{12}{a}_{23}+a_{21}{a}_{32}{a}_{13}-a_{31}{a}_{22}{a}_{13}$
这个矩阵的det用cofactors表示如下:
a11(a22a33−a23a32)+a12()+a13()$a_{11}(a_{22}{a}_{33}-a_{23}{a}_{32})+a_{12}()+a_{13}()$
我们需要把注意力集中在括号里的式子，来看看选择a11$a_{11}$的情况:

a11对应的cofactor即剩下的2(n - 1)行2(n - 1)列的det(还要添加上符号，根据选择元素的i和j之和为奇数还是偶数确定，若为偶数，则为正号，若为奇数，则为负号)

cofactor的定义如下:

来看一个2*2的例子:

Tridiagonal matrices

得出三对角线矩阵det的规律后，我们可以继续求得|A5$A_5$| = 0,|A6$A_6$| = 1,|A7$A_7$| = 1,它的det是以6为循环的奇妙的组合。