矩阵可以理解为线性变换在一组基下的表示,故矩阵乘法就是两个线性变换结合后的表示。Hoffman把矩阵更一般地作为函数,故其给矩阵乘法的引入是:矩阵乘法是对矩阵的行进行线性组合时产生的东西。在定义和后面的例子(Example 10)中,都体现了矩阵乘法的这一个思想:A某一行的每一个数与B的行向量进行线性组合,得到AB的这一行的结果。特别提到了矩阵乘法是not commutative的。 列矩阵(column matrix)引致一个有用的notation(可以从定义证得):如果有矩阵Bn×p,其中列向量为B1,…,Bp,则B=[B1,…,Bp]并且AB=[AB1,…,ABp]。 Theorem 8 给出了矩阵乘法满足结合律的证明。其后说明了矩阵的幂是什么,以及A(BC)=(AB)C说明linear combinations of linear combinations of the rows of C are again linear combinations of the rows of C,我认为这个诠释很有意思,Hoffman时刻在提示运算背后的思想。 最后就是要引入elementary matrix,与1.3节介绍的elementary row operations密切相关,elementary matrix的定义是从I进行一次elementary row operations得到的矩阵,故其必是和I同阶的方阵。Theorem 9 说明,如果用e表示一个elementary row operation,则e(A)=e(I)A,即行变换等于左乘对应初等矩阵。其推论显示:A,B两矩阵row equivalent的充要条件是B=PA,其中P是一堆初等矩阵乘积。
A=⎣⎡123−100111⎦⎤,B=⎣⎡214−234⎦⎤ Verify directly that A(AB)=A2B Solution: A(AB)=⎣⎡123−100111⎦⎤⎝⎛⎣⎡123−100111⎦⎤⎣⎡214−234⎦⎤⎠⎞=⎣⎡123−100111⎦⎤⎣⎡5810−10−2⎦⎤=⎣⎡72025−3−4−5⎦⎤A2B=⎝⎛⎣⎡123−100111⎦⎤⎣⎡123−100111⎦⎤⎠⎞⎣⎡214−234⎦⎤=⎣⎡256−1−2−3134⎦⎤⎣⎡214−234⎦⎤=⎣⎡72025−3−4−5⎦⎤ thus A(AB)=A2B.
3.Find two different 2×2 matrices A such that A2=0 but A=0.
Solution:A1=[0010],A2=[0100]
4.For the matrix A of Exercise 2, find elementary matrices E1,E2,…,Ek such that
Ek…E2E1A=I Solution: ⎣⎡123−100111⎦⎤add -2×(1) to (2), add -3×(1) to (3)(2)×(1/2)add −3×(2) to (3)(3)×−2add 1/2×(3) to (2), add −1×(3) to (1)add (2) to (1)⎣⎡100−1231−1−2⎦⎤⎣⎡100−1131−1/2−2⎦⎤⎣⎡100−1101−1/2−1/2⎦⎤⎣⎡100−1101−1/21⎦⎤⎣⎡100−110001⎦⎤⎣⎡100010001⎦⎤ so the corresponding elementary matrices are E1=⎣⎡1−20010001⎦⎤,E2=⎣⎡10−3010001⎦⎤,E3=⎣⎡10001/20001⎦⎤,E4=⎣⎡10001−3001⎦⎤E5=⎣⎡10001000−2⎦⎤,E6=⎣⎡10001001/21⎦⎤,E7=⎣⎡100010−101⎦⎤,E8=⎣⎡100110001⎦⎤
5. Let
A=⎣⎡121−120⎦⎤,B=[3−414]
Is there a matrix C such that CA=B?
Solution: One possible choice of C is C=[12130−12]
6. Let A be an m×n matrix and B and n×k matrix. Show that the columns of C=AB are linear combinations of the columns of A. If α1,…,αn are the columns of A and γ1,…,γk are the columns of C, then
γj=r=1∑nBrjαr Solution: We write A=[α1,…,αn], and C=[γ1,…,γk], then as C=AB, for 1≤i≤m, we have Cij=r=1∑nAirBrj=r=1∑nBrjAir,1≤i≤m thus γj=⎣⎢⎡C1j⋮Cmj⎦⎥⎤=⎣⎢⎡∑r=1nBrjA1r⋮∑r=1nBrjAmr⎦⎥⎤=B1j⎣⎢⎡A11⋮Am1⎦⎥⎤+⋯+Bnj⎣⎢⎡A1n⋮Amn⎦⎥⎤=B1jα1+⋯+Bnjαn
7.Let A and B be 2×2 matrices such that AB=I. Prove that BA=I.
Solution: We have ABA=IA=A, thus A(BA−I)=0, assume BA=I, then the system AX=0 has non-trivial solutions, thus A is row-equivalent to a row-reduced echelon matrix which is not I, so there’s P which is product of elementary matrices such that PA=[1000] this means the second row of P=PAB is 0, a contradiction.
8. Let
C=[C11C21C12C22]
be a 2×2 matrix. We inquire when it is possible to find 2×2 matrices A and B such that C=AB−BA. Prove that such matrices can be found if and only if C11+C22=0.
Solution: Let A=[acbd],B=[egfh], then AB=[ae+bgce+dgaf+bhcf+dh],BA=[ea+fcag+hceb+dfgb+dh], thus AB−BA=[bg−cfce+dg−ag−hcaf+bh−eb−dfcf−bg] if C=AB−BA, then C11+C22=bg−cf+cf−bg=0. Conversely, if C11+C22=0, then C11=−C22, thus solve the equation bg−cfaf+bh−eb−df=(a−d)f+b(h−e)ce+dg−ag−hc=(d−a)g+c(e−h)=C11=C12=C21 It’s a system of 3 equations with 8 unknowns, fix a=1,d=0,h=0,e=1, the system becomes bg−cff−bg−c=C11=C12=C21 thus b(C21+c)−c(C12+b)=C11, or bC21=C11+cC12, if C21=0, let c=1; if C21=0, let c=−C11/C12 and b=1, all other unkowns can be solved.