《线性代数与解析几何》复习要点

1、 1221()12( 1)nnjjjjjnjNa aa2211()12( 1)nniiNiii nia aa 运算前提条件定义性质加法A + BA与B是同类型的对应元素相加A + B = B + A; (A + B) + C = A + (B + C);A + O = A; A + (A) = O数乘kAk是一个数用k乘A的每一个元素k(lA) = (kl)A; (k + l)A = kA + lA;k(A + B) = kA + kB; (1)A = A乘法ABA的列数 = B的行数(aij)ml(bij)ln = (cij)mn cij = (AB)C = A(BC); A(B+C) =A

2、B+AC;(A+B)C =AC+BC; (kA)B = k(AB)幂 AmA是方阵, m是正整数A1 = A, Ak+1 = AkAAkAl = Ak+l; (Ak)l = Akl转置AT无(aij)ml T = (aji)lm(AT)T = A; (A + B)T = AT + BT;(kA)T = kAT; (AB)T = BTAT多项式f(A)A是一个方阵,f(x) = asxs + a1x + a0f(A) = asAs +a1A+a0IA = ( )f(A) = f( ) ,A = ( ), f(A) = O f( ) = 0行列式|A|A是一个方阵, |A1| = |A|1逆矩阵A

3、1A是一个方阵且|A|0若AB = BA = I则B = A1唯一性, (A1)1 = A, (A1)m = (Am)1,(AT)1 = (A1)T, (kA)1 = k1A1, (AB)1 = B1A1, 满秩, 特征值01nikkjka b: : : : 3 1 01 3 00 0 4A,2/ 1 0 2/ 12/ 1 0 2/ 1 0 1 0 Q.4 0 00 4 00 0 2T1 AQQAQQ;1 1 232111311 1 1|,2222332 .6/ 1 3/ 1 2/ 16/ 1 3/ 1 2/ 1 6/2 3/ 1 0 ),(321 qqqQ msssmmaaaaaaaaaA2

4、12221212111, : mnmmnnccccccccc 212222111211 msmmssaaaaaaaaa 212222111211 snssnnbbbbbbbbb 212222111211 mnnnmmcccccccccC212221212111, : 212222111211snssnnbbbbbbbbb 212222111211mnmmnnccccccccc mnmmnnccccccccc 212222111211 msmmssaaaaaaaaa 212222111211 snssnnbbbbbbbbb 212222111211 mnmmnnccccccccc 2122221

5、11211 msmmssaaaaaaaaa 212222111211 snssnnbbbbbbbbb 212222111211 mnmmnnaaaaaaaaaA 212222111211 mnmmnnbbbbbbbbbB 212222111211 mnmmnnaaaaaaaaaA 212222111211 mnmmnnbbbbbbbbbB 2122221112110 10 1A0 00 1B0 01 1C0 00 1B 0 37702 3520 432143214321xxxxxxxxxxxx 1 3 7 7 2 3 5 21 1 1 1 0 0 0 07/4 7/5 1 07/3 7/2 0

6、 1,1 0 7/47/3 ,01 7/57/22 1 ).,( ,1 0 7/47/301 7/57/2112 14321Rccccxxxx ,11 ,1143 xx).,( ,1 1 7/ 1 7/ 11 1 7/97/511214321Rccccxxxx ,7/ 1 7/ 1 ,7/97/521 xx,1 1 7/ 1 7/ 1 ,1 1 7/97/521 1 3 7 7 2 3 5 21 1 1 1 0 0 0 0 0 1 3 4 1 0 2 5,23 1 0 ,5 40 1 21 ).,( ,23 1 0 5 40 1 21214321Rccccxxxx 21421325 34 xxxxxx 22222211100pprrnfyyyyyy 且规范形是唯一的且规范形是唯一