代入4得a21a221
22
1显然若需矩阵0
01与矩阵10
0合同则数域F应是复数域101与矩阵100不是合同的1
1即是说1在实数域R上,矩阵012在复数域C上,矩阵01或令P00也可i
01与矩阵10
0是合同的1
习题821求可逆矩阵P使PAP为对角形矩阵:11A0302131;102A23
222T
201
313
解:显然有fx1x2x3x12x2x36x1x32x2x31x13x32x2
2
x32
2
212
x3
2
y1x13x3x1y13y3xy可作线性替换:y2x223即x2y223yxxy33331即P0002031T1stPAP020102000212
f2显然有fx1x2x3
x1y1y33x4x1x26x1x32x2x3,令x2y1y2x2y33
21222
fx1x2x34y1y212y1y312y2y34y1y34y2y312y34y1y34y22y3
22
y1z1z3z1y1y3再令z2y22y3y2z22z3yzz3y3331则有P1011001002001001010011004T0stPAP000040000
2用可逆线性替换化下列二次型为标准型,并写出所做的线性替换(分别用初等变换法和配方法加以完成):1fx1x2x34x1x3x2
2
2fx1x2x3x12x1x22x2x3x3
22
3fx1x2x32x14x1x24x1x35x28x2x35x3
222
4fx1x2x3x4x1x1x2x3x4
2
x1y1y3222解:令x2y2,标准型为fx1x2x34y1y24y31xyy1331P0101014T0PAP001010004
y1x21x23即作了线性变换y2x2x3x1y322x1y1y2yr