线性代数答案第三，四章 课后答案

?0??0?????得基础解系?2??1?取P2??12?．

?1??12?????当?3?1时，解方程(A?E)x?0，由

?1?A?E??0?0?0220??2?2??~?1??0?0?0100??1? 0??0?0???????得基础解系?3???1?取P3???12?，于是正交变换为

?1??12??????x1??1????x2???0?x??0?3??(2)

011二

22??y1????222?12??y2? 且有f?2y1?5y2?y3．

??12???y3?次

型

矩

阵

为

0?1??1A??0????111?100?111?1??0?1???1?21??A??E?10?111???100?11??1?1011???(??1)(??3)(??1)，故A的特征值为?1??1,?2?3,?3??4?1

?1????2???1??2?当?1??1时，可得单位特征向量P1?，

?1?????2?1???2?

?1????2??1??2?当?2?3时，可得单位特征向量P2?，

?1?????2?1????2????当?3??4?1时，可得单位特征向量P3??????1??0???1?2???0?2?．于是正交变换为

，P4??1??0??1?2????0?2???1??2?x1?????1?x2??2?x???1?3????x??2?4??1??2??1???????12121212120120???0??1???2????0?????1??2?y1??y2?2222且有f??y1?3y2?y3?y4． y3??y4??13．证明 A为实对称矩阵，则有一正交矩阵T，使得

TAT?1?2??????B成立． ??n??其中?1,?2,?,?n为A的特征值，不妨设?1最大，

T为正交矩阵，则T则f?xT?1?TT且T?1，故A?TT?1TTB?TB

TAx?xTTTBTx?yBy??1y1??2y2????nyn．

222其中y?Tx

当y?Tx?Tx?x?1时，

即

y1?y2???yn?1即y1?y2???yn?1

22222222f最大?(?1y1????nyn)最大??2?14． 解 (1) A??1?1?1?60y1?1??1．故得证．

1??0?, ?4???21?6010?4 ??38?0故f为负定．

a11??2?0，

?211?6?11?0，11?1???1(2) A??2???11?12?1302?130?3209?61???3?1，a11?1?0，?6??1??19??13?4?0,

0?6?0,A?24?0．故f为正定． 9?a11?15．证明 设U????a?n1a12?an2???T??a1n??????(a1,a2,?,an)，x???ann????x1??x1?， ???xn??f?xTAx?xTUTUx?(Ux)(Ux)

?(a11x1???a1nxn,a21x1???a2nxn,?,an1x1???annxn) ?a11x1???a1nxn??a21x1???a2nxn?????ax???axnnn?n11???22?(ax???ax)?(ax???ax) 1111nn2112nn????2???(an1x1???annxn)?0．

?a11x1???a1nxn?0?若“?0”成立，则?成立． ??ax???ax?0nnn?n11???即对任意x?????x1??x1?使?1x1??2x2????nxn?0成立． ???xn??则?1,?2,?,?n线性相关,U的秩小于n，则U不可逆，与题意产生矛盾．于是f?0成立．故f?xAx为正定二次型．

16．证明 A正定，则矩阵A满秩，且其特征值全为正． 不妨设?1,?,?n为其特征值，?i?0由定理8知，存在一正交矩阵P

Ti?1,?,n

??1??T使PAP????????????????2????? ??n???1?2??????????????n????1?1?2????? ??n??又因P为正交矩阵，则P可逆，P所以A?PQQPTTT?PT．

T?PQ?(PQ)．

T令(PQ)?U，U可逆,则A?U

U.