黄克智版张量分析习题解析

1.3 求证矢量的非退化性。即：若矢量 v 与它所属的矢量空间 中的任意矢量 u 都正交，即：u·v=0，则矢量 v=0。
证明：因为 u 为任意，所以可取 u1，u2，u3，使得
u1x u1y u1z det U u2x u2 y u2z 0
uy
uz
vy wz vz wy vz wx vxwz vxwy vy wx
uy vxwy vywx uz vzwx vxwz i uz vywz vzwy ux vxwy vywx j ux vzwx vxwz uy vywz vzwy k
Ax Cy Dz Cz Dy Ay Cz Dx CxDz Az CxDy Cy Dx
Bxi By j Bzk
Bx Cy Dz Cz Dy By Cz Dx CxDz Bz CxDy Cy Dx
Axi Ay j Azk
Bx Ay Cz Dx CxDz Az CxDy Cy Dx Ax By Cz Dx CxDz Bz CxDy Cy Dx i By Ax Cy Dz Cz Dy Az CxDy Cy Dx Ay Bx Cy Dz Cz Dy Bz CxDy Cy Dx j Bz Ax Cy Dz Cz Dy Ay Cz Dx CxDz Az Bx Cy Dz Cz Dy By Cz Dx CxDz k
i jk
C D Cx Cy Cz Dx Dy Dz
Cy Dz Cz Dy i Cz Dx CxDz j CxDy Cy Dx k
i jk B C Bx By Bz
Cx Cy Cz
ByCz BzCy i BzCx BxCz j BxCy ByCx k
u • wv uxwx uywy uzwz vxi vy j vzk
u • vw uxvx uyvy uzvz wxi wy j wzk
u • wv u • vw
uxwx uywy uzwz vxi vy j vzk uxvx uyvy uzvz wxi wy j wzk
uyvz uzvy i uzvx uxvz j uxvy uyvx k
i
j
k
u v w uyvz uzvy uzvx uxvz uxvy uyvx
wx
wy
wz
wz uzvx uxvz wy uxvy uyvx i wx uxvy uyvx wz uyvz uzvy j
wy uyvz uzvy wx uzvx uxvz k
uv w uv w
1.2 求证： (A×B) ×(C×D)=B(A·C×D) －A(B·C×D) =C(A·B×D) －D(A·B×C)
证明：
i jk
A B Ax Ay Az Bx By Bz
Ay Bz Az By i Az Bx AxBz j AxBy Ay Bx k
Cxi Cy j Czk
Ax ByCz BzCy Ay BzCx BxCz Az BxCy ByCx
Dxi Dy j Dzk
Cx Ax By Dz Bz Dy Bx Az Dy Ay Dz Dx Ax ByCz BzCy Bx AzCy AyCz i Cy Ay Bz Dx BxDz By AxDz Az Dx Dy Ay BzCx BxCz By AxCz AzCx j Cz Az BxDy By Dx Bz Ay Dx AxDy Dz Az BxCy ByCx Bz AyCx AxCy k
1.1 求证：u×(v×w)=(u·w)v－(u·v) w 并问： u×(v×w) 与 (u×v)×w 是否相等？u，v，w 为矢量。
证明：
i jk
v w vx vy vz wx wy wz
vywz vzwy i vzwx vxwz j vxwy vywx k
i
j
k源自文库
u v w ux
A • B D Ax By Dz Bz Dy Ay Bz Dx Bx Dz Az Bx Dy Cy Dx A • B C Ax ByCz BzCy Ay BzCx BxCz Az BxCy ByCx
CA• B D DA• BC
Ax By Dz Bz Dy Ay Bz Dx BxDz Az Bx Dy By Dx
uy vxwy vywx uz vzwx vxwz i uz vywz vzwy ux vxwy vywx j ux vzwx vxwz uy vywz vzwy k
uv w u • wv u • vw
i jk
uv ux uy uz vx vy vz
i jk B D Bx By Bz
Dx Dy Dz
By Dz Bz Dy i Bz Dx BxDz j Bx Dy By Dx k
i
A B C D Ay Bz Az By
Cy Dz Cz Dy
j Az Bx Ax Bz Cz Dx Cx Dz
k Ax By Ay Bx Cx Dy Cy Dx
A • C D Ax Cy Dz Cz Dy Ay Cz Dx Cx Dz Az Cx Dy Cy Dx B • C D Bx Cy Dz Cz Dy By Cz Dx Cx Dz Bz Cx Dy Cy Dx
BA • C D AB • C D