向量外积问题

lihpb

引理 2 设 $\{u_1, u_2, \cdots, u_n\}$ 是 $E^n$ 中线性无关的向量组，$\Phi\in \bigwedge^n(E^n)$. 则\[
\Phi\left(u_2 \wedge u_3 \cdots \wedge u_n, u_1 \wedge u_3 \cdots \wedge u_n, \cdots, u_1 \wedge u_2 \cdots \wedge u_{n-1}\right)=(-1)^{\lfloor n/2\rfloor} \Phi(u_1, u_2, \cdots, u_n)^{n-1}
\]

hbghlyj

设$D=\det(u_1,\dots,u_n),D\ne0$.
$v_1=u_2 \wedge u_3 \cdots \wedge u_n$,
$v_2=u_1 \wedge u_3 \cdots \wedge u_n$,
$\cdots\cdots$
$v_n=u_1 \wedge u_2 \cdots \wedge u_{n-1}$, 则$v_i^Tu_j=\begin{cases}0&i\ne j\\(-1)^{i-1}D&i=j\end{cases}$

把行向量$v_i^T$排成一个n×n矩阵$\pmatrix{v_1^T\\\vdots\\v_n^T}$，把列向量$u_i$排成一个n×n矩阵$\pmatrix{u_1&\cdots&u_n}$，
因为当$i\ne j$时$v_i^Tu_j=0$，所以这两个矩阵之积为对角矩阵$$\pmatrix{v_1^T\\\vdots\\v_n^T}\pmatrix{u_1&\cdots&u_n}=\operatorname{diag}(D,-D,D,-D,\dots,(-1)^{n-1}D)$$
（右边的$D,-D,D,-D,\dots$中，$-D$共出现了$\lfloor n/2\rfloor$次）
这三个矩阵都是n×n矩阵，两边取$\det$得
$$\det(v_1,\dots,v_n)\det(u_1,\dots,u_n)=(-1)^{\lfloor n/2\rfloor}D^n$$
代入$D=\det(u_1,\dots,u_n)$，
$$\det(v_1,\dots,v_n)D=(-1)^{\lfloor n/2\rfloor}D^n$$
即
$$\det(v_1,\dots,v_n)=(-1)^{\lfloor n/2\rfloor}\det(u_1,\dots,u_n)^{n-1}$$

hbghlyj

$n=2$，$(-1)^{\lfloor n/2\rfloor}=-1$，验证：

1. Simplify[Det[{{x2,y2},{x1,y1}}]==-Det[{{x1,y1},{x2,y2}}]]^1

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$n=3$，$(-1)^{\lfloor n/2\rfloor}=-1$，验证：

1. Simplify[Det[{Cross[{x2,y2,z2},{x3,y3,z3}],Cross[{x1,y1,z1},{x3,y3,z3}],Cross[{x1,y1,z1},{x2,y2,z2}]}]==-Det[{{x1,y1,z1},{x2,y2,z2},{x3,y3,z3}}]^2]

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$n=4$，$(-1)^{\lfloor n/2\rfloor}=1$，验证：

1. Simplify[Det[{Cross[{x2,y2,z2,w2},{x3,y3,z3,w3},{x4,y4,z4,w4}],Cross[{x1,y1,z1,w1},{x3,y3,z3,w3},{x4,y4,z4,w4}],Cross[{x1,y1,z1,w1},{x2,y2,z2,w2},{x4,y4,z4,w4}],Cross[{x1,y1,z1,w1},{x2,y2,z2,w2},{x3,y3,z3,w3}]}]==Det[{{x1,y1,z1,w1},{x2,y2,z2,w2},{x3,y3,z3,w3},{x4,y4,z4,w4}}]^3]

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$n=5$，$(-1)^{\lfloor n/2\rfloor}=1$，验证：

1. Simplify[Det[{Cross[{x2,y2,z2,w2,u2},{x3,y3,z3,w3,u3},{x4,y4,z4,w4,u4},{x5,y5,z5,w5,u5}],Cross[{x1,y1,z1,w1,u1},{x3,y3,z3,w3,u3},{x4,y4,z4,w4,u4},{x5,y5,z5,w5,u5}],Cross[{x1,y1,z1,w1,u1},{x2,y2,z2,w2,u2},{x4,y4,z4,w4,u4},{x5,y5,z5,w5,u5}],Cross[{x1,y1,z1,w1,u1},{x2,y2,z2,w2,u2},{x3,y3,z3,w3,u3},{x5,y5,z5,w5,u5}],Cross[{x1,y1,z1,w1,u1},{x2,y2,z2,w2,u2},{x3,y3,z3,w3,u3},{x4,y4,z4,w4,u4}]}]==Det[{{x1,y1,z1,w1,u1},{x2,y2,z2,w2,u2},{x3,y3,z3,w3,u3},{x4,y4,z4,w4,u4},{x5,y5,z5,w5,u5}}]^4]

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lihpb

hbghlyj 发表于 2024-11-15 03:55
设$D=\det(u_1,\dots,u_n),D\ne0$.
$v_1=u_2 \wedge u_3 \cdots \wedge u_n$,
$v_2=u_1 \wedge u_3 \cdots  ...

kuing.cjhb.site/forum.php?mod=viewthread&tid=12577

帮我看看这个，也是行列式的问题