- #1
Mandelbroth
- 611
- 24
Consider ##\vec{a}=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}## and ##\vec{b}=\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}##.
Is any part of the following NOT true?
$$\vec{a}\wedge\vec{b}=\frac{1}{2}(\vec{a}\otimes\vec{b}-\vec{b}\otimes\vec{a}) = \frac{1}{2}\begin{bmatrix} 0 & a_1b_2-a_2b_1 & a_1b_3-a_3b_1 \\ a_2b_1-a_1b_2 & 0 & a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 & a_3b_2-a_2b_3 & 0 \end{bmatrix}$$
(Edited for accuracy to make me feel better )
I wasn't sure if that is actually an equality or not. If it is an equality, it makes defining the cross product rather easy. It would just be the Hodge dual of the wedge product of a with b, because the inner product of the cross product of a and b with any basis of the matrix would be 0...
Is any part of the following NOT true?
$$\vec{a}\wedge\vec{b}=\frac{1}{2}(\vec{a}\otimes\vec{b}-\vec{b}\otimes\vec{a}) = \frac{1}{2}\begin{bmatrix} 0 & a_1b_2-a_2b_1 & a_1b_3-a_3b_1 \\ a_2b_1-a_1b_2 & 0 & a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 & a_3b_2-a_2b_3 & 0 \end{bmatrix}$$
(Edited for accuracy to make me feel better )
I wasn't sure if that is actually an equality or not. If it is an equality, it makes defining the cross product rather easy. It would just be the Hodge dual of the wedge product of a with b, because the inner product of the cross product of a and b with any basis of the matrix would be 0...
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