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How would I use the properties of determinants to show that for any two vectors $\bf u$ and $\bf v$ in $ℝ^3$

$$\bf u \times v = (-v) \times u$$

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First you would need a definition of cross product. – GEdgar Sep 22 '12 at 2:13

You can write the cross product as

$${\bf u\times v}=\left(\left|\begin{matrix} u_2&u_3\\ v_2&v_3\end{matrix}\right|,\left|\begin{matrix} u_3&u_1\\ v_3&v_1\end{matrix}\right|,\left|\begin{matrix} u_1&u_2\\ v_1&v_2\end{matrix}\right|\right)$$ Now use $$\left|\begin{matrix} c&d\\ a&b\end{matrix}\right|=-\left|\begin{matrix} a&b\\ c&d\end{matrix}\right|$$


$$\left|\begin{matrix} -a&-b\\ c&d\end{matrix}\right|=-\left|\begin{matrix} a&b\\ c&d\end{matrix}\right|$$

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If you use the determinant definition of cross product $${\bf u\times v}=\left|\begin{matrix} \bf {i} & \bf {j} & \bf {k} \\ u_x&u_y&u_z\\ v_x&v_y&v_z \end{matrix}\right|$$ it is immediate because interchanging two rows of the determinant changes the sign.

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Peter and Ross have already answered your question. They used determinants. I won't.

Let $\mathrm{u} := (u_1, u_2, u_3)$ and $\mathrm{v} := (v_1, v_2, v_3)$. The cross product of $\mathrm{u}$ and $\mathrm{v}$ is defined as follows

$$\mathrm{u} \times \mathrm{v} := \left[\begin{array}{c} u_2 v_3 - u_3 v_2\\ u_3 v_1 - u_1 v_3\\ u_1 v_2 - u_2 v_1\end{array}\right]$$

Let us now introduce the following skew-symmetric matrices

$$Q_1 := \left[\begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\\\end{array}\right]$$

$$Q_2 := \left[\begin{array}{ccc} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0\\\end{array}\right]$$

$$Q_3 := \left[\begin{array}{ccc} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\\\end{array}\right]$$

Note that the $i$-th entry of $\mathrm{u} \times \mathrm{v}$ is the sum of all $9$ entries of the entrywise product of $Q_i$ and

$$\mathrm{u} \mathrm{v}^T = \left[\begin{array}{ccc} u_1 v_1 & u_1 v_2 & u_1 v_3\\ u_2 v_1 & u_2 v_2 & u_2 v_3\\ u_3 v_1 & u_3 v_2 & u_3 v_3\\\end{array}\right]$$

Using the trace instead of entrywise product, we also have that the $i$-th entry of $\mathrm{u} \times \mathrm{v}$ is

$$\mathrm{tr}(Q_i^T\mathrm{u} \mathrm{v}^T) = \mathrm{tr}(\mathrm{v}^T Q_i^T\mathrm{u}) = \mathrm{tr}(\mathrm{u}^T Q_i\mathrm{v})$$

where I used the property $\mathrm{tr}(A B C) = \mathrm{tr}(C A B)$ and also property $\mathrm{tr}(A^T) = \mathrm{tr}(A)$. Therefore

$$\mathrm{u} \times \mathrm{v} = \left[\begin{array}{c} \mathrm{tr}(\mathrm{u}^T Q_1\mathrm{v})\\ \mathrm{tr}(\mathrm{u}^T Q_2\mathrm{v})\\ \mathrm{tr}(\mathrm{u}^T Q_3\mathrm{v})\end{array}\right]$$

Finally, we have that

$$\mathrm{(-v)} \times \mathrm{u} = -\left[\begin{array}{c} \mathrm{tr}(\mathrm{v}^T Q_1\mathrm{u})\\ \mathrm{tr}(\mathrm{v}^T Q_2\mathrm{u})\\ \mathrm{tr}(\mathrm{v}^T Q_3\mathrm{u})\end{array}\right] = - \left[\begin{array}{c} \mathrm{tr}(\mathrm{u}^T Q_1^T\mathrm{v})\\ \mathrm{tr}(\mathrm{u}^T Q_2^T\mathrm{v})\\ \mathrm{tr}(\mathrm{u}^T Q_3^T\mathrm{v})\end{array}\right] = \left[\begin{array}{c} \mathrm{tr}(\mathrm{u}^T Q_1\mathrm{v})\\ \mathrm{tr}(\mathrm{u}^T Q_2\mathrm{v})\\ \mathrm{tr}(\mathrm{u}^T Q_3\mathrm{v})\end{array}\right] = \mathrm{u} \times \mathrm{v}$$

because the $Q_i$ matrices are skew-symmetric (i.e., $Q_i^T = -Q_i$).

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Note that the question was explicitly about using the properties of the determinant for proving it (otherwise the most obvious proof would be to just calculate both and compare). – celtschk Sep 22 '12 at 9:31
@celtschk: Yeah, but I wanted to write the cross product in terms of the trace. – Rod Carvalho Sep 22 '12 at 9:39
Which raises the question: Why? – celtschk Sep 22 '12 at 9:41
@celtschk: Because 1) I want to. 2) it makes it easier to implement in MATLAB for example. 3) I hate determinants. – Rod Carvalho Sep 22 '12 at 9:43
Easier to implement than the trivial component-wise formula (which I guess would even be more efficient because there are no unnecessary multiplications with $1$)? – celtschk Sep 22 '12 at 9:53

Several nice answers have been given. I would like to explain it from another point of view.

The fact $u\times v= (-v)\times u$ is a very basic property for cross product. There are of course several ways to prove it. But I think the best way is to explain and prove it using the definition of cross product. Otherwise, if we need to calculate a lot to prove it, I don't think the calculation can help you to understand it better, and the fundamental thing may be concealed by the calculation.

Hint: In order to see $u\times v= (-v)\times u$, we need first show $u\times v= -v\times u$ and $-v\times u= (-v)\times u$. The best way is to use the definition (see wiki) $u\times v=\|u\|\|v\|\sin\theta\mathbf{n}$.

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