Let $u=(u_1, u_2, u_3), v=(v_1, v_2, v_3)\in\mathbb{R}^3$. Define the cross (vector) product of $u$ and $v$ as the unique vector $u\wedge v\in\mathbb{R}^3$ characterized by $$ (u\wedge v)\cdot w = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \\ \end{vmatrix} $$ for all $w=(w_1,w_2,w_3)\in\mathbb{R}^3$.

I just want to show that this indeed uniquely defines a vector $u\wedge v$, but this is quite a strange definition to me and I don't know how to start. Usually for proving uniqueness we suppose there is another element satisfying the definition and we work out calculations to show that they're the same element. But this defines an element involving another two fixed elements so I'm a bit confused in this situation.

  • 2
    $\begingroup$ The determinant is a linear functional of $w$. Every linear functional is of the form $\square\cdot w$ for some vector $\square$. This is because you can write linear transformations as matrix transformations wrt a basis, and doing this in the case the codomain is 1-dim gives you basically the dot product. $\endgroup$
    – anon
    Aug 26, 2021 at 4:24

2 Answers 2


For any finite-dimensional real vector space $V$, if $g$ is an inner product on $V$ (i.e $g:V\times V\to\Bbb{R}$, is bilinear, symmetric and positive definite), then we get an induced linear mapping $g^{\flat}:V\to V^*$ (read as 'gee-flat'), and the definition is \begin{align} g^{\flat}(v):=g(v,\cdot). \end{align} The meaning is that $g$ is a function of two variables, taking values in $\Bbb{R}$, so what we can do is fill up one of the entries to get $g(v,\cdot)$, and this can eat another vector $w$ to product a real number, so $g^{\flat}(v)=g(v,\cdot)\in V^*$.

Now, the mapping $g^{\flat}$ is injective, because if $v\in \ker(g^{\flat})$, then it means $g^{\flat}(v):=g(v,\cdot)=0$, which explicitly means for all $w\in V$, $g(v,w)=0$. So, in particular, $g(v,v)=0$, hence by positive-definiteness, $v=0$. Therefore, $\ker(g^{\flat})=\{0\}$, hence we have an injection. Recall that on a finite-dimensional space, $V$ and $V^*$ have the same dimension, so an injection is automatically an isomorphism. The inverse mapping $(g^{\flat})^{-1}:V^*\to V$ is typically denoted $g^{\sharp}:V^*\to V$.

We're going to apply this logic in the specific case $V=\Bbb{R}^3$ and $g(x,y)=\sum_{i=1}^3x_iy_i$ being the standard inner product. So, we have an isomorphism $g^{\flat}:\Bbb{R}^3\to(\Bbb{R}^3)^*$. Now, the definition of the cross product is that given $u,v\in\Bbb{R}^3$, we recall that the determinant is a multilinear function of the rows/columns, so the function $\det(u,v,\cdot):\Bbb{R}^3\to\Bbb{R}$ is linear. This means $\det(u,v,\cdot)\in (\Bbb{R}^3)^*$. Thus, the definition of the cross product is \begin{align} u\times v&=g^{\sharp}\bigg(\det(u,v,\cdot)\bigg)\in \Bbb{R}^3 \end{align}

So, it is the fact that $\Bbb{R}^3\cong (\Bbb{R}^3)^*$ via the inner product which makes it possible to define the cross product in this way.

Anyway, if all you want to know is how to calculate the cross product, then recall that in general, given any vector $\xi\in \Bbb{R}^3$, we can take any orthonormal basis, for example the standard basis $\{e_1,e_2, e_3\}$, and write $\xi=g(\xi,e_1)e_1+g(\xi,e_2)e_2+g(\xi,e_3)e_3$ (i.e dot the vector into the basis to get the projections onto the axis, and then multiply by the unit vector).

So, for the cross product, we can also do the same: \begin{align} u\times v&=\det(u,v,e_1)\cdot e_1+\det(u,v,e_2)\cdot e_2 + \det(u,v,e_3)\cdot e_3 \end{align} You may have seen this summarized mnemonically as \begin{align} u\times v&= \det \begin{pmatrix} u_1&u_2&u_3\\ v_1&v_2&v_3\\ \hat{x}&\hat{y}&\hat{z} \end{pmatrix} \end{align}


One way to understand this definition is as a compact way of specifying the components of the cross product. For instance, the formula should apply for the case $w=e_1$. Then the definition requires

$$(u\wedge v)\cdot e_1 = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ 1 & 0 & 0 \\ \end{vmatrix}=\begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}. $$ But $(u\wedge v)\cdot e_1$ is the $1$st component of $u\wedge v$, so we have determined the first component of the cross product. Taking $w=e_2,e_3$ yields the other two. (And this is certainly unique: Once $u,v$ have been specified, then so too have the components of $u\wedge v$.)


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