Editing Cross product (section)

=== Conversion to matrix multiplication ===
The vector cross product also can be expressed as the product of a [[skew-symmetric matrix]] and a vector:<ref name=Liu>{{cite journal |title=Hadamard, Khatri-Rao, Kronecker and other matrix products |journal=Int J Information and Systems Sciences |volume=4 |pages=160–177 |year=2008 |publisher=Institute for scientific computing and education |url=https://www.researchgate.net/publication/251677036 |author1=Shuangzhe Liu |author2=Gõtz Trenkler |issue=1 }}</ref>
<math display="block">\begin{align}
  \mathbf{a} \times \mathbf{b} = [\mathbf{a}]_{\times} \mathbf{b}
    &= \begin{bmatrix}\,0&\!-a_3&\,\,a_2\\ \,\,a_3&0&\!-a_1\\-a_2&\,\,a_1&\,0\end{bmatrix}\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix} \\
  \mathbf{a} \times \mathbf{b} = {[\mathbf{b}]_\times}^\mathrm{\!\!T} \mathbf{a}
    &= \begin{bmatrix}\,0&\,\,b_3&\!-b_2\\ -b_3&0&\,\,b_1\\\,\,b_2&\!-b_1&\,0\end{bmatrix}\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix},
\end{align}</math>
where superscript {{math|{{sup|T}}}} refers to the [[transpose]] operation, and ['''a''']<sub>×</sub> is defined by:
<math display="block">[\mathbf{a}]_{\times} \stackrel{\rm def}{=} \begin{bmatrix}\,\,0&\!-a_3&\,\,\,a_2\\\,\,\,a_3&0&\!-a_1\\\!-a_2&\,\,a_1&\,\,0\end{bmatrix}.</math>

The columns ['''a''']<sub>×,i</sub> of the skew-symmetric matrix for a vector '''a''' can be also obtained by calculating the cross product with [[unit vectors]]. That is,
<math display="block">[\mathbf{a}]_{\times, i} = \mathbf{a} \times \mathbf{\hat{e}_i}, \; i\in \{1,2,3\} </math>
or
<math display="block">[\mathbf{a}]_{\times} = \sum_{i=1}^3\left(\mathbf{a} \times \mathbf{\hat{e}_i}\right)\otimes\mathbf{\hat{e}_i},</math>
where <math>\otimes</math> is the [[outer product]] operator.

Also, if '''a''' is itself expressed as a cross product:
<math display="block">\mathbf{a} = \mathbf{c} \times \mathbf{d}</math>
then
<math display="block">[\mathbf{a}]_{\times} = \mathbf{d}\mathbf{c}^\mathrm{T} - \mathbf{c}\mathbf{d}^\mathrm{T} .</math>

{{math proof|title=Proof by substitution
|proof=Evaluation of the cross product gives
<math display="block"> \mathbf{a} = \mathbf{c} \times \mathbf{d} = \begin{pmatrix} 
c_2 d_3 - c_3 d_2 \\
c_3 d_1 - c_1 d_3 \\
c_1 d_2 - c_2 d_1 \end{pmatrix}
</math>
Hence, the left hand side equals
<math display="block"> [\mathbf{a}]_{\times} = \begin{bmatrix} 
        0         & c_2 d_1 - c_1 d_2 & c_3 d_1 - c_1 d_3 \\
c_1 d_2 - c_2 d_1 &         0         & c_3 d_2 - c_2 d_3 \\
c_1 d_3 - c_3 d_1 & c_2 d_3 - c_3 d_2 &         0 \end{bmatrix}
</math>
Now, for the right hand side,
<math display="block"> \mathbf{c} \mathbf{d}^{\mathrm T} = \begin{bmatrix}
c_1 d_1 & c_1 d_2 & c_1 d_3 \\
c_2 d_1 & c_2 d_2 & c_2 d_3 \\
c_3 d_1 & c_3 d_2 & c_3 d_3 \end{bmatrix}
</math>
And its transpose is
<math display="block"> \mathbf{d} \mathbf{c}^{\mathrm T} = \begin{bmatrix}
c_1 d_1 & c_2 d_1 & c_3 d_1 \\
c_1 d_2 & c_2 d_2 & c_3 d_2 \\
c_1 d_3 & c_2 d_3 & c_3 d_3 \end{bmatrix}
</math>
Evaluation of the right hand side gives
<math display="block"> \mathbf{d} \mathbf{c}^{\mathrm T} - 
\mathbf{c} \mathbf{d}^{\mathrm T} = \begin{bmatrix} 
0 & c_2 d_1 - c_1 d_2 & c_3 d_1 - c_1 d_3 \\
 c_1 d_2 - c_2 d_1 & 0 & c_3 d_2 - c_2 d_3 \\
c_1 d_3 - c_3 d_1 & c_2 d_3 - c_3 d_2 & 0 \end{bmatrix}
</math>
Comparison shows that the left hand side equals the right hand side.
}}

This result can be generalized to higher dimensions using [[geometric algebra]]. In particular in any dimension bivectors can be identified with skew-symmetric matrices, so the product between a skew-symmetric matrix and vector is equivalent to the grade-1 part of the product of a bivector and vector.<ref name="lounesto2001">{{cite book
  | author = Lounesto, Pertti
  | title = Clifford algebras and spinors
  | url = https://archive.org/details/cliffordalgebras00loun
  | url-access = limited
  | publisher = Cambridge: Cambridge University Press
  | year = 2001
  | isbn = 978-0-521-00551-7
  | pages = [https://archive.org/details/cliffordalgebras00loun/page/n200 193]
}}</ref>  In three dimensions bivectors are [[Hodge dual|dual]] to vectors so the product is equivalent to the cross product, with the bivector instead of its vector dual. In higher dimensions the product can still be calculated but bivectors have more degrees of freedom and are not equivalent to vectors.<ref name="lounesto2001"/>

This notation is also often much easier to work with, for example, in [[epipolar geometry]].

From the general properties of the cross product follows immediately that
<math display="block">[\mathbf{a}]_{\times} \, \mathbf{a} = \mathbf{0}</math> &nbsp; and &nbsp; <math display="block">\mathbf{a}^\mathrm T \, [\mathbf{a}]_{\times} = \mathbf{0}</math>
and from fact that ['''a''']<sub>×</sub> is skew-symmetric it follows that
<math display="block">\mathbf{b}^\mathrm T \, [\mathbf{a}]_{\times} \, \mathbf{b} = 0. </math>

The above-mentioned triple product expansion (bac–cab rule) can be easily proven using this notation.

As mentioned above, the Lie algebra '''R'''<sup>3</sup> with cross product is isomorphic to the Lie algebra '''so(3)''', whose elements can be identified with the 3×3 skew-symmetric matrices. The map '''a''' → ['''a''']<sub>×</sub> provides an isomorphism between '''R'''<sup>3</sup> and '''so(3)'''. Under this map, the cross product of 3-vectors corresponds to the [[commutator]] of 3x3 skew-symmetric matrices.

:{| class="toccolours collapsible collapsed" width="70%" style="text-align:left"
!Matrix conversion for cross product with canonical base vectors
|-
|Denoting with <math>\mathbf{e}_i \in \mathbf{R}^{3 \times 1}</math> the <math>i</math>-th canonical base vector, the cross product of a generic vector <math>\mathbf{v} \in \mathbf{R}^{3 \times 1}</math> with <math>\mathbf{e}_i</math> is given by: <math>\mathbf{v} \times \mathbf{e}_i = \mathbf{C}_i \mathbf{v}</math>, where

<math display="block"> \mathbf{C}_1 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix}, \quad
\mathbf{C}_2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}, \quad
\mathbf{C}_3 = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
</math>

These matrices share the following properties:
* <math>\mathbf{C}_i^\textrm{T} = -\mathbf{C}_i</math> (skew-symmetric);
* Both trace and determinant are zero;
* <math>\text{rank}(\mathbf{C}_i) = 2</math>;
* <math>\mathbf{C}_i \mathbf{C}_i^\textrm{T} = \mathbf{P}^{ ^\perp}_{\mathbf{e}_i}</math> (see below);

The [[projection (linear algebra)#Orthogonal projection|orthogonal projection matrix]] of a vector <math>\mathbf{v} \neq \mathbf{0}</math> is given by <math>\mathbf{P}_{\mathbf{v}} = \mathbf{v}\left(\mathbf{v}^\textrm{T} \mathbf{v}\right)^{-1} \mathbf{v}^T</math>. The projection matrix onto the [[orthogonal complement]] is given by <math>\mathbf{P}^{ ^\perp}_{\mathbf{v}} = \mathbf{I} - \mathbf{P}_{\mathbf{v}}</math>, where <math>\mathbf{I}</math> is the identity matrix. For the special case of <math>\mathbf{v} = \mathbf{e}_i</math>, it can be verified that

<math display="block"> \mathbf{P}^{^\perp}_{\mathbf{e}_1} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad
\mathbf{P}^{ ^\perp}_{\mathbf{e}_2} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad
\mathbf{P}^{ ^\perp}_{\mathbf{e}_3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}
</math>

For other properties of orthogonal projection matrices, see [[projection (linear algebra)]].
|}