During my course in linear algebra, the instructor stated that A cross B is the same as the "skew symmetric matrix" of A times B. So, first of all, can someone clarify or provide sources about skew symmetric matrices? Secondly, I can't really comprehend the idea of how a single column vector crossed with another could be represented by a matrix.

Anyhow, thanks in advance!


The skew-symmetric tensor product of two vectors with components $A_i$ and $B_i$ is the tensor represented by the matrix with components $S_{ij}=A_iB_j - A_jB_i$. It is skew-symmetric (antisymmetric) because $S_{ij}=-S_{ji}$.

The advantage of this representation is that unlike the vector cross product, which is specific to three dimensions, the skew-symmetric product generalizes the concept to arbitrary dimensions.

Explicitly (in three dimensions),


  • $\begingroup$ This is what I don't get: Sij=AiBj−AjBi How can A and B vector have 2 components as in i and j? $\endgroup$ – Abu Bakr Apr 23 '17 at 18:10
  • $\begingroup$ Let me give an example. Say, $A=[1,2,3]$, $B=[4,5,6]$. So, $A_1=1$, $A_2=2$, $A_3=3$, $B_1=4$, $B_2=5$, $B_3=6$. Then $S_{12}=A_1B_2-A_2B_1 = 5-8=-3$. $S_{23}=A_2B_3-A_3B_2=12-15=-3$. And so on. So the $i$ and $j$ indices just cycle through all possible values between 1 and $D$ (for $D$ dimensions) individually, and independently of each other. $\endgroup$ – Viktor Toth Apr 23 '17 at 18:16

Imagine a column vector ${\bf A} = (A_1, A_2, A_3)$ and define the matrix

$$ A_\times = \left(\begin{array}{ccc} 0 & -A_3 & A_2 \\ A_3 & 0 & -A_1 \\ -A_2 & A_1 & 0 \end{array}\right) $$

Note that if ${\bf B}$ is another column vector, then

$$ A_\times {\bf B} = {\bf A}\times {\bf B} $$


$$ {\rm Transpose}(A_\times) = -A_\times $$

  • 2
    $\begingroup$ I don't really get this part: A×B=A×B , can you clarify? $\endgroup$ – Abu Bakr Apr 23 '17 at 18:24
  • $\begingroup$ I think that the "St Andrew cross" should be as an index (see the way I write it in (math.stackexchange.com/q/2239153)) $\endgroup$ – Jean Marie Apr 23 '17 at 19:24
  • $\begingroup$ @JeanMarie Thanks for the suggestion $\endgroup$ – caverac Apr 23 '17 at 22:00
  • $\begingroup$ @AbuBakr Please see JeanMarie's comment $\endgroup$ – caverac Apr 23 '17 at 22:01
  • $\begingroup$ $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} $$ the 3×3 skew symmetrix matrix above is the linear algebra representation of the cross product operator. $\endgroup$ – ja72 Apr 23 '17 at 23:15

We can substitute vector product $ \mathbf{a} \times \mathbf{b}$ by multiplying the vector $\mathbf{b}$ by a matrix because skew-symmetric matrix corresponding to the first vector $\mathbf{a}$ is defined as

$S(\mathbf{a})=[\mathbf{a} \times \mathbf{i} \ \ \mathbf{a} \times \mathbf{j} \ \ \mathbf{a} \times \mathbf{k} ]$,
where $\mathbf{i},\mathbf{j},\mathbf{k}$ are standard basis vectors forming as columns identity matrix $ {I} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 &0 \\ 0 & 0 &1 \end{bmatrix}$.

This gives formula presented above by caverac (you can notice for example that columns are (easy to check) orthogonal vectors to both $\mathbf{a}$ and appropriate standard basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ - also lengths of $S(\mathbf{a})$ columns are coherent with properties of cross product for this case).

In this case we have below formula with the use of multiplication the vector by the matrix interpreted as the sum of products of vector columns of matrix by components of vector (scalars):

$S(\mathbf{a})\mathbf{b}= (\mathbf{a} \times \mathbf{i})b_x + (\mathbf{a} \times \mathbf{j})b_y + (\mathbf{a} \times \mathbf{k}) b_z =\mathbf{a} \times (b_x\mathbf{i} + b_y\mathbf{j} + b_z\mathbf{k} )=\mathbf{a} \times \mathbf{b} $,

$b_x , b_y , b_z$ are coordinates of $\mathbf{b}$ vector.


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