What is the relationship between vector and its associated skew symmetric matrix?

Question

This is my first post in this forum, so hello everyone!

I am working with geometries (i.e. areas, volumes and inertias of polygons and polyhedrons in 3D space). For doing that, I to use both the Cross Product and the Parallel Axis Theorem, among others.

According to wikipedia, these can be defined, by using the

[...] skew symmetric matrix associated with the position vector [...]

and then operating on them (see links)

So, my question is:

What is the connection between $ \vec{r} = (x,y,z)$ and $ [r]=\left( \begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{array} \right)$ ?

After googling and looking for an answer for this for a while, I have not been able to find an answer. I guess it may have something to do with the eigenvectors, but I am just not sure.

Any help will be appreciated!

Thanks!

Answer 1

You seem to have exchanged $y$ and $z$ in your formula, but $A=\begin{pmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{pmatrix}$ is the matrix such that $A\vec v=\vec r\times\vec v$ for all $\vec v$.

The fact that this matrix is skew symmetric corresponds to the fact that $\vec r\times \vec v$ is always perpendicular to $\vec v$ -- the skew-symmetric matrices are exactly those where $\vec v\cdot A\vec v=0$ for all column vectors $\vec v$.

It is somewhat interesting that every $3\times 3$ skew-symmetric matrix arises in this way -- so whenever you have a linear transformation $\mathbb R^3\to\mathbb R^3$ where the output is always perpendicular to the input, that transformation is necessarily the same as crossing with an appropriately chosen $\vec r$. In dimensions other than $3$ this don't work because the vector space of skew-symmetric matrices has a different dimension than the space itself -- this is in fact one of the reasons why there is no straightforward generalization of the cross product to dimensions other than $3$. (Various such generalizations do exist but can't be claimed to be straightforward).

Answer 2

If you want only vector definition for skew symmetric matrix assume $r$ is a vector $= [r_x r_y r_z]^T$ in the frame described by versors $i, j, k$ Then skew-symetric matrix $S(r)$ assigned to the vector $r$ is constructed this way $$S(r) = \begin{pmatrix} r \times i \ \ r \times j \ \ r \times k \end{pmatrix}$$

As you see in the form of matrix you have written down in the first column $r_x$ is missing, in the second $r_y$ , in the third $r_z$ so columns are perpendicular to $i, j, k$ as to $r$ as well. This is the only matrix (up to the scale of columns) that has such property.