With $ a^2 + b^2 + c^2 = 1$, what type of matrix is $A$?
$A = \begin{bmatrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{bmatrix}$
So far I've tested $A$ for several types, I know that $A$ is non-orthogonal, obviously skew-symmetric and singular, as it's determinant equals $0$ and there is no chance to get an inverse.
But nothing is related to the $ a^2 + b^2 + c^2 = 1$ condition, any ideas?
On
Looks like $A$ is the matrix of the linear transformation $\vec{x}\mapsto \vec{x}\times (c,b,a)$. Therefore it is the so called infinitesimal generator for the group of rotations around the axis $\vec{n}=(c,b,a)$ (a unit vector!). We get those rotation matrices (in the right handed direction) using the exponential $$R(\theta)=e^{\theta A},$$ where matrix exponential is defined by the power series $$ e^X=I_3+\sum_{n=1}^\infty\frac1{n!}X^n. $$ Of course, $\theta$ is the angle of rotation. If $\vec{n}$ were not a unit vector, the rotations would go at angular speed $\omega=||\vec{n}||$.
Don't know if this makes the matrices special in any way :-(
Basically you are looking at the Lie algebra $\mathfrak{so}(3,\Bbb{R})$. This is in a sense the reason cross products appear frequently when one studies the mechanics of rotations.
If you have not read as much algebra to make sense of Jyrkis answer: in other words it is "rotation around an axis" (vector cross product) by the vector $(c,b,a)$. If you remember from linear algebra $$(x,y,z)^T \times (c,b,a)^T$$ Now try calculating $$\begin{bmatrix}0&a&-b\\-a&0&c\\b&-c&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}$$ Unless I am tired and wrote it wrong, the results should be the same!
In other words it is a matrix representation of cross product by a vector.