Given $a^2+b^2+c^2 = 1$ what type is this matrix?

Question

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?

@mick thanks for the quick reply, but I don't get your question?, is it about the nature of my question? why should this matrix belong to a "special" type? — Hans, Aug 27 '17 at 20:34
Yes. Not Every matrix has an important property. Not sure about this one though. — mick, Aug 27 '17 at 20:38
Well, it's a unit vector in the 3d subspace of skew symmetric matrices - equipped with the given norm - among all $3\times 3$ matrices. — Berci, Aug 27 '17 at 20:41
By @dxiv comment, it means minimum polynomial is $x^3+x=x(x^2+1)=0$ that means it's eigenvalues are $0,i,-i$. Now we can say it's a special matrix! — Amin235, Aug 27 '17 at 20:46
You can see that using Jyrki's argument, too, since the eigenvalues of a rotation matrix are $1=e^{0},e^{i\theta}$, and $e^{-i\theta}$. @Amin235 (Just a neat agreement between two points of view.) — Thomas Andrews, Aug 27 '17 at 20:59
@ThomasAndrews we have $A^{3^n}=(-1)^n,A$. How does it interpret by Jyrki's argument. Thanks — Amin235, Aug 27 '17 at 21:46
One special property is $A^{n+2} = -A^n$, making it a periodic matrix, i.e. $A^5=A$. Also the Drazin and Penrose inverses are given by the transpose $A^+=A^T$. Its eigenvalues are ${-i, 0, +i}$ with corresponding eigenvectors ${(A^2-iA)v, a, (A^2+iA)v}$ where $v$ can be any vector such that $Av\ne 0$ — greg, Aug 27 '17 at 22:14
@Hans That follows directly from $,A=-A^T,$, and does not require $a^2+b^2+c^2=1,$. — dxiv, Aug 28 '17 at 02:47
@Amin235 When $A^{n+2}=-A^n$ for all $n>0$ it means that $$ (\theta A)+\frac{(\theta A)^3}{3!}+\frac{(\theta A)^5}{5!}+\cdots=A(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots)=A\sin\theta.$$ Similarly the even degree terms give something like $A^2\cos\theta $ except for an error in $A^0$. — Jyrki Lahtonen, Aug 28 '17 at 13:01
@dxiv Oh, sure thing. Haven't noticed, thanks for your feedback. — Hans, Aug 29 '17 at 04:15

score 10 · Answer 1 · answered Aug 27 '17 at 20:38

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.

score 4 · Accepted Answer · answered Aug 27 '17 at 21:22

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.

Given $a^2+b^2+c^2 = 1$ what type is this matrix?

2 Answers2