$A^3$ When $A$ Is Skew-Symmetric

Skew-symmetric matrices are square matrices for which $A^T = -A$ is true. In the following lecture note,

there is a reference to a property of a skew-symmetric matrix; let's say for $A$, which is the skew symmetric matrix form of the vector $a$ (the paper calls them $\omega_x$ and $\omega$),

$$A^3 = -(a^T a) \cdot A$$

The vector would be $(a_1,a_2,a_3)$, it's skew-symmetric matrix

$$\left[\begin{array}{rrr} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{array}\right]$$

and, as is known, cross product between vectors can be converted to dot product, through the use of skew-symmetric form, i.e. $a \times b = A \cdot b$.

For $A^3$, when I write down the elements of the vector / matrix and perform the multiplication, the property checks out. Is there any other way to prove this algebraically? And would this proof be valid for all dimensions?

• What do you understand by the skew matrix form of a vector in 4 dimensions? – Justpassingby Dec 17 '15 at 9:45
• Let's ignore the other dimensions other than 3; I need to grok that first :) – BB_ML Dec 17 '15 at 11:14
• I have the impression that it is essential that we consider real quantities and a real inner product. – Urgje Dec 17 '15 at 14:06

The skew symmetric matrix form $A$ of a $3$-vector $a$ is the matrix of the linear transformation that corresponds to the vector product $x\mapsto a\times x.$ So we are really looking at the matrix of

$$x\mapsto a\times(a\times (a\times x)))$$

which is easily seen to be perpendicular to both $a$ and $x$ (hence a scalar multiple of $Ax=a\times x$).

If $a=e$ is a unit vector then the following vectors have equal norms:

$$\|e\times(e\times (e\times x)))\|=\|e\times (e\times x))\| =\|e\times x\|\hbox{ (but not necessarily \|x\|)}$$

because we are each time taking the vector product with an orthogonal unit vector. For general $a=\|a\|e$ we then have

\eqalign{\|a\times(a\times (a\times x)))\|&=\|a\|^3\|e\times(e\times (e\times x)))\|\\ &=\|a\|^3\|e\times x\|\\ &=\|a\|^2\|a\times x\|\\ &=(a^Ta)\|a\times x\|}

Because $A^3x$ and $Ax$ are parallel (see above), this gives

$$A^3x=\pm(a^Ta)Ax.$$

Finally, the minus sign can be worked out with your right hand.

In higher dimensions skew symmetric matrices have more degrees of freedom than a single vector and it is not generally true that $A^3x$ is a scalar multiple of $Ax.$ It remains true that $A^3x$ is orthogonal to both $A^2x$ and $x$ itself (the latter because $A^3$ is also skew symmetric).

If you want to work out the exponential map (one-parameter subgroup generated by a skew symmetric matrix) then it may be useful to know that in an appropriate orthonormal coordinate system $A$ becomes block diagonal and its effect on each of the two-dimensional invariant subspaces is a rotation by 90 degrees followed by multiplication by a scalar; so on each of these subspaces $A^2$ has the effect of multiplication by a negative scalar (but the scalars could be different for different components).

• For the 3D case; I understand the equal norm argument - how can I tie that to $A^3 = -(a^Ta)A$? – BB_ML Dec 17 '15 at 11:13
• If $a$ is a general vector (i.e., not necessarily unit) then the norm on the extreme left is $\|a\|^2=a^Ta$ times the norm on the extreme right. Tie that to the fact that the vectors are parallel and you have your identity, possibly up to sign (which is why you need your right hand). – Justpassingby Dec 17 '15 at 11:18
• Got it! Thanks. – BB_ML Dec 17 '15 at 11:33