$A^3$ When $A$ Is Skew-Symmetric Skew-symmetric matrices are square matrices for which $A^T = -A$ is true. In the following lecture note, 
http://ethaneade.com/lie.pdf
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? 
 A: 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).
