Dtjytyk

Let $S$ be a $3 times 3$ matrix which satisfies $S^T = S$ and $trace(S)=0$. Show for any $a,b in mathbb{R}^3$ that
$$ S( a times b) +(Sa) times b + a times (Sb) =0 $$

This can obviously done by a (long) brute force approach, but is there a more interesting way to approach this? Thank you!

This is the approach that I was saying.

The form
$$g(a,b)=S(atimes b)+Satimes b+atimes Sb$$

is linear in $a$, linear in $b$, and anti-symmetric $g(a,b)=-g(b,a)$.

By the symmetry of $S$, and rotating those triple products.
$$begin{align}
acdot g(a,b)&=acdot S(atimes b)+acdot(Satimes b) + acdot(atimes Sb)\
&=Sacdot(atimes b)+bcdot(atimes Sa) + Sbcdot(atimes a)\
&=Sacdot(atimes b)+Sacdot(btimes a) + 0\
&=Sacdot(atimes b)-Sacdot(atimes b)\
&=0
end{align}$$

Also $$bcdot g(a,b)=-bcdot g(b,a)=0$$

Intermission:

The multiplication $(atimes b)cdot g(a,b)$ is essentially computing the trace of $S$ in the basis $a,b,atimes b$. Now, it seemed to me that it is even faster to just go to compute this part for the basis $e_1,e_2,e_3$, only since $S$ is already a matrix in this basis, which is moreover orthogonal.

Likewise, the computation above could just be done for $e_1,e_2,e_3$, which is all we need. But since the general case is as easy, I left it that way.

So we do,
$$begin{align}
e_3cdot g(e_1,e_2)&=e_3cdot Se_3+e_3cdot (Se_1times e_2)+e_3cdot (e_1times Se_2)\
&=e_3cdot Se_3+e_1cdot Se_1+e_2cdot Se_2\
&=0 text{ because this is the trace of }S
end{align}$$

From the two computations above $e_kcdot g(e_i,e_j)$ is zero for all $i,j,kin{1,2,3}$.

Therefore, $g(e_i,e_j)=0$ for all $i,jin{1,2,3}$, since it is orthogonal to all elements of the frame $e_1,e_2,e_3$.

Hence $g(a,b)=0$ for all $a,b$, since the bilinear form is zero at all elements of the basis $e_1,e_2,e_3$.

The problem becomes a little easier if we transform orthogonally all matrices and vectors to the basis where symmetric matrix $S$ becomes diagonal matrix $D$, vectors $a,b$ are still any vectors but expressed with reference to this basis, denote them $m,n$.

In this basis

$$ D( m times n) +(Dm) times n + m times (Dn) =0 $$

Vector product can be replaced with a use of skew-symmetric matrix, denote it $K$ .... $K(m)$ means here the skew-symmetric matrix assigned to the vector $m$ according to the rules in Wikipedia.

$$ D K(m) n +K(Dm) n + K(m) Dn =0 $$
For any $n$
$$ (D K(m) +K(Dm) + K(m) D)n =0 $$
hence we need to prove that for any $m$

$$ D K(m) +K(Dm) + K(m) D =0 $$

Denote $D=begin{bmatrix} d_1 & 0 & 0 \
0 & d_2 & 0 \
0 & 0 & d_3\ end{bmatrix}$ and $m=begin{bmatrix} x & y & z end{bmatrix}^T $

Then $Dm=begin{bmatrix} d_1x & d_2y & d_3z end{bmatrix}^T $ and $K(m)= begin{bmatrix} 0 & -z & y \
z & 0 & -x \
-y & x & 0\ end{bmatrix}$.

Consequently $DK(m)+K(m)D=begin{bmatrix} 0 & -d_1z & d_1y \
d_2z & 0 & -d_2x \
-d_3y & d_3x & 0\ end{bmatrix}+begin{bmatrix} 0 & -d_2z & d_3y \
d_1z & 0 & -d_3x \
-d_1y & d_2x & 0\ end{bmatrix} = begin{bmatrix} 0 & -(d_2+d_1)z & (d_3+d_1)y \
(d_1+d_2)z & 0 & -(d_3+d_2)x \
-(d_1+d_3)y & (d_2+d_3)x & 0\ end{bmatrix} $

Under change of basis trace is preserved ( $operatorname{trace} (D) = operatorname{trace}(S)) $ hence we have $d_1+d_2+ d_3=0$ what leads to $$DK(m)+K(m)D= -K(Dm)$$

Therefore for any $m$ we have $$ D K(m) +K(Dm) + K(m) D =0 $$ .