TITLE: Natural explanation for a matrix identity
QUESTION [5 upvotes]: I recently came across this curious fact in some calculations with the strain tensor in fluid mechanics:
Let $A$ be an antisymmetric 3 by 3 matrix and $S$ be a traceless symmetric 3 by 3 matrix. Then it is easy to see that $AS+SA$ is an antisymmetric 3 by 3 matrix.
$A$ corresponds to a 3-dimensional vector $v$ in the following way: for any 3-dimensional vector $w$ we have $Aw=v\times w$. When the same procedure is applied to $AS+SA$, it will produce the vector $-Sv$.
I have two ways to prove this. The first is to write everything in terms of indices, and watch the big chunk mysteriously fizzles away after massive cancellation. The other way is representation-theoretic: it boils down to the fact that, in terms of $SO(3)$-representations, $3\otimes 5=3\oplus5\oplus7$, and everything except the constant factor follows from Schur's Lemma.
The representation-theoretic proof also says that the projection $3\otimes 5\to 3$ is "natural" in some sense. But now I have two descriptions of this projection, so is there a natural way (without indices, but also more concrete than abstract representation-theoretic nonsense) to see that they agree up to the minus sign?

REPLY [10 votes]: Your claim can be restated and generalised slightly as 
$$S(v\times w)+(Sv)\times w+v\times(Sw)=\text{trace}(S)v\times w$$ 
for all symmetric $S$ and all $v,w\in\mathbb{R}^3$.  Just as with the trace-free case, it is straightforward to check the identity by a large and unilluminating calculation.
Another approach is as follows. We have an orthonormal basis of eigenvectors $e_i$ with eigenvalues $s_i$ say and the cross product of any two of them will be the third one up to sign.  It is then easy to check the identity when $v=e_i$ and $w=e_j$ (noting that both sides are zero if $i=j$).  The general case then follows by bilinearity.
My guess is that there is a better, more direct proof but I cannot see one at the moment.  I agree that this is a curious fact, and I am surprised that I have never seen it before.