Is $x\times(T\cdot n)$ a triple product?
That is, does $x\times(T\cdot n) = - n \cdot (T\times x)$ ?
Where $T$ is a matriz (2-rank tensor).
I try the Einstein notation, but got stuck:
$ \epsilon_{ijk}x_jT_{kl}n_l = -\epsilon_{ikj}n_lT_{kl}x_j=-n_l\epsilon_{ikj}T_{kl}x_j $
$ ?= -n_l(T_l \times x)_i \; \dots \; ? $
If we for some strange reason define the "cross product" of a Matrix $T^u_j$ and a vector $n_k$ like this (and please note that this is by no means a vector, rather a (2,0) tensor) $$(T\times n)^{iu}=\varepsilon^{ijk}T^u_jn_k$$ a natural analogy of the usual triple product is (and again please note that this is not a scalar, rather a contravariant vector) $$\left(x\cdot(T\times n)\right)^u=x_i\epsilon^{ijk}T^u_jn_k$$ Now we can use the skew symmetric property $\varepsilon^{ijk}=-\varepsilon^{kji}$ and a simple renaming (swap $i,k$) to get $$x_i\epsilon^{ijk}T^u_jn_k=n_k\epsilon^{ijk}T^u_jx_i=-n_k\epsilon^{kji}T^u_jx_i=-n_i\epsilon^{ijk}T^u_jx_k=\left(-n\cdot(T\times x)\right)^u$$ Hence $$x\cdot(T\times n)=-n\cdot(T\times x)$$