I was given the following problem to consider:
Find the rank II tensor T that satisfies the identity,
$T - Tr(T)\delta = \delta\times\vec{a}$
No information was given for the vector $a$, so we assume it's an arbitrary vector.
The right-hand side of the equation is a vector product between a rank 2 tensor and a vector, but I think since we're working in three dimensions, we can write it as:
$T = Tr(T)\delta_{jk} -\delta_{jm}\epsilon_{mkn}a_n$
I am, unfortunately, stuck at this point. I've tried shuffling around with indices, but I didn't manage to do anything of notice.
Any help or tips are appreciated, thanks in advance.
Consider the skew-symmetric matrix $$\eqalign{ A &= -\varepsilon\cdot a &= I\times a \cr A^T &= +\varepsilon\cdot a &= -A \cr }$$ and calculate its trace $$\eqalign{ {\rm tr}(A) &= A:I = A^T:I^T = -A:I = -{\rm tr}(A) \cr {\rm tr}(A) &= 0 \cr }$$ Therefore $\,T=A\,$ is a solution to the problem.
NB: In matrix notation, a colon denotes the double-dot product, i.e. $$A:B = A_{ij}B_{ij}$$