The current $J_i$ due to an electric field $E_i$ is given by $J_i=\sigma_{ij}E_j$, where $\sigma_{ij}$ is the conductivity tensor. In a certain coordinate system, $$(\sigma_{ij})=\begin{pmatrix} 2&{-1}&{-1}\\ {-1}&2&{-1}\\ {-1}&{-1}&2\end{pmatrix}$$
Show that there is a direction along which no current flows.
Honestly I'm not really sure what I'm meant to show, obviously that there is a direction along which no current flows, but I'm not sure what that 'means', I understand the sentence, but I know what it corresponds to showing, if that makes any sense.
If possible I would ask for a full solution, just so I can study the answer to help me approach the other question, but anything is helpful.
Any help is appreciated, thank you.
To sum up the exchange in comments:
We want to find a direction along which there's no current flow, regardless of the particular electric field. But the component of $J$ along some direction vector $n$ is given by the scalar product $n^T J$. So we want $n^T J=n^T \sigma E=0$; for this to be true regardless of the electric field $E$, we require $n^T \sigma=0$ i.e. $n$ should be an eigenvector of $\sigma$ with eigenvalue $0$. (Strictly it's a left-eigenvector; but $\sigma$ is symmetric, so this distinction doesn't matter.) By inspection, one sees that $n=(1,1,1)^T$ is such an eigenvector and we conclude that there is no current flow in this direction.
An perhaps more 'physical' approach is to consider a few choices of electric field $E$, and see if there's a common direction with no current flow. For instance, if $E=(1,0,0)^T$ then $J=(2,-1,-1)^T$; if $E=(0,1,0)$ then $J=(-1,2,-1)^T$. In both cases, there will be no current flow in a direction perpendicular to $J$. Hence the cross product of these two resulting currents, $(2,-1,-1)\times(-1,2,-1)=(3,3,3)$, will be perpendicular to both currents.
So there's no current flow in the direction of $(3,3,3)$ for either of these electric fields. To confirm that this holds regardless of $E$, we note that when $E=(0,0,1)^T$ we have $J=(-1,-1,2)^T$ which is perpendicular to $(3,3,3)^T$. So $(3,3,3)$ is perpendicular to $J$ whenever $E$ is any basis vector, and by linearity we conclude that it's true for any $E$. So we conclude that there's no flow in the direction defined by $(3,3,3)^T$, which is the same as that defined by $(1,1,1)^T$.