How to solve this tensorial equation $ g_{ij} x^ix^j=0$ , Einstein notation for summation is used.
$ i, j \in \{1,2,3,4\} , x=\{x^1, x^2, x^3, x^4\} $
Given $ g_{ij} $ is a non singular $ 4$ x $4$ matrix, and symmetric $ g_{ij} =g_{ji} $ How to find non trivial solutions $ x \neq 0$?
Comment : I know how to solve this $g_{ij} y^i_ny^j_k=\delta^n_k $ but I can't see how that helps.
I suspect you intend each $g_{ij},\,x^i$ to be real. Then, since $g$ is symmetric, diagonalizing it obtains a matrix with real diagonal entries. We're given that none of these are zero, and if they're all positive the only solution is $x=0$. But if some eigenvalues are positive and others are negative, nontrivial solutions exist. A similarity transformation can set each eigenvalue of $g$ to be $\pm 1$ with the $+1$s grouped together, so we can rewrite the problem as $x_\uparrow^2=x_\downarrow^2$ (say).