I am trying to deduce the Minkowski's dot product for two dimentional space: $$g=x^1y^1-c^2t_xt_y$$ If $f$ denote the Lorentz's transformation for two dimentional case: $$\begin{array}{rcll} f:&\mathbb{R}&\longrightarrow &\mathbb{R}\\ &x&\longmapsto &f(x)=\Lambda x \end{array}\qquad \text{ with }\ \Lambda = \gamma\left(\begin{array}{cc} 1 &-v\\ -v/c^2 & 1 \end{array}\right)$$ we have to find a matrix: $$G=\left(\begin{array}{cc} g_{11} & g_{12}\\ g_{21} & g_{22} \end{array}\right)$$ such that $\chi^tG\chi= (\Lambda \chi)^tG(\Lambda \chi)$ or equivalently: $$G=\Lambda ^tG\Lambda$$ we get the system: $$\begin{array}{rlr} \frac{v^2}{c^2}g_{11}-\frac{v}{c^2}g_{12}-\frac{v}{c^2}g_{21}+\frac{v^2}{c^4}g_{22}&=0&(1)\\ -vg_{11}+\frac{v^2}{c^2}g_{12}+\frac{v^2}{c^2}g_{21}-\frac{v}{c^2}g_{22}&=0& (2)\\ -vg_{11}+\frac{v^2}{c^2}g_{12}+\frac{v^2}{c^2}g_{21}-\frac{v}{c^2}g_{22}&=0& (3)\\ v^2g_{12}-vg_{12}-vg_{21}+\frac{v^2}{c^2}g_{22}&=0&(4) \end{array}$$ which is equivalent to: $$\begin{array}{rll} vg_{11}-g_{12}-g_{21}+\frac{v}{c^2}g_{22}&=0\\ c^2g_{11}-vg_{12}-vg_{21}+g_{22}&=0 \end{array}$$ if we denote $g_{21}=\lambda$ and $g_{22}=\mu$, it's easy to prove that: $$G=\left(\begin{array}{cc} -\mu/c^2& -\lambda\\ \lambda &\mu \end{array}\right)$$ As $G$ must be symmetric, $\lambda=0$; so then: $$G=\left(\begin{array}{cc} -\mu/c^2& 0\\ 0 &\mu \end{array}\right)$$ My question is:
How we can get that $\mu$ must be $-c^2$?
