Let $A$ be any real $4$ by $4$ invertible matrix such that $A^t\eta A=\lambda\eta$.
Here $\eta=diag(-1,1,1,1)$. Then I have to show that $\lambda$ is positve.
I tried the determinant but it gives $\lambda^4$ is positive. I tried the trace but there is no guarantee that $A^t\eta A$ has a positve trace. I tried to find a counterexample but there doesn't seem any...
It is so frustrating....Is there some magic way to show that $\lambda$ is positive?
It is an immediate consequence of Sylvester's theorem: the signature of a quadratic form does not change under congruence transformations. The signature of $\lambda \eta$, i.e, $$ (-sign(\lambda),sign(\lambda),sign(\lambda),sign(\lambda))$$ is the same as the one of $\eta$ $$(-,+,+,+)$$ if and only if $\lambda >0$.
Notice that it would by false in $\mathbb{R}^2$.