If $A \in O(3,1)$, why does $A^* gA = g$?

Question

Let $O(3,1)$ denote the Lorentz group, i.e. the group of all linear transformations on $\mathbb{R}^4$ that preserve the following inner product: $$\Lambda(x,y) = x^0y^0 - x^1y^1 - x^2y^2 - x^3y^3.$$ Further, let $$g = \begin{bmatrix} 1 &0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{bmatrix}$$

I am following a book on mathematical physics which states that if $A \in O(3,1)$, it follows that $$A^*gA = g$$ where $A^*$ is the adjoint of $A$.

Based on the wording of the text it seems this should be obvious, but I cannot see why this holds. How would one prove this?

Answer 1

If $A\in O(3,1)$, then, for any two vectors $x,y\in\Bbb R^4$, $\Lambda(Ax,Ay)=\Lambda(x,y)$. But $\Lambda(x,y)=x^*gy$, and$$\Lambda(Ax,Ay)=(Ax)^*g(Ay)=x^*A^*gAy.$$So,$$\left(\forall x,y\in\Bbb R^4\right):x^*A^*gAy=x^*gy,$$which means that $A^*gA=g$.