Suppose that $A\in O(2)$. I.e. $A$ is an invertible real 2 by 2 matrix such that $A^T=A^{-1}$. We can define a map $T_A:\mathbb{R}^{2,2}\to \mathbb{R}^{2,2}$ by $T_A(B)=ABA^T$. Linearity is clear, and so we can consider the transformation matrix $[T_A]$. This is a 4 by 4 matrix. I was wondering whether we could say $[T_A]\in O(4)$?
I guess one approach would be to take a basis of $\mathbb{R}^{2,2}$ and apply $T_A$ to each basis vector to construct the matrix $[T_A]$, and then verify from there. However this seems very tedious (especially if it is not true!). Is there a more elegant way? (or is it even true?)
Thanks in advance
In the same way orthogonal matrices $A$ preserve angles ($\langle v, w \rangle = \langle Av, Aw \rangle$) with respect to the Euclidean inner product, you can consider the analogous inner product on $\mathbb{R}^{2,2}$.
This inner product on $\mathbb{R}^{2,2}$ can be written as $\langle B, C \rangle := B_{11} C_{11} + B_{12} C_{12} + B_{21} C_{21} + B_{22} C_{22}$, or more succinctly as $\langle B, C \rangle = \text{Tr}(B^\top C)$.
Using properties of trace and the orthogonality of $A$, we have $$\langle T_A(B), T_A(C) \rangle = \langle ABA^\top, ACA^\top\rangle = \text{Tr}((ABA^\top)^\top ACA^\top) = \text{Tr}(AB^\top C A^\top) = \text{Tr}(B^\top C) = \langle B, C \rangle$$ so $T_A$ does preserve the inner product on $\mathbb{R}^{2,2}$.