Show that $\mathfrak{so}(p,q,\mathbb{R}) = I_{p,q} \mathfrak{so}(n,\mathbb{R})$.

Question

I found this statement in another stackexchange post but couldn't get the $\mathfrak{so}(p,q,\mathbb{R}) \subseteq I_{p,q} \mathfrak{so}(n,\mathbb{R})$ direction. For the reverse, I could show that $\left(I_{p,q} x\right)^t I_{p,q} = - I_{p,q} \left(I_{p,q} x\right)$ (assuming $x\in \mathfrak{so}(n,\mathbb{R})$), but I'm stuck on getting from $x^t I_{p,q} = -I_{p,q} x$ to $x^t = -x$.

For reference, $I_{p,q}$ is defined as

$$\begin{pmatrix} I_p & 0_{p\times q} \\ 0_{q\times p} & -I_q\end{pmatrix}.$$

Any help would be appreciated.

Answer 1

You don’t need to show the other inclusion. $I_{p,q}$ is an isomorphism so you have that the dimensions of the two vector spaces are the same, and it is a fact that if $ U\subset V$ and their dimensions are equal then $U=V$

Edit to add;

Let $X\in so(p,q)$, and $\eta$ be the pseudo Euclidean norm of signature $(p,q)$, Then by definition: $$X^T\eta+\eta X=0$$ Now note that $\eta X$ is by definition in the Lie algebra $so(n)$. It follows that $ X=\eta^2X=\eta(\eta X)$ hence lies in $\eta so(n).$