Let's fix notation and define the bilinear form $\eta: \mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}$ by:

$\eta((x,y,z,t),(x',y',z',t')) = xx'+yy'+zz'- tt'$

Given a map $T:\mathbb{R}^4 \to \mathbb{R}^4$ which fixes $0$ and preserves $\nu$ we want to show that $T$ is linear.

Let $e_1,e_2,e_3,e_4$ be the canonical basis of $\mathbb{R}^4$. The first observation is that for any four vectors $v_1,v_2,v_3,v_4$ such that $\eta(v_i,v_j) = \eta(e_i,e_j)$ for all $i,j$ the linear map sending each $v_i$ to $e_i$ is invertible and preserves $\eta$.

Hence by composing $T$ with a linear invertible $\eta$ preserving map we may assume that $Te_i = e_i$ for $i = 1,2,3,4$.

Now we have for any $v \in \mathbb{R}^4$ that $\eta(v,e_i) = \eta(Tv,Te_i) = \eta(Tv, e_i)$ this implies that $T$ is the identity (since we can get each coordinate of $Tv$).

Let's fix notation and define the bilinear form $\eta: \mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}$ by:

$\eta((x,y,z,t),(x',y',z',t')) = xx'+yy'+zz'- tt'$

Given a map $T:\mathbb{R}^4 \to \mathbb{R}^4$ which fixes $0$ and preserves $\eta$ we want to show that $T$ is linear.

Let $e_1,e_2,e_3,e_4$ be the canonical basis of $\mathbb{R}^4$. The first observation is that for any four vectors $v_1,v_2,v_3,v_4$ such that $\eta(v_i,v_j) = \eta(e_i,e_j)$ for all $i,j$ the linear map sending each $v_i$ to $e_i$ is invertible and preserves $\eta$.

Hence by composing $T$ with a linear invertible $\eta$ preserving map we may assume that $Te_i = e_i$ for $i = 1,2,3,4$.

Now we have for any $v \in \mathbb{R}^4$ that $\eta(v,e_i) = \eta(Tv,Te_i) = \eta(Tv, e_i)$ this implies that $T$ is the identity (since we can get each coordinate of $Tv$).

Let's fix notation and define the bilinear form $\eta: \mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}$ by:

$\eta((x,y,z,t),(x',y',z',t')) = xx'+yy'+zz'- tt'$

Given a map $T:\mathbb{R}^4 \to \mathbb{R}^4$ which fixes $0$ and preserves $\nu$ we want to show that $T$ is linear.

Let $e_1,e_2,e_3,e_4$ be the canonical basis of $\mathbb{R}^4$. The first observation is that for any four vectors $v_1,v_2,v_3,v_4$ such that $\eta(v_i,v_j) = \eta(e_i,e_j)$ for all $i,j$ the linear map sending each $v_i$ to $e_i$ is invertible and preserves $\nu$.

Hence by composing $T$ with a linear invertible $\nu$ preserving map we may assume that $Te_i = e_i$ for $i = 1,2,3,4$.

Now we have for any $v \in \mathbb{R}^4$ that $\eta(v,e_i) = \eta(Tv,Te_i) = \eta(Tv, e_i)$ this implies that $T$ is the identity (since we can get each coordinate of $Tv$).

Let's fix notation and define the bilinear form $\eta: \mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}$ by:

$\eta((x,y,z,t),(x',y',z',t')) = xx'+yy'+zz'- tt'$

Given a map $T:\mathbb{R}^4 \to \mathbb{R}^4$ which fixes $0$ and preserves $\nu$ we want to show that $T$ is linear.

Let $e_1,e_2,e_3,e_4$ be the canonical basis of $\mathbb{R}^4$. The first observation is that for any four vectors $v_1,v_2,v_3,v_4$ such that $\eta(v_i,v_j) = \eta(e_i,e_j)$ for all $i,j$ the linear map sending each $v_i$ to $e_i$ is invertible and preserves $\eta$.

Hence by composing $T$ with a linear invertible $\eta$ preserving map we may assume that $Te_i = e_i$ for $i = 1,2,3,4$.

Now we have for any $v \in \mathbb{R}^4$ that $\eta(v,e_i) = \eta(Tv,Te_i) = \eta(Tv, e_i)$ this implies that $T$ is the identity (since we can get each coordinate of $Tv$).