Let $M$ be a 4-vector space over $\mathbb R$. Since $M$ is linearly isomorphic to $\mathbb R^4$, the linear isomorphism is also a diffeomorphism between $M$ and $\mathbb R^4$. For each $p\in M$, we define a Minkowski metric tensor on $T_pM$ is a bilinear, symmetric and non-degenerate form $\,\langle,\rangle_p:\ T_pM\times T_pM\longrightarrow\mathbb R\, $ with signature $(+,+,+,-)$. The space $\big(M,\langle,\rangle\big)$ then is called Minkowski space.
Lemma 1. There exists an orthonormal basis for $T_pM\ $ (orthonormal with respect to $\langle,\rangle_p$). Lemma 2. For any basis $\mathcal B=(e_1,e_2,e_3,e_4)$ of $T_pM$, \begin{align} \langle u,v\rangle_p\,=\,\big[u\big]^\top_{\mathcal B}\,\langle,\rangle_{p\mathcal B} \,\big[v\big]_{\mathcal B},\ \forall\,u,v\in T_pM. \end{align} where $\,\langle,\rangle_{p\mathcal B}=\big(\langle e_i,e_j\rangle\big)_{ij}$. Thus, we have an orthonormal basis of $T_pM$ in which our Minkowski metric tensor has the form \begin{align} \langle u,v\rangle_p\,=\,u_1v_1+u_2v_2+u_3v_3-u_4v_4,\ \forall\,u,v\in T_pM.\tag1 \end{align}
Let $\eta=$ diag$(I_3,-1)$. A Lorentz transformation is a linear map $\,T:\,\mathbb R^4\longrightarrow\mathbb R^4\,$ satisfies \begin{align} \big[T\big]^\top\eta\big[T\big]\,=\,\eta. \end{align} It is the transformation that preserves the Minkowski structure of $M$. More specifically, the matrix of each Lorentz transformation is given by \begin{align*} \begin{bmatrix} y_1\\ y_2\\ y_3 \\ ct' \end{bmatrix} =\begin{bmatrix} 1+(\gamma-1)\displaystyle\frac{v_1^2}{v^2} & (\gamma-1)\displaystyle\frac{v_1v_2}{v^2} & (\gamma-1)\displaystyle\frac{v_1v_3}{v^2} & -\displaystyle\frac{\gamma v_1}{c} \\ (\gamma-1)\displaystyle\frac{v_1v_2}{v^2} & 1+(\gamma-1)\displaystyle\frac{v_2^2}{v^2} & (\gamma-1)\displaystyle\frac{v_2v_3}{v^2} & -\displaystyle\frac{\gamma v_2}{c} \\ (\gamma-1)\displaystyle\frac{v_1v_3}{v^2} & (\gamma-1)\displaystyle\frac{v_2v_3}{v^2} & 1+(\gamma-1)\displaystyle\frac{v_3^2}{v^2} & \displaystyle\frac{\gamma v_3}{c} \\ -\displaystyle\frac{\gamma v_1}{c} & -\displaystyle\frac{\gamma v_2}{c} & -\displaystyle\frac{\gamma v_3}{c} & \gamma \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \\ ct \end{bmatrix} \end{align*} where \begin{align*} c=1,\ \ \ \gamma=\sqrt{1-\frac{v^2}{c^2}}^{-1}=\frac{1}{\sqrt{1-v^2}},\ \ \ v^2=\|v\|^2=\sqrt{v_1^2+v_2^2+v_3^2}. \end{align*}
I found some solution for obtaining the Lorentz transformation with physics and geometry approach. My question is how can one derrive the formula for the Lorentz transformation $T$ with linear algebra approach ? Apparently, we can solve this matrix equation in term of linear system equation, but I wonder if there is other ways better ?
I hope someone will help me with some hints for this problem. Thanks.