= Solution
Now use $(ct,x)^T$ and $\eta_2=\operatorname{diag}(1,-1)$. Write
$$
L=\begin{pmatrix}a&b\\d&e\end{pmatrix}.
$$
Preservation of the <Minkowski metric> says that the first column has squared norm $1$, the second has squared norm $-1$, and the columns are orthogonal for the <Minkowski inner product>:
$$
a^2-d^2=1,\qquad b^2-e^2=-1,\qquad ab-de=0.
$$
Orthochronousness gives $a\ge1$. From the orthogonality equation, $b=de/a$; substituting in the second norm identity gives $e^2=a^2$. Hence $(b,e)=\varepsilon(d,a)$ with $\varepsilon=\pm1$. The <determinant> is then $\varepsilon(a^2-d^2)=\varepsilon$, so properness forces $\varepsilon=1$. Thus $b=d$ and $e=a$.
Set $\beta=d/a$. The first norm identity implies $|\beta|<1$ and $a=(1-\beta^2)^{-1/2}$. With $u=c\beta$, this gives
$$
\boxed{L=\begin{pmatrix}\gamma&\gamma u/c\\\gamma u/c&\gamma\end{pmatrix},
\qquad\gamma=(1-u^2/c^2)^{-1/2},\quad |u|<c.}
$$
Conversely, the displayed <matrix> has $\gamma^2-(\gamma u/c)^2=1$, equal off-diagonal entries, <determinant> one, and positive time-time entry. Direct multiplication therefore gives $L^T\eta_2L=\eta_2$, proving both directions. The plus signs are the <Lorentz boost> convention used in the paper; reversing the boost direction replaces $u$ by $-u$.
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