Problem with the proof that for every timelike vector there exists an inertial coordinate system in which its spatial coordinates are zero I am reading lecture notes on special relativity and I have a problem with the proof of the following proposition.
Proposition. If $X$ is timelike, then there exists an inertial coordinate system in which $X^1 = X^2 = X^3 = 0$.
The proof states that as $X$ is timelike, it has components of the form $(a, p\,\mathbf{e})$, where $\mathbf{e}$ is a unit spatial vector and $\lvert a \rvert > \lvert p \rvert$. Then one considers the following four four-vectors:
\begin{align*}
\frac{1}{\sqrt{a^2 - p^2}}(a, p\,\mathbf{e}) & & \frac{1}{\sqrt{a^2 - p^2}}(p, a\,\mathbf{e}) & & (0, \mathbf{q}) & & (0, \mathbf{r})\,,
\end{align*}
where $\mathbf{q}$ and $\mathbf{r}$ are chosen so that $(\mathbf{e}, \mathbf{q}, \mathbf{r})$ form an orthonormal triad in Euclidean space. Then the proof concludes that these four-vectors define an explicit Lorentz transformation and stops there.
For me this explicit Lorentz transformation is represented by the following matrix.
\begin{bmatrix}
\frac{1}{\sqrt{a^2 - p^2}} a & \frac{p}{\sqrt{a^2 - p^2}} & 0 & 0 \\
\frac{p}{\sqrt{a^2 - p^2}} e^1 & \frac{a}{\sqrt{a^2 - p^2}} e^1 & q^1 & r^1 \\
\frac{p}{\sqrt{a^2 - p^2}} e^2 & \frac{a}{\sqrt{a^2 - p^2}} e^2 & q^2 & r^2 \\
\frac{p}{\sqrt{a^2 - p^2}} e^3 & \frac{a}{\sqrt{a^2 - p^2}} e^3 & q^3 & r^3 \\
\end{bmatrix}
However, multiplying the column vector $(X^0, X^1, X^2, X^3)$ by the matrix above does not seem to yield a column vector whose spatial components are zero.
What did I miss?
 A: Actually, your matrix can be greatly simplified as
$$
M =
\begin{bmatrix}
\frac{1}{\sqrt{a^2 - p^2}} a & \frac{p}{\sqrt{a^2 - p^2}} & 0 & 0 \\
\frac{p}{\sqrt{a^2 - p^2}} & \frac{a}{\sqrt{a^2 - p^2}} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
since $(\mathbf{e}, \mathbf{q}, \mathbf{r})$ forms an orthonormal triad, hence $e_1 = q_2 = r_3 = 1$ and the other coefficients are zero.
Now, as pointed out by Valter Moretti in his comment, the matrix you're looking for is the inverse of this matrix. An easy calculation gives the inverse.
$$
M^{-1} =
\frac{1}{(a^2 - p^2)^{\frac{3}{2}}} 
\begin{bmatrix}
a(a^2 - p^2) & -p(a^2 - p^2) & 0 & 0 \\
-p(a^2 - p^2) & a(a^2 - p^2) & 0 & 0 \\
0 & 0 & (a^2 - p^2)^{\frac{3}{2}} & 0 \\
0 & 0 & 0 & (a^2 - p^2)^{\frac{3}{2}}
\end{bmatrix}
$$
Finally, one easily checks the result as follows.
$$
M^{-1} \times 
\begin{bmatrix}
a \\
p \\
0 \\
0
\end{bmatrix}
= 
\begin{bmatrix}
\sqrt{a^2 - p^2} \\
0 \\
0 \\
0 
\end{bmatrix}
$$
Note that taking $\mathbf{q}$ and $\mathbf{r}$ such that $(\mathbf{e}, \mathbf{q}, \mathbf{r})$ forms an orthonormal triad is equivalent to doing a spatial rotation such that only $X^1$ is nonvanishing. Thus, a more geometric, less algebraic, proof will start with a spatial rotation then will proceed with a boost along the $x$-axis.
A: 
The 3+1-Lorentz transformation is
\begin{align}                 
    \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=}  \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot}  \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t
\tag{01a}\label{01a}\\
 c\,t^{\boldsymbol{\prime}} & \boldsymbol{=}   \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right)
\tag{01b}\label{01b}\\
\gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{01c}\label{01c} 
\end{align}
and in differential form
\begin{align}                 
    \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=}  \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot}  \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt
\tag{02a}\label{02a}\\
c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=}   \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right)
\tag{02b}\label{02b}      
\end{align}
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$
For time-like vectors
Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace
\begin{equation}
\mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}
\tag{03}\label{03}      
\end{equation}
under the assumption
\begin{equation}
\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c
\tag{04}\label{04}      
\end{equation}

For space-like vectors
Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace
\begin{equation}
\mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}
\tag{05}\label{05}      
\end{equation}
under the assumption
\begin{equation}
\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c
\tag{06}\label{06}      
\end{equation}
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$
ADDENDUM

*

*From \eqref{02a} and \eqref{03} if
\begin{equation}
\mathbf{X}\boldsymbol{=}
\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\\
\mathbf{x}\\
\\
x_4
\end{bmatrix}\,\quad \texttt{with   } 
\left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{>}0
\tag{07}\label{07}      
\end{equation}
is a time-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity
\begin{equation}
\boxed{\:\:\mathbf{u}\boldsymbol{=}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:}
\tag{08}\label{08}      
\end{equation}
its space component is zero
\begin{equation}
\mathbf{X}'\boldsymbol{=}
\begin{bmatrix}
x'_1\\
x'_2\\
x'_3\\
x'_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\\
\mathbf{x}'\\
\\
x'_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\\
\boldsymbol{0}\\
\\
x'_4
\end{bmatrix}
\,\quad \texttt{with   } 
x'^{2}_4\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3
\tag{09}\label{09}      
\end{equation}


*From \eqref{02b} and \eqref{05} if
\begin{equation}
\mathbf{X}\boldsymbol{=}
\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\\
\mathbf{x}\\
\\
x_4
\end{bmatrix}\,\quad \texttt{with   } 
\left\Vert\mathbf{X}\right\Vert^2\boldsymbol{=}x^2_4\boldsymbol{-}x^2_1\boldsymbol{-}x^2_2\boldsymbol{-}x^2_3\boldsymbol{<}0
\tag{10}\label{10}      
\end{equation}
is a space-like 4-vector in an inertial frame $\color{blue}{\mathbf{S}}$, then in any inertial frame $\color{blue}{\mathbf{S}'}$ moving with velocity
\begin{equation}
\boxed{\:\:\mathbf{u}\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\dfrac{\mathbf{x}}{x_4}c\boldsymbol{=}\left\Vert\dfrac{\mathbf{x}}{x_4}\right\Vert^{\boldsymbol{-}2}\left(\dfrac{x_1}{x_4},\dfrac{x_2}{x_4},\dfrac{x_3}{x_4} \right)c \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:}
\tag{11}\label{11}      
\end{equation}
its time component is zero
\begin{equation}
\mathbf{X}'\boldsymbol{=}
\begin{bmatrix}
x'_1\\
x'_2\\
x'_3\\
x'_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\vphantom{x'_1}\\
\mathbf{x}'\\
\vphantom{x'_3}\\
x'_4
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\vphantom{x'_1}\\
\mathbf{x}'\\
\vphantom{x'_3}\\
0\vphantom{x'_4}
\end{bmatrix}
\,\quad \texttt{with   } 
\Vert\mathbf{x}'\Vert^2\boldsymbol{=}x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3\boldsymbol{-}x^2_4
\tag{12}\label{12}      
\end{equation}
A: The matrix you wrote will take the standard basis $\{(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)\}$ into the basis constructed from $\mathbf{X}$. Therefore, to take $\mathbf{X}$ into something proportional to $(1,0,0,0)$, you need to use the inverse matrix.
