<s>[INST] 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?
 [/INST] 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.
</s>