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?