# Show that any proper homogeneous Lorentz transformation may be expressed as the product of a boost times a rotation

ANSWER - Parts I & II

$$\texttt{C O N T E N T S}$$

$$\texttt{Abstract}$$

$$\boldsymbol\S\texttt{ A. Proper homogeneous Lorentz transformations : }\Lambda$$

$$\boldsymbol\S\texttt{ B. Lorentz boosts : }\mathrm L\left(\mathbf{v}\right)$$

$$\boldsymbol\S\texttt{ C. The decomposition : }\Lambda = \rm L\left(\mathbf{v}\right)\mathcal R$$

$$\boldsymbol\S\texttt{ D. The uniqueness of the decomposition:}\rm L\left(\mathbf{v}'\right)\mathcal R'\!=\!\rm L\left(\mathbf{v}\right)\mathcal R \implies \mathbf{v}'\!=\!\mathbf{v},\mathcal R'\!=\!\mathcal R$$

$$\boldsymbol\S\texttt{ E. Decomposition with Lorentz boost first : }\Lambda = \rm \mathcal Q\, L\left(\mathbf u \right)$$

Abstract

It will be proved that a proper homogeneous Lorentz transformation ($$\Lambda$$) can be decomposed by one and only one way in a Lorentz boost ($$\rm L$$) with velocity $$\mathbf v$$ and a rotation in space ($$\mathcal R$$) $$$$\Lambda = \rm L\,\mathcal R \tag{a}\label{a}$$$$ in this order. The components of the velocity $$\mathbf v$$ and the characteristics of the rotation $$\mathcal R$$ (matrix elements, axis and angle) will be given as expressions of the matrix elements of $$\Lambda$$.

It must be noted that the transformation $$\mathcal R$$ is assumed a priori as spatial in general. That it is a pure rotation is proved a posteriori.

$$\boldsymbol\S$$ A. Proper homogeneous Lorentz transformations

For completeness we note the most important properties of Lorentz transformations as found in many textbooks and the Web. So let a 4-vector in Minkowski space-time
$$$$\mathbf X = \begin{bmatrix} \vphantom{\dfrac{a}{b}}\\ \mathbf x\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ c\,t\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} x_1\vphantom{\dfrac{a}{b}}\\ x_2\vphantom{\dfrac{a}{b}}\\ x_3\vphantom{\dfrac{a}{b}}\\ x_4\vphantom{\dfrac{a}{b}} \end{bmatrix}\,, \qquad x_k \in \mathbb R \tag{A-01}\label{A-01}$$$$ with norm $$$$\Vert\mathbf X\Vert^2=\left(c\,t\right)^2-\vert\mathbf x \vert^2=x^2_4-\left(x^2_1+x^2_2+x^2_3\right) \tag{A-02}\label{A-02}$$$$ An homogeneous Lorentz transformation
$$$$\mathbf X'= \begin{bmatrix} \vphantom{\dfrac{a}{b}}\\ \mathbf x'\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ c\,t'\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} x'_1\vphantom{\dfrac{a}{b}}\\ x'_2\vphantom{\dfrac{a}{b}}\\ x'_3\vphantom{\dfrac{a}{b}}\\ x'_4\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} \Lambda_{11} & \Lambda_{12} & \Lambda_{13} & \Lambda_{14}\vphantom{\dfrac{a}{b}}\\ \Lambda_{21} & \Lambda_{22} & \Lambda_{23} & \Lambda_{24}\vphantom{\dfrac{a}{b}}\\ \Lambda_{31} & \Lambda_{32} & \Lambda_{33} & \Lambda_{34}\vphantom{\dfrac{a}{b}}\\ \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} x_1\vphantom{\dfrac{a}{b}}\\ x_2\vphantom{\dfrac{a}{b}}\\ x_3\vphantom{\dfrac{a}{b}}\\ x_4\vphantom{\dfrac{a}{b}} \end{bmatrix} =\Lambda\mathbf X\,, \quad \Lambda_{ij} \in \mathbb R \tag{A-03}\label{A-03}$$$$ is one by which the norm \eqref{A-02} remains invariant. This norm is expressed in a different way as
\begin{align} \Vert\mathbf{X}\Vert^2 &=- \begin{bmatrix} x_1 & x_2 & x_3 & x_4 \end{bmatrix} \begin{bmatrix} +1 & \hphantom{+}0& \hphantom{+}0& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & +1 & \hphantom{+}0 & \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & \hphantom{+}0 & +1& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 &\hphantom{+}0 &\hphantom{+}0 &-1\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} x_1\vphantom{\dfrac{a}{b}}\\ x_2\vphantom{\dfrac{a}{b}}\\ x_3\vphantom{\dfrac{a}{b}}\\ x_4\vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ & =- \begin{bmatrix} x_1\vphantom{\dfrac{a}{b}}\\ x_2\vphantom{\dfrac{a}{b}}\\ x_3\vphantom{\dfrac{a}{b}}\\ x_4\vphantom{\dfrac{a}{b}} \end{bmatrix}^{\boldsymbol{\top}} \begin{bmatrix} +1 & \hphantom{+}0& \hphantom{+}0& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & +1 & \hphantom{+}0 & \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & \hphantom{+}0 & +1& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 &\hphantom{+}0 &\hphantom{+}0 &-1\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} x_1\vphantom{\dfrac{a}{b}}\\ x_2\vphantom{\dfrac{a}{b}}\\ x_3\vphantom{\dfrac{a}{b}}\\ x_4\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-04}\label{A-04} \end{align} that is $$$$\Vert\mathbf X\Vert^2=-\mathbf X^{\boldsymbol{\top}} \eta\,\mathbf X \tag{A-05}\label{A-05}$$$$ where $$$$\eta= \begin{bmatrix} +1 & \hphantom{+}0& \hphantom{+}0& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & +1 & \hphantom{+}0 & \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & \hphantom{+}0 & +1& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 &\hphantom{+}0 &\hphantom{+}0 &-1\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-06}\label{A-06}$$$$ From the invariant space-time interval
\begin{align} &\Vert\mathbf X'\Vert^2=\Vert\mathbf X\Vert^2\:\Longrightarrow\:-\mathbf X'^{\boldsymbol{\top}} \eta\,\mathbf X'=-\mathbf X^{\boldsymbol{\top}} \eta\,\mathbf X\:\Longrightarrow \nonumber\\ & \left(\Lambda\mathbf X\right)^{\boldsymbol{\top}} \eta\,\left(\Lambda\mathbf X\right)=\mathbf X^{\boldsymbol{\top}} \eta\,\mathbf X\:\Longrightarrow\:\mathbf X^{\boldsymbol{\top}}\Lambda^{\boldsymbol{\top}}\eta\,\Lambda\mathbf X=\mathbf X^{\boldsymbol{\top}} \eta\,\mathbf X \nonumber \end{align} so for any $$\mathbf X$$ $$$$\mathbf X^{\boldsymbol{\top}}\left(\Lambda^{\boldsymbol{\top}}\eta\,\Lambda-\eta\right)\mathbf X=0 \nonumber$$$$ hence $$$$\boxed{\:\:\Lambda^{\boldsymbol{\top}}\eta\,\Lambda =\eta\vphantom{\dfrac{a}{b}}\:\:} \tag{A-07}\label{A-07}$$$$ written explicitly

$$$$\begin{bmatrix} \Lambda_{11} & \Lambda_{21} & \Lambda_{31} & \Lambda_{41}\vphantom{\dfrac{a}{b}}\\ \Lambda_{12} & \Lambda_{22} & \Lambda_{32} & \Lambda_{42}\vphantom{\dfrac{a}{b}}\\ \Lambda_{13} & \Lambda_{23} & \Lambda_{33} & \Lambda_{43}\vphantom{\dfrac{a}{b}}\\ \Lambda_{14} & \Lambda_{24} & \Lambda_{34} & \Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} +1 & \hphantom{+}0& \hphantom{+}0& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & +1 & \hphantom{+}0 & \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & \hphantom{+}0 & +1& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 &\hphantom{+}0 &\hphantom{+}0 &-1\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \Lambda_{11} & \Lambda_{12} & \Lambda_{13} & \Lambda_{14}\vphantom{\dfrac{a}{b}}\\ \Lambda_{21} & \Lambda_{22} & \Lambda_{23} & \Lambda_{24}\vphantom{\dfrac{a}{b}}\\ \Lambda_{31} & \Lambda_{32} & \Lambda_{33} & \Lambda_{34}\vphantom{\dfrac{a}{b}}\\ \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} +1 & \hphantom{+}0& \hphantom{+}0& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & +1 & \hphantom{+}0 & \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & \hphantom{+}0 & +1& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 &\hphantom{+}0 &\hphantom{+}0 &-1\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-08}\label{A-08}$$$$

or $$$$\begin{bmatrix} \Lambda_{11} & \Lambda_{21} & \Lambda_{31} & -\Lambda_{41}\vphantom{\dfrac{a}{b}}\\ \Lambda_{12} & \Lambda_{22} & \Lambda_{32} & -\Lambda_{42}\vphantom{\dfrac{a}{b}}\\ \Lambda_{13} & \Lambda_{23} & \Lambda_{33} & -\Lambda_{43}\vphantom{\dfrac{a}{b}}\\ \Lambda_{14} & \Lambda_{24} & \Lambda_{34} & -\Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \Lambda_{11} & \Lambda_{12} & \Lambda_{13} & \Lambda_{14}\vphantom{\dfrac{a}{b}}\\ \Lambda_{21} & \Lambda_{22} & \Lambda_{23} & \Lambda_{24}\vphantom{\dfrac{a}{b}}\\ \Lambda_{31} & \Lambda_{32} & \Lambda_{33} & \Lambda_{34}\vphantom{\dfrac{a}{b}}\\ \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} +1 & \hphantom{+}0& \hphantom{+}0& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & +1 & \hphantom{+}0 & \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & \hphantom{+}0 & +1& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 &\hphantom{+}0 &\hphantom{+}0 &-1\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-09}\label{A-09}$$$$
Hence the first three column 4-vectors of $$\Lambda$$ are space-like with norm squared $$+1$$ while the fourth column 4-vector is time-like with norm squared $$-1$$ as in the following 4 equations \begin{align} \texttt{column 1 vector : }\Lambda^2_{11}+\Lambda^2_{21} +\Lambda^2_{31}-\Lambda_{41}^2 & =+1 \tag{A-10.1}\label{A-10.1}\\ \texttt{column 2 vector : }\Lambda^2_{12}+\Lambda^2_{22} +\Lambda^2_{32}-\Lambda_{42}^2 & =+1 \tag{A-10.2}\label{A-10.2}\\ \texttt{column 3 vector : }\Lambda^2_{13}+\Lambda^2_{23} +\Lambda^2_{33}-\Lambda_{43}^2 & =+1 \tag{A-10.3}\label{A-10.3}\\ \texttt{column 4 vector : }\Lambda^2_{14}+\Lambda^2_{24} +\Lambda^2_{34}-\Lambda_{44}^2 & =-1 \tag{A-10.4}\label{A-10.4} \end{align} Also the four column 4-vectors are mutually pseudo-orthogonal to each other. These properties are expressed through the following 6 equations \begin{align} \texttt{column 1 by column 2 : }\Lambda_{11}\Lambda_{12}+\Lambda_{21}\Lambda_{22} +\Lambda_{31}\Lambda_{32}-\Lambda_{41}\Lambda_{42} & =0 \tag{A-10.5}\label{A-10.5}\\ \texttt{column 1 by column 3 : }\Lambda_{11}\Lambda_{13}+\Lambda_{21}\Lambda_{23} +\Lambda_{31}\Lambda_{33}-\Lambda_{41}\Lambda_{43} & =0 \tag{A-10.6}\label{A-10.6}\\ \texttt{column 1 by column 4 : }\Lambda_{11}\Lambda_{14}+\Lambda_{21}\Lambda_{24} +\Lambda_{31}\Lambda_{34}-\Lambda_{41}\Lambda_{44} & =0 \tag{A-10.7}\label{A-10.7}\\ \texttt{column 2 by column 3 : }\Lambda_{12}\Lambda_{13}+\Lambda_{22}\Lambda_{23} +\Lambda_{32}\Lambda_{33}-\Lambda_{42}\Lambda_{43} & =0 \tag{A-10.8}\label{A-10.8}\\ \texttt{column 2 by column 4 : }\Lambda_{12}\Lambda_{14}+\Lambda_{22}\Lambda_{24} +\Lambda_{32}\Lambda_{34}-\Lambda_{42}\Lambda_{44} & =0 \tag{A-10.9}\label{A-10.9}\\ \texttt{column 3 by column 4 : }\Lambda_{13}\Lambda_{14}+\Lambda_{23}\Lambda_{24} +\Lambda_{33}\Lambda_{34}-\Lambda_{43}\Lambda_{44} & =0 \tag{A-10.10}\label{A-10.10} \end{align} The 16 elements of $$\Lambda$$ must satisfy the 10 equations \eqref{A-10.1}-\eqref{A-10.10}. Hence, irrelevant here, the set of all transformations $$\Lambda$$ with property \eqref{A-07} must be 6-parametric.

From equation \eqref{A-07} $$$$\det\left(\Lambda^{\boldsymbol{\top}}\eta\,\Lambda\right) =\det\eta \quad \Longrightarrow \quad \det\Lambda^{\boldsymbol{\top}}\cdot\det\eta\cdot\det\Lambda=\det\eta\quad \Longrightarrow \nonumber$$$$ $$$$\left(\det\Lambda\right)^2=+1 \quad \texttt{or} \quad \det\Lambda=\pm 1 \tag{A-11}\label{A-11}$$$$

Now, the element $$\Lambda_{44}$$ is the factor relating the time variables $$t,t'$$ $$$$t'=\Lambda_{44}\,t \cdots \tag{A-12}\label{A-12}$$$$ To exclude the case of time inversion it's necessary this factor to be positive. But from \eqref{A-10.4} $$$$\Lambda_{44}^2=1+\left(\Lambda^2_{14}+\Lambda^2_{24}+\Lambda^2_{34}\right)\ge +1 \tag{A-13}\label{A-13}$$$$ So we must have $$$$\boxed{\:\:\Lambda_{44} \ge +1 \qquad \texttt{(orthochronus)}\vphantom{\dfrac{a}{b}}\:\:} \tag{A-14}\label{A-14}$$$$ On the other hand to exclude space inversion we must have from \eqref{A-11} $$$$\boxed{\:\:\det\Lambda=+1\qquad \texttt{(space inversion exclusion)}\vphantom{\dfrac{a}{b}}\:\:} \tag{A-15}\label{A-15}$$$$ An homogeneous Lorentz transformation that beyond condition \eqref{A-07} satisfies also the orthochronus condition \eqref{A-14} and the excluding space inversion condition \eqref{A-15} is called proper homogeneous Lorentz transformation.

Note also that applying $$\eta$$ on the two sides of \eqref{A-07} we have $$$$\eta\,\Lambda^{\boldsymbol{\top}}\eta\,\Lambda =\eta^2=\mathrm I\quad \Longrightarrow \quad \left(\eta\,\Lambda^{\boldsymbol{\top}}\eta\right)\Lambda =\mathrm I \nonumber$$$$ so $$$$\Lambda^{-1} =\eta\,\Lambda^{\boldsymbol{\top}}\eta \tag{A-16}\label{A-16}$$$$ explicitly $$$$\Lambda^{-1} = \begin{bmatrix} +1 & \hphantom{+}0& \hphantom{+}0& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & +1 & \hphantom{+}0 & \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & \hphantom{+}0 & +1& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 &\hphantom{+}0 &\hphantom{+}0 &-1\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \Lambda_{11} & \Lambda_{21} & \Lambda_{31} & \Lambda_{41}\vphantom{\dfrac{a}{b}}\\ \Lambda_{12} & \Lambda_{22} & \Lambda_{32} & \Lambda_{42}\vphantom{\dfrac{a}{b}}\\ \Lambda_{13} & \Lambda_{23} & \Lambda_{33} & \Lambda_{43}\vphantom{\dfrac{a}{b}}\\ \Lambda_{14} & \Lambda_{24} & \Lambda_{34} & \Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} +1 & \hphantom{+}0& \hphantom{+}0& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & +1 & \hphantom{+}0 & \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 & \hphantom{+}0 & +1& \hphantom{-}0\vphantom{\dfrac{a}{b}}\\ \hphantom{+}0 &\hphantom{+}0 &\hphantom{+}0 &-1\vphantom{\dfrac{a}{b}} \end{bmatrix}\quad \Longrightarrow \nonumber$$$$ $$$$\Lambda^{-1} = \begin{bmatrix} \hphantom{-}\Lambda_{11} & \hphantom{-}\Lambda_{21} & \hphantom{-}\Lambda_{31} & -\Lambda_{41}\vphantom{\dfrac{a}{b}}\\ \hphantom{-}\Lambda_{12} & \hphantom{-}\Lambda_{22} & \hphantom{-}\Lambda_{32} & -\Lambda_{42}\vphantom{\dfrac{a}{b}}\\ \hphantom{-}\Lambda_{13} & \hphantom{-}\Lambda_{23} & \hphantom{-}\Lambda_{33} & -\Lambda_{43}\vphantom{\dfrac{a}{b}}\\ -\Lambda_{14} & -\Lambda_{24} & -\Lambda_{34} & \hphantom{-}\Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-17}\label{A-17}$$$$ Note that $$\Lambda^{-1}$$ is also a proper homogeneous Lorentz transformation since

$$\boldsymbol\S$$ B. Lorentz boosts

A Lorentz boost with velocity $$\mathbf v$$ $$$$\mathbf v = \begin{bmatrix} v_1 \vphantom{\dfrac{a}{b}}\\ v_2 \vphantom{\dfrac{a}{b}}\\ v_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{B-01}\label{B-01}$$$$ has in general the following form $$$$\mathrm L\left(\mathbf{v}\right) = \begin{bmatrix} \mathrm I+\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \mathbf v ^{\boldsymbol{\top}} & -\dfrac{\gamma_v}{c}\mathbf v \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ -\dfrac{\gamma_v}{c}\mathbf v^{\boldsymbol{\top}} & \hphantom{-}\gamma_v \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{B-02}\label{B-02}$$$$ where $$$$\gamma_v=\left(1-\dfrac{v^2}{c^2}\right)^{-\frac12} \tag{B-03}\label{B-03}$$$$ Note that $$$$\mathbf v \mathbf v ^{\boldsymbol{\top}}= \begin{bmatrix} v_1 \vphantom{\dfrac{a}{b}}\\ v_2 \vphantom{\dfrac{a}{b}}\\ v_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} v_1 \vphantom{\dfrac{a}{b}}\\ v_2 \vphantom{\dfrac{a}{b}}\\ v_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} ^{\boldsymbol{\top}} = \begin{bmatrix} v_1 \vphantom{\dfrac{a}{b}}\\ v_2 \vphantom{\dfrac{a}{b}}\\ v_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} v_1 & v_2 & v_3\vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} v^2_1 & v_1v_2 & v_1v_3 \vphantom{\dfrac{a}{b}}\\ v_2v_1 & v^2_2 & v_2v_3 \vphantom{\dfrac{a}{b}}\\ v_3v_1 & v_3v_2 & v^2_3\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{B-04}\label{B-04}$$$$ A Lorentz boost is a special kind of proper homogeneous Lorentz transformation with two additional properties : it is symmetric with inverse $$$$\mathrm L^{-1}=\bigl[\mathrm L\left(\mathbf{v}\right)\bigr]^{-1} =\mathrm L\left(-\mathbf{v}\right) = \begin{bmatrix} \mathrm I+\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \mathbf v ^{\boldsymbol{\top}} & +\dfrac{\gamma_v}{c}\mathbf v \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ +\dfrac{\gamma_v}{c}\mathbf v^{\boldsymbol{\top}} & \hphantom{-}\gamma_v \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{B-05}\label{B-05}$$$$

$$\boldsymbol\S$$ C. The decomposition

The question is if a proper homogeneous Lorentz transformation

$$$$\Lambda = \begin{bmatrix} \Lambda_{11} & \Lambda_{12} & \Lambda_{13} & \Lambda_{14}\vphantom{\dfrac{a}{b}}\\ \Lambda_{21} & \Lambda_{22} & \Lambda_{23} & \Lambda_{24}\vphantom{\dfrac{a}{b}}\\ \Lambda_{31} & \Lambda_{32} & \Lambda_{33} & \Lambda_{34}\vphantom{\dfrac{a}{b}}\\ \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix}\quad \Lambda^{\boldsymbol{\top}}\eta\,\Lambda =\eta\,,\quad \Lambda_{44}\ge +1\,,\quad \det\Lambda=+1 \tag{C-01}\label{C-01}$$$$ could be decomposed in a space transformation $$\mathcal R$$ followed by a Lorentz boost $$\rm L$$ with velocity $$\mathbf v$$ $$$$\Lambda = \rm L\left(\mathbf{v}\right)\mathcal R \tag{C-02}\label{C-02}$$$$ Consider that the space transformation $$\mathcal R$$ is represented by the following $$4\times 4-$$matrix $$$$\mathcal R= \begin{bmatrix} \hphantom{=}\rm R\hphantom{^{\boldsymbol{\top}}} & \hphantom{====}\boldsymbol 0\hphantom{=} \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \hphantom{=}\boldsymbol 0^{\boldsymbol{\top}} & \hphantom{====}1\hphantom{=} \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{C-03}\label{C-03}$$$$
where $$$$\mathrm R = \begin{bmatrix} R_{11} & R_{12} & R_{13} \vphantom{\dfrac{a}{b}}\\ R_{21} & R_{22} & R_{23}\vphantom{\dfrac{a}{b}}\\ R_{31} & R_{32} & R_{33} \end{bmatrix}\qquad R_{ij} \in \mathbb R \tag{C-04}\label{C-04}$$$$ From equations \eqref{B-02},\eqref{C-03} we have \begin{align} \rm L\left(\mathbf{v}\right)\mathcal R & = \begin{bmatrix} \mathrm I+\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \mathbf v ^{\boldsymbol{\top}} & -\dfrac{\gamma_v}{c}\mathbf v \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ -\dfrac{\gamma_v}{c}\mathbf v^{\boldsymbol{\top}} & \hphantom{-}\gamma_v \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \hphantom{=}\rm R\hphantom{^{\boldsymbol{\top}}} & \hphantom{====}\boldsymbol 0\hphantom{=} \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \hphantom{=}\boldsymbol 0^{\boldsymbol{\top}} & \hphantom{====}1\hphantom{=} \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix}\quad \Longrightarrow \nonumber\\ \rm L\left(\mathbf{v}\right)\mathcal R & = \begin{bmatrix} \mathrm R+\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} & -\dfrac{\gamma_v}{c}\mathbf v \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ -\dfrac{\gamma_v}{c}\left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} & \hphantom{-}\gamma_v \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{C-05}\label{C-05} \end{align}

For decomposition \eqref{C-02} to be valid the following equality must be satisfied if we insert expressions \eqref{C-01},\eqref{C-05} for $$\Lambda$$ and $$\rm L\left(\mathbf{v}\right)\mathcal R$$ respectively
$$$$\underbrace{ \begin{bmatrix} \begin{array}{ccc|c} \Lambda_{11} & \Lambda_{12} & \Lambda_{13} & \Lambda_{14}\vphantom{\dfrac{a}{b}}\\ \Lambda_{21} & \Lambda_{22} & \Lambda_{23} & \Lambda_{24}\vphantom{\dfrac{a}{b}}\\ \Lambda_{31} & \Lambda_{32} & \Lambda_{33} & \Lambda_{34}\vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \hline \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{array} \end{bmatrix}}_{\Lambda} = \underbrace{ \begin{bmatrix} \begin{array}{ccc|c} & & & \vphantom{\dfrac{a}{b}}\\ & \mathrm R+\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} & & -\dfrac{\gamma_v}{c}\mathbf v\vphantom{\dfrac{a}{b}}\\ & & & \vphantom{\dfrac{a}{b}}\\ \hline & -\dfrac{\gamma_v}{c}\left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} & & \gamma_v \vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{array} \end{bmatrix}}_{\rm L\left(\mathbf{v}\right)\mathcal R} \tag{C-06}\label{C-06}$$$$ Equating the fourth column of the matrix in the left to the fourth column of the matrix in the right we determine the boost velocity $$\mathbf v$$ and the $$\gamma-$$factor in terms of the elements $$\Lambda_{ij}$$ $$$$\begin{bmatrix} \begin{array}{c} \Lambda_{14} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \hline \Lambda_{44}\vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c} \vphantom{\dfrac{a}{b}}\\ -\dfrac{\gamma_v}{c}\mathbf v \vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \hline \gamma_v\vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{array} \end{bmatrix}\qquad \Longrightarrow \nonumber$$$$

$$$$\mathbf v=\begin{bmatrix} v_1 \vphantom{\dfrac{a}{b}}\\ v_2 \vphantom{\dfrac{a}{b}}\\ v_3 \vphantom{\dfrac{a}{b}} \end{bmatrix}=-\dfrac{c}{\Lambda_{44}} \begin{bmatrix} \Lambda_{14} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34} \vphantom{\dfrac{a}{b}} \end{bmatrix}\quad \texttt{and} \quad \gamma_v = \Lambda_{44} \tag{C-07}\label{C-07}$$$$ Equation \eqref{C-07} is a great advantage because we can express the Lorentz boost $$\rm L\left(\mathbf{v}\right)$$ and its inverse $$\mathrm L^{-1}=\bigl[\mathrm L\left(\mathbf{v}\right)\bigr]^{-1} =\mathrm L\left(-\mathbf{v}\right)$$ in terms of the elements $$\Lambda_{ij}$$. Since from \eqref{C-02} $$$$\mathcal R=\bigl[\mathrm L\left(\mathbf{v}\right)\bigr]^{-1}\Lambda=\mathrm L\left(-\mathbf{v}\right)\Lambda \tag{C-08}\label{C-08}$$$$ above equation has in the lhs the unknown matrix $$\mathcal R$$ while the rhs is an expression in terms of the elements $$\Lambda_{ij}$$. So from above equation \eqref{C-08} we could determine completely the matrix $$\mathcal R$$ what we will do in the following.

To express the inverse of the Lorentz boost \eqref{B-05} in terms of the elements $$\Lambda_{ij}$$ we have, based on equation \eqref{C-07} \begin{align} \dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \mathbf v ^{\boldsymbol{\top}} & =\dfrac{\Lambda^2_{44}}{c^2 \left(\Lambda_{44}+1\right)} \left(-\dfrac{c}{\Lambda_{44}} \begin{bmatrix} \Lambda_{14} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34} \vphantom{\dfrac{a}{b}} \end{bmatrix}\right) \left(-\dfrac{c}{\Lambda_{44}} \begin{bmatrix} \Lambda_{14} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34} \vphantom{\dfrac{a}{b}} \end{bmatrix}\right) ^{\boldsymbol{\top}} \nonumber\\ & =\dfrac{1}{\Lambda_{44}+1} \begin{bmatrix} \Lambda_{14} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \Lambda_{14} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34} \vphantom{\dfrac{a}{b}} \end{bmatrix} ^{\boldsymbol{\top}}= \dfrac{1}{\Lambda_{44}+1} \begin{bmatrix} \Lambda_{14} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \Lambda_{14} & \Lambda_{24} & \Lambda_{34}\vphantom{\dfrac{a}{b}} \end{bmatrix} \: \Longrightarrow \nonumber \end{align} $$$$\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \mathbf v ^{\boldsymbol{\top}} = \dfrac{1}{\Lambda_{44}+1} \begin{bmatrix} \Lambda^2_{14} & \Lambda_{14}\Lambda_{24} & \Lambda_{14}\Lambda_{34} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24}\Lambda_{14} & \Lambda^2_{24} & \Lambda_{24}\Lambda_{34} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34}\Lambda_{14} & \Lambda_{34}\Lambda_{24} & \Lambda^2_{34}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{C-09}\label{C-09}$$$$ Hence from \eqref{B-05} $$$$\mathrm L\left(-\mathbf{v}\right)= \begin{bmatrix} 1+\dfrac{\Lambda^2_{14}}{\Lambda_{44}+1} & \dfrac{\Lambda_{14}\Lambda_{24}}{\Lambda_{44}+1} & \dfrac{\Lambda_{14}\Lambda_{34}}{\Lambda_{44}+1} & - \Lambda_{14}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\Lambda_{24}\Lambda_{14}}{\Lambda_{44}+1} & 1+\dfrac{\Lambda^2_{24}}{\Lambda_{44}+1} & \dfrac{\Lambda_{24}\Lambda_{34}}{\Lambda_{44}+1} & - \Lambda_{24}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\Lambda_{34}\Lambda_{14}}{\Lambda_{44}+1} & \dfrac{\Lambda_{34}\Lambda_{24}}{\Lambda_{44}+1} & 1+\dfrac{\Lambda^2_{34}}{\Lambda_{44}+1} & - \Lambda_{34}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ - \Lambda_{14} & - \Lambda_{24} & - \Lambda_{34} & \hphantom{-} \Lambda_{44}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{C-10}\label{C-10}$$$$ and equation \eqref{C-08} yields \begin{align} &\mathcal R = \begin{bmatrix} \:\: & & & \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\: & \mathrm R & & \boldsymbol 0 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\: & & & \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\:\:\:\:\: & \:\:\boldsymbol 0 ^{\boldsymbol{\top}}\:\: & \:\:\:\: & \:\:\:1\:\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}= \begin{bmatrix} \:\:R_{11} & R_{12} & R_{13} & 0 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\:R_{21} & R_{22} & R_{23} & 0 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\:R_{31} & R_{32} & R_{33} & 0 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\:\:\:0\:\: & \:\:0\:\: & \:\:0\:\: & \:\:\:1\:\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}= \tag{C-11}\label{C-11}\\ &\begin{bmatrix} 1+\dfrac{\Lambda^2_{14}}{\Lambda_{44}+1} & \dfrac{\Lambda_{14}\Lambda_{24}}{\Lambda_{44}+1} & \dfrac{\Lambda_{14}\Lambda_{34}}{\Lambda_{44}+1} & - \Lambda_{14}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\Lambda_{24}\Lambda_{14}}{\Lambda_{44}+1} & 1+\dfrac{\Lambda^2_{24}}{\Lambda_{44}+1} & \dfrac{\Lambda_{24}\Lambda_{34}}{\Lambda_{44}+1} & - \Lambda_{24}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\Lambda_{34}\Lambda_{14}}{\Lambda_{44}+1} & \dfrac{\Lambda_{34}\Lambda_{24}}{\Lambda_{44}+1} & 1+\dfrac{\Lambda^2_{34}}{\Lambda_{44}+1} & - \Lambda_{34}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ - \Lambda_{14} & - \Lambda_{24} & - \Lambda_{34} & \hphantom{-} \Lambda_{44}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \Lambda_{11} & \Lambda_{12} & \Lambda_{13} & \Lambda_{14}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{21} & \Lambda_{22} & \Lambda_{23} & \Lambda_{24}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{31} & \Lambda_{32} & \Lambda_{33} & \Lambda_{34}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \nonumber \end{align} Elaborating the matrix product of the rhs and using the properties \eqref{A-10.1}-\eqref{A-10.10} we have $$$$\mathrm R= \begin{bmatrix} R_{11} & R_{12} & R_{13} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{21} & R_{22} & R_{23}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{31} & R_{32} & R_{33} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}= \begin{bmatrix} \Lambda_{11}-\dfrac{\Lambda_{14}\Lambda_{41}}{\Lambda_{44}+1} & \Lambda_{12}-\dfrac{\Lambda_{14}\Lambda_{42}}{\Lambda_{44}+1} & \Lambda_{13}-\dfrac{\Lambda_{14}\Lambda_{43}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{21}-\dfrac{\Lambda_{24}\Lambda_{41}}{\Lambda_{44}+1} & \Lambda_{22}-\dfrac{\Lambda_{24}\Lambda_{42}}{\Lambda_{44}+1} & \Lambda_{23}-\dfrac{\Lambda_{24}\Lambda_{43}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{31}-\dfrac{\Lambda_{34}\Lambda_{41}}{\Lambda_{44}+1} & \Lambda_{32}-\dfrac{\Lambda_{34}\Lambda_{42}}{\Lambda_{44}+1} & \Lambda_{33}-\dfrac{\Lambda_{34}\Lambda_{43}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{C-12}\label{C-12}$$$$ or in an equation only $$$$\boxed{\:\:R_{ij} =\Lambda_{ij}-\dfrac{\Lambda_{i4}\Lambda_{4j}}{\Lambda_{44}+1}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{C-13}\label{C-13}$$$$

Proof that $$\rm R$$ represents a pure rotation and expressions for its axis and angle are given in ANSWER - Part II.

(to be continued in ANSWER - Part II)

It is not so readily seen, to be honest. It goes in the literature by the name "polar decomposition".

The shortest argument is the one by H. Urbantke "Elementary Proof of Moretti’s Polar Decomposition Theorem for Lorentz Transformations" (here), which is a not so strightforward simplification of prof. Valter Moretti's argument here. You need Sexl & Urbantke's famous book to follow Urbantke.

The following text is adapted from Florian Scheck's "Mechanics: From Newton's Laws to Deterministic Chaos (Sixth Edition)". Please refer to Section 4.5 of the book for a detailed proof.

Every restricted Lorentz transformation $$\Lambda \in L_{+}^\uparrow$$ can be written, in a unique way, as the product of a pure rotation followed by a pure boost,

$$\Lambda = B(\mathbf{v}) R,$$

where the parameters of the two transformations being given by

$$v^i/c = \frac{{\Lambda^i}_0}{{\Lambda^0}_0},\quad {R^i}_k = {\Lambda^i}_k - \frac{1}{1+{\Lambda^0}_0} {\Lambda^i}_0 {\Lambda^0}_k,$$

or as the product of a pure boost followed by a pure rotation,

$$\Lambda = R B(\mathbf{w}),$$

where the vector $$\mathbf{w}$$ is given by

$$w^i/c = \frac{{\Lambda^0}_i}{{\Lambda^0}_0},$$

and $$R$$ is the same rotation as above.

You may verify the above statement by direct calculation. You may also notice that $$\mathbf{v} = R\mathbf{w}$$. This is not surprising because

$$B(\mathbf{v}) = R B(\mathbf{w}) R^{-1} = B(R\mathbf{w}).$$