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

I am trying to read Weinberg's book Gravitation and Cosmology. In which he derives the Lorentz transformation matrix for boost along arbitrary direction, (equations 2.1.20 and 2.1.21):

$$\Lambda^i_{\,\,j}=\delta_{ij}+v_i v_j\frac{\gamma-1}{\mathbf{v\cdot v}}$$ $$\Lambda^0_{\,\,j}=\gamma v_j$$

Immediately after that there is a statement, "It can be easily seen that any proper homogeneous Lorentz transformation may be expressed as the product of a boost times a rotation".

How to show that mathematically? It'd be better if someone answers using similar notations as used by Weinberg.

• This is implied by the Cartan-Dieudonné theorem (but I don't know of any easy proof of that theorem, so I can't turn this comment into a good answer). Oct 23, 2020 at 2:25
• I think the usual way is to consider infinitesimal Lorentz transformations. There are six generator matrices, three of which generate rotations and three generate boosts. I don’t understand why Weinberg says it us “easily seen”. Oct 23, 2020 at 5:21
• Another argument: Lorentz transformations preserve $x^2+y^2+z^2+(ict)^2$ and thus are “obviously” just 4D rotations in $x,y,z,ict$ space. The rotations in the $x-ict$, etc. planes can be shown to be boosts, via relationships between sin/cos of imaginary argument and sinh/cosh of real argument. Oct 23, 2020 at 5:27

ANSWER - Parts I & II

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

$$\texttt{Abstract}$$

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

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

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

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

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

$$\boldsymbol\S\texttt{ F. Figures for }\Lambda \texttt{ and decompositions }\rm L\left(\mathbf{v}\right)\mathcal R,\:\mathcal R\,L\left(\mathbf{u}\right)$$

# ANSWER - Part I

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$$.

$$\boldsymbol\S$$ 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)

• one does recognize your distinctive style of answer, my friend. +1. Feb 22, 2021 at 1:37

(continued from ANSWER - Part I)

# ANSWER - Part II

For the three column 3-vectors of the matrix $$\rm R$$ in \eqref{C-12} \begin{align} \boldsymbol \rho_1 & = \begin{bmatrix} R_{11} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{21} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{31} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}= \begin{bmatrix} \Lambda_{11}-\dfrac{\Lambda_{14}\Lambda_{41}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{21}-\dfrac{\Lambda_{24}\Lambda_{41}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{31}-\dfrac{\Lambda_{34}\Lambda_{41}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{C-14.1}\label{C-14.1}\\ \boldsymbol \rho_2 & = \begin{bmatrix} R_{12} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{22} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{32} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}= \begin{bmatrix} \Lambda_{12}-\dfrac{\Lambda_{14}\Lambda_{42}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{22}-\dfrac{\Lambda_{24}\Lambda_{42}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{32}-\dfrac{\Lambda_{34}\Lambda_{42}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{C-14.2}\label{C-14.2}\\ \boldsymbol \rho_3 & = \begin{bmatrix} R_{13} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{23} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{33} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}= \begin{bmatrix} \Lambda_{13}-\dfrac{\Lambda_{14}\Lambda_{43}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{23}-\dfrac{\Lambda_{24}\Lambda_{43}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{33}-\dfrac{\Lambda_{34}\Lambda_{43}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{C-14.3}\label{C-14.3} \end{align} we could prove using again the properties \eqref{A-10.1}-\eqref{A-10.10} that \begin{align} &\vert\boldsymbol\rho_1\vert^2 =\vert\boldsymbol \rho_2\vert^2=\vert\boldsymbol\rho_3\vert^2=1 \tag{C-15.1}\label{C-15.1}\\ &\boldsymbol\rho_1\boldsymbol\cdot\boldsymbol\rho_2 =\boldsymbol\rho_2\boldsymbol\cdot\boldsymbol\rho_3 =\boldsymbol\rho_3\boldsymbol\cdot\boldsymbol\rho_1 =0 \tag{C-15.2}\label{C-15.2}\\ & \boldsymbol\rho_1\boldsymbol\cdot\left(\boldsymbol\rho_2\boldsymbol\times\boldsymbol\rho_3\right) =\boldsymbol +1= \det\rm R \tag{C-15.3}\label{C-15.3} \end{align} So the $$3\times 3-$$matrix $$\rm R$$ in \eqref{C-12} represents a pure rotation and the $$4\times 4-$$matrix $$\mathcal R$$ in \eqref{C-03} is its 4-dimensional representation.

Now, if a $$3\times 3-$$matrix like $$\rm R$$ represents a pure rotation of the coordinate system by an angle $$\theta$$ around a unit vector $$\mathbf n=(n_1,n_2,n_3)$$ then

\begin{align} \mathrm R & = \begin{bmatrix} \cos\theta+(1-\cos\theta)n_1^2&(1-\cos\theta)n_1n_2+\sin\theta n_3&(1-\cos\theta)n_1n_3-\sin\theta n_2\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ (1-\cos\theta)n_2n_1-\sin\theta n_3&\cos\theta+(1-\cos\theta)n_2^2&(1-\cos\theta)n_2n_3+\sin\theta n_1\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ (1-\cos\theta)n_3n_1+\sin\theta n_2&(1-\cos\theta)n_3n_2-\sin\theta n_1&\cos\theta+(1-\cos\theta)n_3^2\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \nonumber \\ &= \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} \tag{C-16}\label{C-16} \end{align}

hence $$$$\cos\theta=\dfrac{\left(R_{11}+R_{22}+R_{33}\right)-1}{2}=\dfrac{\text{trace}(\mathrm R)-1}{2} \tag{C-17}\label{C-17}$$$$ and $$$$\sin\theta\, \mathbf n =\sin\theta \begin{bmatrix} \:\:n_1\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ n_2\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ n_3 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} =\frac{1}{2} \begin{bmatrix} R_{23}-R_{32} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{31}-R_{13} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ R_{12}-R_{21} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{C-18}\label{C-18}$$$$ Inserting from \eqref{C-12} the expressions of $$R_{ij}$$ in terms of the elements $$\Lambda_{ij}$$ we have $$$$\boxed{\:\:\cos\theta=\dfrac{\left(\Lambda_{11}+\Lambda_{22}+\Lambda_{33}-1\right)}{2}-\dfrac{\left(\Lambda_{14}\Lambda_{41}+\Lambda_{24}\Lambda_{42}+\Lambda_{34}\Lambda_{43}\right)}{2\left(\Lambda_{44}+1\right)}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{C-19}\label{C-19}$$$$ and $$$$\sin\theta\, \mathbf n =\sin\theta \begin{bmatrix} \:\:n_1\:\: \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ n_2\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ n_3 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} = \begin{bmatrix} \dfrac{\left(\Lambda_{23}-\Lambda_{32}\right)}{2}-\dfrac{\left(\Lambda_{24}\Lambda_{43}-\Lambda_{34}\Lambda_{42}\right)}{2\left(\Lambda_{44}+1\right)} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\left(\Lambda_{31}-\Lambda_{13}\right)}{2}-\dfrac{\left(\Lambda_{34}\Lambda_{41}-\Lambda_{14}\Lambda_{43}\right)}{2\left(\Lambda_{44}+1\right)} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\left(\Lambda_{12}-\Lambda_{21}\right)}{2}-\dfrac{\left(\Lambda_{14}\Lambda_{42}-\Lambda_{24}\Lambda_{41}\right)}{2\left(\Lambda_{44}+1\right)} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{C-20}\label{C-20}$$$$

$$\boldsymbol\S$$ D. The uniqueness of the decomposition

Suppose now that there exists another pair of Lorentz boost-rotation $$\left[\rm L\left(\mathbf{v}'\right),\mathcal R' \right]$$ decomposition of $$\Lambda$$

$$$$\Lambda = \rm L\left(\mathbf{v}'\right)\mathcal R' \tag{D-01}\label{D-01}$$$$ hence $$$$\rm L\left(\mathbf{v}'\right)\mathcal R' = \rm L\left(\mathbf{v}\right)\mathcal R \tag{D-02}\label{D-02}$$$$ Taking advantage of the block form \eqref{C-05} we must have $$$$\begin{bmatrix} \mathrm R'+\dfrac{\gamma^2_{v'}}{c^2 \left(\gamma_{v'}+1\right)}\mathbf v' \left(\mathrm R'^{\boldsymbol{\top}}\mathbf v'\right)^{\boldsymbol{\top}} & -\dfrac{\gamma_{v'}}{c}\mathbf v' \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ -\dfrac{\gamma_{v'}}{c}\left(\mathrm R'^{\boldsymbol{\top}}\mathbf v'\right)^{\boldsymbol{\top}} & \hphantom{-}\gamma_{v'} \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} = \begin{bmatrix} \mathrm R+\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} & -\dfrac{\gamma_v}{c}\mathbf v \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ -\dfrac{\gamma_v}{c}\left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} & \hphantom{-}\gamma_v \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{D-03}\label{D-03}$$$$ This induces the following 4 equations \begin{align} \gamma_{v'}& =\gamma_{v} \tag{D-04a}\label{D-04a}\\ -\dfrac{\gamma_{v'}}{c}\mathbf v'& = -\dfrac{\gamma_{v}}{c}\mathbf v \tag{D-04b}\label{D-04b}\\ -\dfrac{\gamma_{v'}}{c}\left(\mathrm R'^{\boldsymbol{\top}}\mathbf v'\right)^{\boldsymbol{\top}} & = -\dfrac{\gamma_{v}}{c}\left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} \tag{D-04c}\label{D-04c}\\ \mathrm R'+\dfrac{\gamma^2_{v'}}{c^2 \left(\gamma_{v'}+1\right)}\mathbf v' \left(\mathrm R'^{\boldsymbol{\top}}\mathbf v'\right)^{\boldsymbol{\top}} & = \mathrm R+\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\mathbf v \left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} \tag{D-04d}\label{D-04d} \end{align} From equations \eqref{D-04a},\eqref{D-04b} we have $$$$\boxed{\:\:\mathbf v'=\mathbf v\vphantom{\dfrac{a}{b}}\:\:} \tag{D-05}\label{D-05}$$$$ If we think equation \eqref{D-04b} as between one column $$3\times 1-$$matrices and equation \eqref{D-04c} as between one row $$1\times 3-$$matrices then matrix multiplication side-by-side yields the equation below between $$3\times 3-$$matrices $$$$\dfrac{\gamma^2_{v'}}{c^2 }\mathbf v' \left(\mathrm R'^{\boldsymbol{\top}}\mathbf v'\right)^{\boldsymbol{\top}} = \dfrac{\gamma^2_v}{c^2}\mathbf v \left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} \tag{D-06}\label{D-06}$$$$ hence $$$$\dfrac{\gamma^2_{v'}}{c^2\left(\gamma_{v'}+1\right) }\mathbf v' \left(\mathrm R'^{\boldsymbol{\top}}\mathbf v'\right)^{\boldsymbol{\top}} = \dfrac{\gamma^2_v}{c^2\left(\gamma_v+1\right)}\mathbf v \left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} \tag{D-07}\label{D-07}$$$$ From equations \eqref{D-04d},\eqref{D-07} we have $$\rm R'= R$$ so $$$$\boxed{\:\:\mathcal R'=\mathcal R\vphantom{\dfrac{a}{b}}\:\:} \tag{D-08}\label{D-08}$$$$ Equations \eqref{D-05} and \eqref{D-08} prove the uniqueness of the decomposition.

$$\boldsymbol\S$$ E. Decomposition with Lorentz boost first

An equivalent question would be if a proper homogeneous Lorentz transformation as in equation \eqref{C-01}, repeated here for convenience

$$$$\Lambda = \begin{bmatrix} \Lambda_{11} & \Lambda_{12} & \Lambda_{13} & \Lambda_{14}\vphantom{\dfrac{a}{b}}\\ \Lambda_{21} & \Lambda_{22} & \Lambda_{23} & \Lambda_{24}\vphantom{\dfrac{a}{b}}\\ \Lambda_{31} & \Lambda_{32} & \Lambda_{33} & \Lambda_{34}\vphantom{\dfrac{a}{b}}\\ \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{a}{b}} \end{bmatrix}\quad \Lambda^{\boldsymbol{\top}}\eta\,\Lambda =\eta\,,\quad \Lambda_{44}\ge +1\,,\quad \det\Lambda=+1 \tag{E-01}\label{E-01}$$$$ could be decomposed inversely in a Lorentz boost $$\rm L$$ with velocity $$\mathbf u$$ followed by a space transformation $$\mathcal Q$$ $$$$\Lambda =\mathcal Q\, \rm L\left(\mathbf u\right) \tag{E-02}\label{E-02}$$$$ Consider that the space transformation $$\mathcal Q$$ is represented by the following $$4\times 4-$$matrix $$$$\mathcal Q= \begin{bmatrix} \hphantom{=}\rm Q\hphantom{^{\boldsymbol{\top}}} & \hphantom{====}\boldsymbol 0\hphantom{=} \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \hphantom{=}\boldsymbol 0^{\boldsymbol{\top}} & \hphantom{====}1\hphantom{=} \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{E-03}\label{E-03}$$$$
where $$$$\mathrm Q = \begin{bmatrix} Q_{11} & Q_{12} & Q_{13} \vphantom{\dfrac{a}{b}}\\ Q_{21} & Q_{22} & Q_{23}\vphantom{\dfrac{a}{b}}\\ Q_{31} & Q_{32} & Q_{33} \end{bmatrix}\qquad Q_{ij} \in \mathbb R \tag{E-04}\label{E-04}$$$$ By processing the problem as in $$\boldsymbol\S$$ C we have \begin{align} \mathcal Q\, \rm L\left(\mathbf u\right) & = \begin{bmatrix} \hphantom{=}\rm Q\hphantom{^{\boldsymbol{\top}}} & \hphantom{====}\boldsymbol 0\hphantom{=} \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \hphantom{=}\boldsymbol 0^{\boldsymbol{\top}} & \hphantom{====}1\hphantom{=} \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \mathrm I+\dfrac{\gamma^2_u}{c^2 \left(\gamma_u+1\right)}\mathbf u \mathbf u ^{\boldsymbol{\top}} & -\dfrac{\gamma_u}{c}\mathbf u \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ -\dfrac{\gamma_u}{c}\mathbf u^{\boldsymbol{\top}} & \hphantom{-}\gamma_u \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \quad \Longrightarrow \nonumber\\ \mathcal Q\, \rm L\left(\mathbf u\right) & = \begin{bmatrix} \mathrm Q+\dfrac{\gamma^2_u}{c^2 \left(\gamma_u+1\right)}\left(\mathrm Q\mathbf u\right)\mathbf u^{\boldsymbol{\top}} & -\dfrac{\gamma_u}{c}\mathrm Q\mathbf u \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ -\dfrac{\gamma_u}{c}\mathbf u^{\boldsymbol{\top}} & \hphantom{-}\gamma_u \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{E-05}\label{E-05} \end{align} and $$$$\underbrace{ \begin{bmatrix} \begin{array}{ccc|c} \Lambda_{11} & \Lambda_{12} & \Lambda_{13} & \Lambda_{14}\vphantom{\dfrac{a}{b}}\\ \Lambda_{21} & \Lambda_{22} & \Lambda_{23} & \Lambda_{24}\vphantom{\dfrac{a}{b}}\\ \Lambda_{31} & \Lambda_{32} & \Lambda_{33} & \Lambda_{34}\vphantom{\dfrac{a}{\tfrac{a}{b}}}\\ \hline \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{array} \end{bmatrix}}_{\Lambda} = \underbrace{ \begin{bmatrix} \begin{array}{ccc|c} & & & \vphantom{\dfrac{a}{b}}\\ & \mathrm Q+\dfrac{\gamma^2_u}{c^2 \left(\gamma_u+1\right)}\left(\mathrm Q\mathbf u\right)\mathbf u^{\boldsymbol{\top}} & & -\dfrac{\gamma_u}{c}\mathrm Q\mathbf u\vphantom{\dfrac{a}{b}}\\ & & & \vphantom{\dfrac{a}{b}}\\ \hline & -\dfrac{\gamma_u}{c}\mathbf u^{\boldsymbol{\top}} & & \gamma_u \vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{array} \end{bmatrix}}_{\mathcal Q\rm L\left(\mathbf u\right)} \tag{E-06}\label{E-06}$$$$ Equating the fourth row of the matrix in the left to the fourth row of the matrix in the right we determine the boost velocity $$\mathbf u$$ and the $$\gamma-$$factor in terms of the elements $$\Lambda_{ij}$$ $$$$\begin{bmatrix} \begin{array}{ccc|c} \Lambda_{41} & \Lambda_{42} & \Lambda_{43} & \Lambda_{44}\vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{ccc|c} \hphantom{====}& -\dfrac{\gamma_u}{c}\mathbf u^{\boldsymbol{\top}} & \hphantom{====} & \hphantom{==}\gamma_u \hphantom{==}\vphantom{\dfrac{\tfrac{a}{b}}{b}} \end{array} \end{bmatrix}\qquad \Longrightarrow \nonumber$$$$

$$$$\mathbf u=\begin{bmatrix} u_1 \vphantom{\dfrac{a}{b}}\\ u_2 \vphantom{\dfrac{a}{b}}\\ u_3 \vphantom{\dfrac{a}{b}} \end{bmatrix}=-\dfrac{c}{\Lambda_{44}} \begin{bmatrix} \Lambda_{41} \vphantom{\dfrac{a}{b}}\\ \Lambda_{42} \vphantom{\dfrac{a}{b}}\\ \Lambda_{43} \vphantom{\dfrac{a}{b}} \end{bmatrix}\quad \texttt{and} \quad \gamma_u = \Lambda_{44} \tag{E-07}\label{E-07}$$$$ Now it follows without any further explanation the set of equations \eqref{E-08}-\eqref{E-13} corresponding to \eqref{C-08}-\eqref{C-13}.

At first $$$$\mathcal Q= \Lambda\bigl[\mathrm L\left(\mathbf u\right)\bigr]^{-1}=\Lambda\,\mathrm L\left(-\mathbf u\right) \tag{E-08}\label{E-08}$$$$ The inverse of the Lorentz boost is $$$$\bigl[\mathrm L\left(\mathbf u\right)\bigr]^{-1}=\mathrm L\left(-\mathbf u\right)= \begin{bmatrix} \mathrm I+\dfrac{\gamma^2_u}{c^2 \left(\gamma_u+1\right)}\mathbf u \mathbf u ^{\boldsymbol{\top}} & +\dfrac{\gamma_u}{c}\mathbf u \vphantom{\dfrac{\gamma}{c}\boldsymbol{\upsilon}^{\boldsymbol{\top}}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ +\dfrac{\gamma_u}{c}\mathbf u^{\boldsymbol{\top}} & \hphantom{-}\gamma_u \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{E-08a}\label{E-08a}$$$$ To express it in terms of the elements $$\Lambda_{ij}$$ we have, based on equation \eqref{E-07} \begin{align} \dfrac{\gamma^2_u}{c^2 \left(\gamma_u+1\right)}\mathbf u \mathbf u ^{\boldsymbol{\top}} & =\dfrac{\Lambda^2_{44}}{c^2 \left(\Lambda_{44}+1\right)} \left(-\dfrac{c}{\Lambda_{44}} \begin{bmatrix} \Lambda_{41} \vphantom{\dfrac{a}{b}}\\ \Lambda_{42} \vphantom{\dfrac{a}{b}}\\ \Lambda_{43} \vphantom{\dfrac{a}{b}} \end{bmatrix}\right) \left(-\dfrac{c}{\Lambda_{44}} \begin{bmatrix} \Lambda_{41} \vphantom{\dfrac{a}{b}}\\ \Lambda_{42} \vphantom{\dfrac{a}{b}}\\ \Lambda_{43} \vphantom{\dfrac{a}{b}} \end{bmatrix}\right) ^{\boldsymbol{\top}} \nonumber\\ & =\dfrac{1}{\Lambda_{44}+1} \begin{bmatrix} \Lambda_{41} \vphantom{\dfrac{a}{b}}\\ \Lambda_{42} \vphantom{\dfrac{a}{b}}\\ \Lambda_{43} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \Lambda_{41} \vphantom{\dfrac{a}{b}}\\ \Lambda_{42} \vphantom{\dfrac{a}{b}}\\ \Lambda_{43} \vphantom{\dfrac{a}{b}} \end{bmatrix}^{\boldsymbol{\top}}= \dfrac{1}{\Lambda_{44}+1} \begin{bmatrix} \Lambda_{41} \vphantom{\dfrac{a}{b}}\\ \Lambda_{42} \vphantom{\dfrac{a}{b}}\\ \Lambda_{43} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \Lambda_{41} & \Lambda_{42} & \Lambda_{43}\vphantom{\dfrac{a}{b}} \end{bmatrix} \: \Longrightarrow \nonumber \end{align} $$$$\dfrac{\gamma^2_u}{c^2 \left(\gamma_u+1\right)}\mathbf u \mathbf u ^{\boldsymbol{\top}} = \dfrac{1}{\Lambda_{44}+1} \begin{bmatrix} \Lambda^2_{41} & \Lambda_{41}\Lambda_{42} & \Lambda_{41}\Lambda_{43} \vphantom{\dfrac{a}{b}}\\ \Lambda_{42}\Lambda_{41} & \Lambda^2_{42} & \Lambda_{42}\Lambda_{43} \vphantom{\dfrac{a}{b}}\\ \Lambda_{43}\Lambda_{41} & \Lambda_{43}\Lambda_{42} & \Lambda^2_{43}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{E-09}\label{E-09}$$$$ hence $$$$\mathrm L\left(-\mathbf u\right)= \begin{bmatrix} 1+\dfrac{\Lambda^2_{41}}{\Lambda_{44}+1} & \dfrac{\Lambda_{41}\Lambda_{42}}{\Lambda_{44}+1} & \dfrac{\Lambda_{41}\Lambda_{43}}{\Lambda_{44}+1} & - \Lambda_{41}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\Lambda_{42}\Lambda_{41}}{\Lambda_{44}+1} & 1+\dfrac{\Lambda^2_{42}}{\Lambda_{44}+1} & \dfrac{\Lambda_{42}\Lambda_{43}}{\Lambda_{44}+1} & - \Lambda_{42}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\Lambda_{43}\Lambda_{41}}{\Lambda_{44}+1} & \dfrac{\Lambda_{43}\Lambda_{42}}{\Lambda_{44}+1} & 1+\dfrac{\Lambda^2_{43}}{\Lambda_{44}+1} & - \Lambda_{43}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ - \Lambda_{41} & - \Lambda_{42} & - \Lambda_{43} & \hphantom{-} \Lambda_{44}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{E-10}\label{E-10}$$$$ and equation \eqref{E-08} yields \begin{align} &\mathcal Q = \begin{bmatrix} \:\: & & & \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\: & \mathrm Q & & \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} \:\:Q_{11} & Q_{12} & Q_{13} & 0 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\:Q_{21} & Q_{22} & Q_{23} & 0 \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \:\:Q_{31} & Q_{32} & Q_{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{E-11}\label{E-11}\\ &\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} \begin{bmatrix} 1+\dfrac{\Lambda^2_{41}}{\Lambda_{44}+1} & \dfrac{\Lambda_{41}\Lambda_{42}}{\Lambda_{44}+1} & \dfrac{\Lambda_{41}\Lambda_{43}}{\Lambda_{44}+1} & - \Lambda_{41}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\Lambda_{42}\Lambda_{41}}{\Lambda_{44}+1} & 1+\dfrac{\Lambda^2_{42}}{\Lambda_{44}+1} & \dfrac{\Lambda_{42}\Lambda_{43}}{\Lambda_{44}+1} & - \Lambda_{42}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \dfrac{\Lambda_{43}\Lambda_{41}}{\Lambda_{44}+1} & \dfrac{\Lambda_{43}\Lambda_{42}}{\Lambda_{44}+1} & 1+\dfrac{\Lambda^2_{43}}{\Lambda_{44}+1} & - \Lambda_{43}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ - \Lambda_{41} & - \Lambda_{42} & - \Lambda_{43} & \hphantom{-} \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
$$$$\mathrm Q= \begin{bmatrix} Q_{11} & Q_{12} & Q_{13} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ Q_{21} & Q_{22} & Q_{23}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ Q_{31} & Q_{32} & Q_{33} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix}= \begin{bmatrix} \Lambda_{11}-\dfrac{\Lambda_{41}\Lambda_{14}}{\Lambda_{44}+1} & \Lambda_{12}-\dfrac{\Lambda_{42}\Lambda_{14}}{\Lambda_{44}+1} & \Lambda_{13}-\dfrac{\Lambda_{43}\Lambda_{14}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{21}-\dfrac{\Lambda_{41}\Lambda_{24}}{\Lambda_{44}+1} & \Lambda_{22}-\dfrac{\Lambda_{42}\Lambda_{24}}{\Lambda_{44}+1} & \Lambda_{23}-\dfrac{\Lambda_{43}\Lambda_{24}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \Lambda_{31}-\dfrac{\Lambda_{41}\Lambda_{34}}{\Lambda_{44}+1} & \Lambda_{32}-\dfrac{\Lambda_{42}\Lambda_{34}}{\Lambda_{44}+1} & \Lambda_{33}-\dfrac{\Lambda_{43}\Lambda_{34}}{\Lambda_{44}+1} \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}} \end{bmatrix} \tag{E-12}\label{E-12}$$$$
or by one stroke $$$$\boxed{\:\:Q_{ij} =\Lambda_{ij}-\dfrac{\Lambda_{4i}\Lambda_{j4}}{\Lambda_{44}+1}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{E-13}\label{E-13}$$$$ Comparing this result with \eqref{C-13} we conclude that the rotations of the two decompositions are identical
$$$$\boxed{\:\:\ \rm L\left(\mathbf v\right)\mathcal R=\Lambda =\mathcal Q\, \rm L\left(\mathbf u\right) \quad\implies\quad \mathrm Q\equiv\mathrm R\,,\,\mathcal Q\equiv\mathcal R \vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:\:\:} \tag{E-14}\label{E-14}$$$$

The uniqueness of the decomposition $$\Lambda =\mathcal Q\, \rm L\left(\mathbf u\right)$$ could be proved following similar steps as in $$\boldsymbol\S$$ D for the uniqueness of the decomposition $$\Lambda = \rm L\left(\mathbf v\right)\mathcal R$$.

Given the uniqueness of the decompositions the following equation \eqref{E-16} proves not only by a different path the equality of the two rotations \eqref{E-14} but moreover uncover this relation between the velocities $$\mathbf v,\mathbf u$$ of the Lorentz boosts $$$$\boxed{\:\: \mathbf v = \mathrm R\, \mathbf u\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\:\:} \tag{E-15}\label{E-15}$$$$ Indeed \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}\qquad \stackrel{\rm R\rm R^{\boldsymbol{\top}}=\rm I}{=\!=\!=\!\Longrightarrow} \nonumber\\ \rm L\left(\mathbf{v}\right)\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} \begin{bmatrix} \mathrm I+\dfrac{\gamma^2_v}{c^2 \left(\gamma_v+1\right)}\left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right) \left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right)^{\boldsymbol{\top}} & -\dfrac{\gamma_v}{c}\left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right) \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} \nonumber\\ & = \mathcal R\, \rm L\left(\mathrm R^{\boldsymbol{\top}}\mathbf v\right) \tag{E-16}\label{E-16} \end{align} so $$\mathbf u = \mathrm R^{\boldsymbol{\top}}\mathbf v= \mathrm R^{-1}\mathbf v$$ proving equation \eqref{E-15}.

$$\boldsymbol\S$$ F. Figures for $$\Lambda$$ and decompositions $$\rm L\left(\mathbf{v}\right)\mathcal R,\:\mathcal R\,L\left(\mathbf{u}\right)$$

In Figure-01 it's shown a proper homogeneous transformation $$\:\Lambda\:$$ from an inertial system $$\:\mathbf S\:$$ to another one $$\:\mathbf S'''$$. The system $$\:\mathbf S'''\:$$ is moving with respect to $$\:\mathbf S\:$$ with velocity $$\:\mathbf u\:$$ as in equation \eqref{E-07} $$$$\mathbf u=\begin{bmatrix} u_1 \vphantom{\dfrac{a}{b}}\\ u_2 \vphantom{\dfrac{a}{b}}\\ u_3 \vphantom{\dfrac{a}{b}} \end{bmatrix}=-\dfrac{c}{\Lambda_{44}} \begin{bmatrix} \Lambda_{41} \vphantom{\dfrac{a}{b}}\\ \Lambda_{42} \vphantom{\dfrac{a}{b}}\\ \Lambda_{43} \vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber$$$$ Note that the vector $$\:\mathbf u\:$$ is the representation of this velocity with respect to the system $$\:\mathbf S$$.

In Figure-02 it's shown the decomposition \eqref{C-02}, $$\Lambda = \rm L\left(\mathbf{v}\right)\mathcal R$$, of the proper homogeneous transformation $$\:\Lambda\:$$ of Figure-01. First the system $$\:\mathbf S\:$$ is rotated by $$\:\mathcal R\:$$ to $$\:\mathbf S'$$ . After that the system $$\:\mathbf S'$$ is transformed by the Lorentz boost $$\:\rm L\left(\mathbf{v}\right)\:$$ to the system $$\:\mathbf S'''$$. The velocity $$\:\mathbf v\:$$ is given by equation \eqref{C-07} $$$$\mathbf v=\begin{bmatrix} v_1 \vphantom{\dfrac{a}{b}}\\ v_2 \vphantom{\dfrac{a}{b}}\\ v_3 \vphantom{\dfrac{a}{b}} \end{bmatrix}=-\dfrac{c}{\Lambda_{44}} \begin{bmatrix} \Lambda_{14} \vphantom{\dfrac{a}{b}}\\ \Lambda_{24} \vphantom{\dfrac{a}{b}}\\ \Lambda_{34} \vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber$$$$

Note that the vector $$\:\mathbf v\:$$ is the representation of this velocity with respect to the system $$\:\mathbf S'$$.

In Figure-03 it's shown the decomposition $$\Lambda = \mathcal R\:\rm L\left(\mathbf{u}\right)$$ of the proper homogeneous transformation $$\:\Lambda\:$$ of Figure-01. First the system $$\:\mathbf S\:$$ is transformed by the Lorentz boost $$\:\rm L\left(\mathbf{u}\right)\:$$ to $$\:\mathbf S''$$. After that the system $$\:\mathbf S''$$ is rotated by $$\:\mathcal R\:$$ to $$\:\mathbf S'''$$.

Note that the vectors $$\:\mathbf u\:$$ and $$\:\mathbf v\:$$ are both represent one and the same velocity : that of the system $$\:\mathbf S'''$$ with respect to system $$\:\mathbf S\:$$. Simply, the components of $$\:\mathbf u\:$$ are those with respect to $$\:\mathbf S\:$$ while the components of $$\:\mathbf v\:$$ are those with respect to $$\:\mathbf S'$$.

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.

• You see, that is the problem! Weinberg keeps on making these cryptic comments and that scares the newbies like me. Oct 28, 2020 at 3:43
• Just a comment, since you quoted that old paper of mine. The decomposition theorem has actually a proof independent of the polar decomposition procedure. Such direct proof appears in Urbantke’s paper and, I think, also in his book, and in Sean’s answer below. My observation just concerned the fact that this decomposition is also the polar one (in spite of an indefinite natural metric) and that it can be extended to infinite dimensional spaces in view of this connection. Feb 21, 2021 at 8:20

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

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

I'd like to try a rather intuitive (thus not rigorous) approach. Let's assume that the homogeneous proper Lorentz transformations are defined as 4x4 transformation matrices which leave the Minkowski metric tensor invariant, that is,

$$\Lambda^\alpha_\beta\eta_{\alpha\gamma}\Lambda^\gamma_\delta = \eta_{\gamma\delta}$$

The expressions on each side of the equation are symmetric under switching indices ($$\gamma\leftrightarrow\delta$$). They have 4 diagonal and 2*6 off-diagonal components, but only 10=4+6 of them are independent due to symmetry (the 6 components above the diagonal are always equal to the 6 components below the diagonal).

Therefore, we know that the 4x4 matrix $$\Lambda$$ has 4²=16 components which are subject to 10 constraints, giving us 16-10=6 independent parameters which characterize $$\Lambda$$.

So we know that the group $$SO(3,1)$$ is 6-dimensional, that is, it has 6 generators. Now we can make the ansatz that infinitesimal boosts along each of the three spatial axes and infinitesimal rotations in each of the three spatial planes are exactly the 6 generators of $$SO(3,1)$$.

So we need to show that they indeed leave the Minkowski metric tensor invariant and that they are linearly independent. It's easy to see that rotation leave the Minkowski metric tensor invariant, and boosts can be written as pseudorotations which also allows to easily see that they leave they leave the Minkowski metric tensor invariant (I can give details if necessary). Intuitively, it is also clear that the set of the six generators is linearly independent. Thus, they indeed generate the whole $$SO(3,1)$$.