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.
 A: 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.
A: 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
$$ B(\mathbf{v}) = R B(\mathbf{w}) R^{-1} = B(R\mathbf{w}). $$
A: 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 $SO(3,1)$.
