I am studying relativity and come across a statement:
$g=\begin{bmatrix} 1&0&0&0 \\ 0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}$
If L satisfies these condition:
1) $L^{-1}=gLg^{T}$
2) $L_{11}>0$ (the most upper left element)
3) $\det(L)=1$
Then there exists rotation matrix H,Q and a real number $\phi$ such that
$L = \begin{bmatrix} 1 &0 \\ 0&H\end{bmatrix} \begin{bmatrix} \cosh\phi &\sinh\phi&0&0 \\ \sinh\phi&\cosh\phi&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} \begin{bmatrix} 1 &0 \\ 0&Q\end{bmatrix}$
Since the matrix $L$ is a normal operator under the scalar product $g(\mathbf{x},\mathbf{y})=\mathbf{x}^Tg\mathbf{y}$.
I know we can diagonalize $L$ by some orthogonormal basis (with respect to g).
Then I was trying to obtain a orthogonal basis in the $R^3$ subspace of $R^4$ with respect to the ordinary dot product but in no vain.
Any help would be appreciate.
The conditions on $L$ identify it as a proper orthochronous Lorentz transformation. It’s a standard result that such transformations can be uniquely decomposed into $BR$, where $R$ is a pure rotation about some axis $\vec k$ and $B$ is a pure boost in some direction $\vec n$. The matrix of a boost in the direction of $\vec e_1$ has the form $$B(\vec e_1,\chi) = \begin{bmatrix}\cosh\chi & \sinh\chi & 0 & 0 \\ \sinh\chi & \cosh\chi & 0 & 0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix},$$ so apply a pure rotation $U$ that aligns $B$’s boost direction with $\vec e_1$ to get $$L = BR = U^T B(\vec e_1,\chi) U R$$ from which $$\begin{bmatrix} 1 & 0 \\ 0 & H \end{bmatrix} = U^T$$ and $$\begin{bmatrix}1 & 0 \\ 0 & Q\end{bmatrix} = UR.$$ You can find a complete proof of this decomposition in Arthur Jaffe’s notes on Lorentz transformations, among other places. I’ll briefly sketch his development here.
We can identify spacetime coordinate vectors with $2\times2$ Hermitian matrices—elements of $\mathbf H_2$—as follows: $$x = (x_0,\vec x) = (x_0,x_1,x_2,x_3) \leftrightarrow \hat x = \begin{bmatrix}x_0+x_3 & x_1 - i x_2 \\ x_1 + i x_2 & x_0-x_3 \end{bmatrix}.$$ With this mapping, we have the Minkowski length $x_M^2 = x_0^2-x_1^2-x_2^2-x_3^2 = \det{\hat x}$. We also define the scalar product $\langle A,B\rangle = \frac12\operatorname{tr}(A^*B)$. The matrices $$\sigma_0 = I \\ \sigma_1 = \begin{bmatrix}0&1\\1&0\end{bmatrix} \\ \sigma_2 = \begin{bmatrix}0&-i\\i&0\end{bmatrix} \\ \sigma_3 = \begin{bmatrix}1&0\\0&-1\end{bmatrix},$$ the images of the four standard basis vectors of $\mathbb R^4$, form an orthonormal basis for $\mathbf H_2$. (You might recognize the latter three as the $2\times2$ Pauli matrices.) For convenience, we define $\sigma = (I,\vec\sigma) = (I,\sigma_1,\sigma_2,\sigma_3)$, and so $\hat x = x\cdot\sigma$.
Now, if $A\in SL(2,\mathbb C)$, then the linear transformation $\bar x \mapsto A\hat x A^*$ preserves $\det{\hat x}$. In fact every endomorphism of $\mathbf H_2$ can be represented in this way, and this representation is unique up to sign (both $A$ and $-A$ produce the same transformation). We can therefore represent proper orthochronous Lorentz transformations with elements of $SL(2,\mathbb C)$ via $$\widehat{Lx} = A\hat x A^*.$$ The Lorentz transformation that corresponds to $A$ has elements given by $$L_{\mu\nu} = \langle \sigma_\mu, A \sigma_\nu A^* \rangle.$$
Finally, every $A\in SL(2,\mathbb C)$ has a unique polar decomposition $A=HU$, where $U$ is a unitary matrix and $H$ is a strictly positive Hermitian matrix: take $H = (AA^*)^{1/2}$ and $U=H^{-1}A$. It’s straightforward to verify that unitary matrices correspond to pure rotations and positive Hermitian matrices to pure boosts, which gives you the decomposition at the beginning of this answer. Moreover, every unitary matrix can be expressed as an exponential $$U = \exp\left(-i\frac\theta2 \vec n\cdot\vec\sigma \right),$$ which corresponds to a pure rotation through an angle of $\theta$ about the axis $\vec n$, and a strictly-positive Hermitian matrix can also be expressed as an exponential: $$H = \exp\left(\frac12 \vec n\cdot\vec\sigma \chi \right),$$ which corresponds to a pure boost in the direction of $\vec n$ with parameter $\chi$. Working in the other direction to compute this decomposition for a given Lorentz transformation $L$ is a bit more involved, although $L_{00}=\cosh\chi$ can give you a starting point.