Special Relativity: Verifying a general boost matrix is in the Lorentz group I'm attempting the problem shown below. Using the hint, I have so far found:
\begin{align}B^T \eta B &= 
\begin{pmatrix}\gamma & -\gamma\beta^j \\
-\gamma \beta_k & \delta_k^j+\frac{(\gamma -1)\beta^j \beta_k}{\beta^2}\end{pmatrix}.
\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}.
\begin{pmatrix}\gamma & -\gamma \beta_n\\
-\gamma\beta^m & \delta_n^m+\frac{(\gamma -1)\beta^m \beta_n}{\beta^2}\end{pmatrix}\\&=
\begin{pmatrix}-\gamma^2(1-\beta^j \beta^m) & \gamma^2 \beta_n-\gamma \beta^j\Big(\delta_n^m + \frac{(\gamma -1)\beta^m \beta_n}{\beta ^2}\Big) \\ 
\gamma^2 \beta_k -\gamma \beta^m \Big(\delta_k^j + \frac{(\gamma -1)\beta^j \beta_k}{\beta^2}\Big) & -\gamma^2 \beta_k \beta_n + \Big(\delta_k^j + \frac{(\gamma -1)\beta^j \beta_k}{\beta^2}\Big)\Big(\delta_n^m + \frac{(\gamma -1)\beta^m \beta_n}{\beta^2}\Big)\end{pmatrix}\end{align} 
However, I don't understand what is meant by time-time, time-space, space-time and space-space components. Some elaboration on what this means would be appreciated. Am I supposed to equate the above matrix components with $$\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}.$$
If so, what am I trying to solve for to truly show this is in the Lorentz group?
Please note, I'm only a novice with the summation notation, so I apologize if I've written anything out incorrectly.

Problem I'm Attempting

 A: Hints :
$$
\text{your matrix :}\:\:
\begin{pmatrix}
-1 & 0 \\ 0 & 1
\end{pmatrix}
\text{is this matrix :}\:\:
\eta=
\begin{bmatrix}
-1 & 0 & 0 & 0\\
0  & 1 & 0 & 0\\
0  & 0 & 1 & 0\\
0  & 0 & 0 & 1
\end{bmatrix}
=
\begin{bmatrix}
-1 & \boldsymbol{0}^{\mathsf{T}} \\
\boldsymbol{0}  & \mathrm{I} 
\end{bmatrix}
\tag{01}
$$
where 
$$
\boldsymbol{0}\equiv 
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}
=\text{the null column vector},
\qquad
\boldsymbol{0}^{\mathsf{T}}\equiv
\begin{bmatrix}
0 & 0 & 0
\end{bmatrix}
=\text{the null row vector} 
\tag{02}
$$

$$
B=
\begin{bmatrix}
\gamma & -\gamma\beta^j \\
-\gamma \beta_k & \delta_k^j+\dfrac{(\gamma -1)\beta^j \beta_k}{\beta^2}
\end{bmatrix}
=
\begin{bmatrix}
\gamma & -\gamma\boldsymbol{\beta}^{\mathsf{T}} \\
-\gamma \boldsymbol{\beta}  & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}
\end{bmatrix}
\tag{03}
$$

$$
\boldsymbol{\beta}=
\begin{bmatrix}
\beta_{1}\\
 \beta_{2}\\
 \beta_{3}
\end{bmatrix},
\qquad
\boldsymbol{\beta}^{\mathsf{T}}=
\begin{bmatrix}
\beta^{1} & \beta^{2} & \beta^{3}
\end{bmatrix},
\qquad
 \mathrm{I}=
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\tag{04}
$$

$$
\boldsymbol{\beta}^{\mathsf{T}}\boldsymbol{\beta}=
\begin{bmatrix}
\beta^{1} & \beta^{2} & \beta^{3}
\end{bmatrix}
\begin{bmatrix}
\beta_{1}\\
 \beta_{2}\\
 \beta_{3}
\end{bmatrix}
=\beta^{m}\beta_{m}
=\boldsymbol{\beta}\boldsymbol{\cdot}\boldsymbol{\beta}=\Vert \boldsymbol{\beta} \Vert ^{2}=\beta^{2}=\dfrac{v^2}{c^2}=\dfrac{\gamma^2-1}{\gamma^2}
\tag{05}
$$

$$
\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}=
\begin{bmatrix}
\beta^{1}\\
 \beta^{2}\\
 \beta^{3}
\end{bmatrix}
\begin{bmatrix}
\beta_{1} & \beta_{2} & \beta_{3}
\end{bmatrix}
=\beta_{m}\beta^{n}=
\begin{bmatrix}
\beta_{1}^{2} & \beta_{1}\beta_{2} & \beta_{1}\beta_{3}\\
\beta_{2}\beta_{1} & \beta_{2}^{2} &   \beta_{2}\beta_{3}\\
 \beta_{3}\beta_{1} & \beta_{3}\beta_{2} &  \beta_{3}^{2} 
\end{bmatrix}
\tag{06}
$$

$$
\left(\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}\right)^{2}=
\left(\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}\right)\left(\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}\right)=
\boldsymbol{\beta}\underbrace{   \left    (\boldsymbol{\beta}^{\mathsf{T}} \boldsymbol{\beta}\right)}_{\Vert \boldsymbol{\beta} \Vert ^{2}}\boldsymbol{\beta}^{\mathsf{T}}=
\Vert \boldsymbol{\beta} \Vert ^{2}\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}=\left(\dfrac{{\gamma}^2-1}{\gamma^2}\right)\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}
\tag{07}
$$

Note that if  $\:\mathbf{n}\:$ is a unit 3-vector then $\:\mathrm{P}_{\mathbf{n}}=\mathbf{n}\mathbf{n}^{\mathsf{T}}\:$ is the projection on its direction, since for every $\:\mathbf{x} \in \mathbb{R}^{3}\:$
$$ 
\mathrm{P}_{\mathbf{n}}\mathbf{x}=\mathbf{n}\mathbf{n}^{\mathsf{T}} \mathbf{x}=
\begin{bmatrix}
n_{1}\\
 n_{2}\\
n_{3}
\end{bmatrix}
\underbrace{
\begin{bmatrix}
n_{1} & n_{2} &n_{3}
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}}_{\left(\mathbf{x} \boldsymbol{\cdot}\mathbf{n} \right)} 
=\left(\mathbf{n} \boldsymbol{\cdot}\mathbf{x} \right)\mathbf{n}
\tag{08}
$$
with the well known property of projections
$$
\mathrm{P}_{\mathbf{n}}^{2}=\mathrm{P}_{\mathbf{n}}
\tag{09}
$$
Defining
$$
\mathbf{n} \equiv \dfrac{\boldsymbol{\beta}}{\Vert \boldsymbol{\beta} \Vert}=\dfrac{\boldsymbol{\beta}}{\beta}
\tag{10}
$$
then
$$
\mathrm{P}_{\mathbf{n}}=\mathbf{n}\mathbf{n}^{\mathsf{T}}=\left( \dfrac{ \boldsymbol{\beta}}{\beta} \right) \left(\dfrac{\boldsymbol{\beta}}{\beta}\right)^{\mathsf{T}}=\dfrac{\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}    }{\beta^{2}}
\tag{11}
$$
and (07) is its property as projection.

EDIT :
\begin{align} 
& B^{\mathsf{T}}\eta B  =\\
&\begin{bmatrix}
\gamma & -\gamma\boldsymbol{\beta}^{\mathsf{T}} \\
-\gamma \boldsymbol{\beta}  & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}
\end{bmatrix}
\begin{bmatrix}
-1 & \boldsymbol{0}^{\mathsf{T}}\\
&\\ 
\boldsymbol{0} & \mathrm{I} 
\end{bmatrix}
\begin{bmatrix}
\gamma & -\gamma\boldsymbol{\beta}^{\mathsf{T}} \\
-\gamma \boldsymbol{\beta}  & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}
\end{bmatrix}=\\
&\begin{bmatrix}
\gamma & -\gamma\boldsymbol{\beta}^{\mathsf{T}} \\
-\gamma \boldsymbol{\beta}  & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}
\end{bmatrix}
\begin{bmatrix}
-\gamma & +\gamma\boldsymbol{\beta}^{\mathsf{T}} \\
-\gamma \boldsymbol{\beta}  & \mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}
\end{bmatrix}=
\begin{bmatrix}
\sigma & \boldsymbol{\rho}^{\mathsf{T}} \\
&\\
\boldsymbol{\rho}  & \mathrm{Z}
\end{bmatrix}\equiv \xi
\tag{12}
\end{align}
where $\:\xi\:$ a real symmetric $\:4\times 4\:$ matrix with elements $\:\sigma,\boldsymbol{\rho},\mathrm{Z}\:$ a real scalar, a real 3-vector and a real symmetric $\:3\times 3\:$ matrix respectively, all to be determined. 
Now,
\begin{align} 
\sigma=-\gamma^{2}\underbrace{\left(1-\boldsymbol{\beta}^{\mathsf{T}}\boldsymbol{\beta}\right)}_{1-\tfrac{v^2}{c^2}=\gamma^{-2}} \quad \Longrightarrow \quad \sigma=-1
\tag{13a}
\end{align}
\begin{align} 
\boldsymbol{\rho} & =\gamma^{2}\boldsymbol{\beta}-\gamma\boldsymbol{\beta}-\dfrac{\gamma(\gamma -1)\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}} \boldsymbol{\beta}}{\beta^2}\\
& =\gamma\left(\gamma-1\right)\boldsymbol{\beta}-\gamma\left(\gamma-1\right)\boldsymbol{\beta}\underbrace{\left(\dfrac{\boldsymbol{\beta}^{\mathsf{T}}\boldsymbol{\beta}}{\beta^2}\right)}_{=1} \quad \Longrightarrow \quad \boldsymbol{\rho}=\boldsymbol{0} 
\tag{13b}
\end{align}
\begin{align} 
 \mathrm{Z} & =-\gamma^{2}\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}+\biggl(\mathrm{I}+\dfrac{(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}\biggr)^{2}\\
& = -\gamma^{2}\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}+\mathrm{I}+\dfrac{2(\gamma -1)\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}+\left(\gamma -1\right)^{2}\overbrace{\biggl(\dfrac{\boldsymbol{\beta} \boldsymbol{\beta}^{\mathsf{T}}}{\beta^2}\biggr)^{2}}^{=\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}/\beta^2}\\
& = \mathrm{I}+    \underbrace{\left[-\gamma^{2}+\dfrac{2(\gamma -1)}{\beta^2}+\dfrac{(\gamma -1)^{2}}{\beta^2}   \right]}_{=0}\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}\quad \Longrightarrow \quad \mathrm{Z}=\mathrm{I}
\tag{13c}
\end{align}
So,
\begin{equation}
B^{\mathsf{T}}\eta B = \xi = 
\begin{bmatrix}
\sigma & \boldsymbol{\rho}^{\mathsf{T}} \\
&\\
\boldsymbol{\rho}  & \mathrm{Z}
\end{bmatrix}=
\begin{bmatrix}
-1 & \boldsymbol{0}^{\mathsf{T}}\\
&\\ 
\boldsymbol{0} & \mathrm{I} 
\end{bmatrix}
\equiv \eta
\tag{14}
\end{equation}
QED.
A: Another, less messy way to do this is as follows. Link the said matrix to the identity matrix by a path defined by:
$$\Lambda:\mathbb{R}\to 
\mathscr{M}_{4\times4};\;\Lambda(\zeta) = \left(\begin{array}{c|c}\cosh\zeta & -\hat{B}^T \,\sinh\zeta\\\hline
-\hat{B}\,\sinh\zeta &\mathrm{id} +\hat{B}\,\hat{B}^T\, (\cosh\zeta-1) \end{array}\right)\tag{1}$$
where $\zeta = \operatorname{artanh}\frac{v}{c}$ is the rapidity of the putative boost. Here $\hat{B} =\frac{1}{v} \begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}$ is the unit vector of direction cosines pointing along the direction of the boost.
Exercise:Check that all matrices of the stated form can be written in the form of (1), so that they all form a smooth path through the identity (where they pass when $\zeta=0$).
Note the useful little formula $\hat{B}^T\,\hat{B} = 1$, so that you can almost manipulate matrices of the form in (1) as though their elements were scalar, aside from that $\hat{B}\,\hat{B}^T$ is left unsimplified. You get things like $(\hat{B}\,\hat{B}^T)^N = \hat{B}\,\hat{B}^T;\,N\geq 1$ (this means that $(\hat{B}\,\hat{B}^T)^N$ is the idempotent projector onto the direction of the boost) which you can use in the following.

Excercise: Prove that 

$$\Lambda(\zeta)\Lambda(-\zeta) = \mathrm{id}\tag{2}$$
whence:
$$\frac{\mathrm{d}\Lambda}{\mathrm{d}\zeta} \Lambda^{-1} = -\left(\begin{array}{c|c}0&\hat{B}^T\\\hline\hat{B}&0\end{array}\right)\tag{3}$$

Now, given (3), we have very simply:
$$\frac{\mathrm{d}}{\mathrm{d}\zeta} \left(\Lambda^T\,\eta\,\Lambda\right) = \Lambda^T\left(-\left(\begin{array}{c|c}0&\hat{B}^T\\\hline\hat{B}&0\end{array}\right)^T\,\eta - \eta \,\left(\begin{array}{c|c}0&\hat{B}^T\\\hline\hat{B}&0\end{array}\right)\right)\,\Lambda=0\tag{4}$$
and, since the sought identity follows trivially for $\zeta=0$, we have, through (4) a Cauchy initial value problem wherein the the derivative is a Lipschitz-continuous function of $\Lambda^T\,\eta\,\Lambda$, therefore $\Lambda^T\,\eta\,\Lambda=\eta$, true for all $\zeta$, is the unique solution to this CIVP and the identity is proven.
You've therefore got yourself a rather useful and compact expression for a general boost in (1), and you can see, in the light of (3), that it is the matrix exponential of $\zeta$ times the simple matrix on the right hand side of (3). The matrix on the right hand side of (3) is sometimes called an infinitessimal boost; all infinitessimal boosts are linear combinations, with the direction cosines as weights, of the three infinitessimal boosts for the three spatial co-ordinate directions.
