line element:
$ds^2 = g_{uv} dx^u dx^v$

ADM metric:
$\alpha$ = lapse
$\beta^i$ = shift
$\gamma_{ij}$ = spatial metric
$g_{uv} = \left[\begin{matrix} -\alpha^2 + \beta^2 & \beta_j \\ \beta_i & \gamma_{ij} \end{matrix}\right]$

where $\beta^t = 0$, so $\beta_i = \gamma_{ij} \beta^j$

linear change of coordinates:
$x^u = {A^u}_{u'} x^{u'}$
$dx^u = {A^u}_{u'} dx^{u'}$
$ds^2 = g_{uv} {A^u}_{u'} {A^v}_{v'} dx^{u'} dx^{v'}$

${A^u}_{u'} = \left[\begin{matrix} 1 & -g_{ti} / g_{tt} \\ 0 & \delta^i_j \end{matrix}\right]$
$= \left[\begin{matrix} 1 & -\beta_i / (-\alpha^2 + \beta^2) \\ 0 & \delta^i_j \end{matrix}\right]$

change of coordinates:
$g_{u'v'} = {A^u}_{u'} g_{uv} {A^v}_{v'}$
$ = \left[\begin{matrix} 1 & 0 \\ -\beta_i / (-\alpha^2 + \beta^2) & \delta^i_j \end{matrix}\right] \left[\begin{matrix} -\alpha^2 + \beta^2 & \beta_j \\ \beta_i & \gamma_{ij} \end{matrix}\right] \left[\begin{matrix} 1 & -\beta_i / (-\alpha^2 + \beta^2) \\ 0 & \delta^i_j \end{matrix}\right]$
$ = \left[\begin{matrix} -\alpha^2 + \beta^2 & \beta_j \\ 0 & \gamma_{ij} - \beta_i \beta_j / (-\alpha^2 + \beta^2) \end{matrix}\right] \left[\begin{matrix} 1 & -\beta_i / (-\alpha^2 + \beta^2) \\ 0 & \delta^i_j \end{matrix}\right]$
$ = \left[\begin{matrix} -\alpha^2 + \beta^2 & 0 \\ 0 & \gamma_{ij} - \beta_i \beta_j / (-\alpha^2 + \beta^2) \end{matrix}\right]$

Tada! Change of coordinate absorbs shift into the metric.

Now let ${{A'}^u}_{u'} = \frac{1}{\sqrt{\alpha^2 - \beta^2}} {A^u}_{u'}$
Then we get:
$g_{u'v'} = {{A'}^u}_{u'} g_{uv} {{A'}^v}_{v'}$
$ = \left[\begin{matrix} -1 & 0 \\ 0 & \gamma_{ij} / (\alpha^2 - \beta^2) + \beta_i \beta_j / (\alpha^2 - \beta^2)^2 \end{matrix}\right]$

Now let ${\gamma'}_{ij} = \frac{1}{(\alpha^2 - \beta^2)} \left( \gamma_{ij} + \frac{1}{(\alpha^2 - \beta^2)} \beta_i \beta_j \right)$
$g_{u'v'} = \left[\begin{matrix} -1 & 0 \\ 0 & {\gamma'}_{ij} \end{matrix}\right]$
Tada.

TODO: inverse of $A$ squared should produce the metric.
$g = e \cdot \eta \cdot e^T$
$e^{-1} \cdot g \cdot e^{-T} = \eta$
so for some $e^{-1}$ which 'boosts' / transforms $g$ into $\eta$, its inverse $e$ applied to $\eta$ will produce (or 'unboost' $\eta$ into) $g$.
And all this is turning into the ADM decomposition of tetrad formalism.