Show the Lorentz Transformation Matrices Have an Inverse Assume the Lorentz transformations obey the relationship
$$g_{uv}\Lambda^u_{p}\Lambda^v_\sigma = g_{p\sigma},$$ where $g_{uv}$ is the metric tensor of special relativity.
How can one show, under that assumption, that the Lorentz matrix $\Lambda^a_b$ has an inverse?
 A: Starting from
$$
g_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma = g_{\rho\sigma}
$$
we contract with $g^{\sigma\tau}$
$$
g_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma g^{\sigma\tau} = g_{\rho\sigma} g^{\sigma\tau} = \delta_\rho^\tau
$$
and reorder the factors
$$
g_{\mu\nu} g^{\sigma\tau} \Lambda^\nu{}_\sigma \cdot \Lambda^\mu{}_\rho = \delta_\rho^\tau
$$
which shows that $\Lambda$ has a left-inverse and is thus injective.
As $\Lambda$ is an endomorphism, this is sufficient to show it's invertible and we have
$$
(\Lambda^{-1})^\tau{}_\mu = g_{\mu\nu} g^{\sigma\tau} \Lambda^\nu{}_\sigma = \Lambda_\mu{}^\tau
$$
A: You may order the matrices like this:
$$ \Lambda_\rho^\mu g_{\mu\nu} \Lambda^\nu_\sigma = g_{\rho\sigma} $$
I suppose all the letters should have been Greek. They're called mu, nu, rho, sigma, good to learn them.
In my form, one may view $\mu$ as the summed over index in the first product on the left hand side and $\nu$ as the summed over index in the second product. So making a convention for a matrix $\Lambda$ so that its components are $\Lambda^\mu_\rho$ where $\rho$ is the row and $\mu$ is the column, the equation above is the matrix equation
$$ \Lambda \cdot g \cdot \Lambda^T = g $$
where $T$ means transposition. The matrix on the right hand side is nonsingular, i.e. it has a nonzero determinant, so the factors on the left hand side must also have a nonzero determinant i.e. be invertible.
A: Assume the Lorentz transformation $\Lambda$ is not invertible. Then it is in particular not injective and there exists $0\not=u\in\ker\Lambda$.
The inner product $g$ is nondegenerate so there's a vector $v$ with $g(u,v)\not=0$ and we end up with the contradiction
$$
0\not=g(u,v)=g(\Lambda u, \Lambda v)=g(0,\Lambda v)=0
$$
where we have used the fact that Lorentz transfomations leave the inner product invariant, which is exactly your starting equation.
A: $$g_{uv}\Lambda^u_{p}\Lambda^v_\sigma=g_{p\sigma} \Longleftrightarrow \Lambda^Tg\Lambda=g$$
where g is the matrix whose entries are $g_{uv}$
$$
\det(\Lambda^Tg\Lambda)=\det(g)
$$
$$
\det(\Lambda^T)\det(g)\det(\Lambda)=\det(g)
$$
Obviously, $\det(g)\neq0$ and $\det(\Lambda^T)=\det(\Lambda)$
Then $$\det(\Lambda)^2=1$$
Since $\det(\Lambda)$ never vanishes, the matrix $\Lambda$ is always invertible.
A: I have come up with the following proof:
Begin with the relationship,
$g_{vu}\Lambda^u_{p}\Lambda^v_\sigma = g_{p\sigma}$
Which is the same as,
$\Lambda_{pv}\Lambda^v_\sigma = g_{p\sigma}$
Now multiply both sides of the equation by $g^{ap}$ to yield,
$g^{ap}\Lambda_{pv}\Lambda^v_\sigma = g^{ap}g_{p\sigma}$
This simplifies to:
$\Lambda^a_{v}\Lambda^v_\sigma = \delta^a_\sigma$
Also, we know that
$(\Lambda_v^a)^{-1}\Lambda^v_\sigma = \delta^a_\sigma$   By definition of inverse.
Therefore,
$(\Lambda_v^a)^{-1} = \Lambda^a_{v}$
