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?
Answer
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.
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