= Solution
A nondegenerate <Lorentzian metric> requires $M\ne0$; take $M>0$ as the radius convention. The original coordinates describe $\lambda>0$, with $\lambda=0$ a candidate boundary rather than an ordinary point of that chart. The change of coordinate gives
$$
dt=dv+\frac{M}{\lambda^2}d\lambda.
$$
Substituting and cancelling the two terms proportional to $d\lambda^2/\lambda^2$ produces
$$
\boxed{
ds^2=-\lambda^2dv^2-2M\,dv\,d\lambda+M^2d\Omega^2.
}
$$
The $(v,\lambda)$ block of this <metric tensor> has <determinant> $-M^2$ and <inverse metric>
$$
\begin{pmatrix}
g^{vv}&g^{v\lambda}\\
g^{\lambda v}&g^{\lambda\lambda}
\end{pmatrix}
=
\begin{pmatrix}
0&-1/M\\
-1/M&\lambda^2/M^2
\end{pmatrix}.
$$
All coefficients are smooth through $\lambda=0$, and the block has one negative and one positive <eigenvalue>. Together with the positive round <two-sphere> factor this is a regular <Lorentzian metric>. It extends to a neighborhood with either sign of $\lambda$. The usual polar-coordinate problem at the poles of the <two-sphere> is removed by separate angular charts. Therefore $\lambda=0$ is only a <coordinate singularity> of the original chart. This is the degenerate-horizon patch of <Bertotti-Robinson spacetime>.
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