Let $(V,b)$ a symmetric bilinear space. An old theorem of Witt says that if $(V,b)$ is regular, then given a subspace $W$ of $V$ and an isometry $\sigma: W \to V$, there exists an isometry $\Sigma: V \to V$ which extends $\sigma$.

Does this necessarily fail if $(V,b)$ is singular - e.g. if the radical $V^\perp$ is one-dimensional?