(n - 1) dimensional submanifold of the manifold R^n

Let $\displaystyle A$ be a symmetric $\displaystyle n \times n$ matrix over $\displaystyle \mathbb{R}$.

Let $\displaystyle 0 \neq b \in \mathbb{R}$.

Show that the surface $\displaystyle M = \{x\in \mathbb{R}^n \mid x^T A x = b\}$ is an $\displaystyle (n - 1)$ dimensional submanifold of the manifold $\displaystyle \mathbb{R}^n$.

Please, help!

Re: (n - 1) dimensional submanifold of the manifold R^n

The differential of the smooth function $\displaystyle f: R^n \rightarrow R, f: x \mapsto x^TAx-b$ is

$\displaystyle f_*: R^n \rightarrow R, f_*: v \mapsto 2x^TAv$

Since $\displaystyle b \ne 0$, $\displaystyle x^tA$ never vanishes on the locus $\displaystyle f=0$ so $\displaystyle f_*$ has full rank.

That is, all points of M are regular points. Hence M is a n-1 dimensional sub-manifold.

Re: (n - 1) dimensional submanifold of the manifold R^n