Let $A$ be a symmetric $n \times n$ matrix over $\mathbb{R}$. Let $0 \neq b \in \mathbb{R}$.
Show that the surface $M = \{x\in \mathbb{R}^n \mid x^T A x = b\}$ is an $(n - 1)$-dimensional submanifold of the manifold $\mathbb{R}^n$.
I was thinking about starting with a basis in $\mathbb{R}^n$ s.t. $ \begin{pmatrix} x & y & z \end{pmatrix} \cdot \begin{pmatrix} a_1 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0& a_3\end{pmatrix} \cdot \begin{pmatrix} x\\ y \\ z\end{pmatrix}$ $ = ax^2+by^2+cz^2 = {\tilde{b}} $,
where $\tilde{b} >0$
Differentiating of $\tilde{b} $ gives us $\begin{pmatrix} 2ax & 2by & 2cz \end{pmatrix}$ which has rank 1.