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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.

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Yes. This was my idea: which part are you stuck on? – Matt Dec 20 '12 at 1:32

You can use the following result which is known:

Let $f:M\longrightarrow N$ be a smooth map where $M$ is $(n + k)$-dimensional and $N$ is $n$-dimensional. If $q = f(p)$ is a regular value, then $f^{-1}(q)$ is a $k$-dimensional smooth submanifold.

In particular deems $f:\mathbb{R}^{n}\longrightarrow \mathbb{R}$, given by $f(x)=x^{T}Ax$.

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