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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: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=549&t=512900 which part are you stuck on?2012-12-20

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