2
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $f(x)=x^{T}Ax$.

$\endgroup$
  • $\begingroup$ @John Lennon You can give ok on my reply? $\endgroup$ – Manoel Jan 30 '17 at 1:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.