Let $A$ be an invertible, symmetric $n\times n$ matrix and define $F:\mathbb R^n\rightarrow\mathbb R$ by $F(x)=x\cdot Ax$
For $k>0$, how can I show that the set $T_k=\{x\in\mathbb R^n:F(x)=k\}$ is either empty or a submanifold of $\mathbb R^n$?
I know that if $rank(\delta F(x))=1$ for all $x\in T_k$, then $T_k$ is a submanifold of $\mathbb R^n$
$\delta F(x)=(\delta_1 F(x)...\delta_n F(x))$
The rank of $\delta F(x)$ is zero iff all the partial derivatives are zero, why does this imply that $T_k$ is empty?
If $T_{k}$ is not empty, then the rank of the differential $d_{x}F$ is $1$ at all points $x\in T_{k}$. Indeed, for $v\in\mathbb{R}^{n}$, the differential $d_{x}F(v)$ is given by $$ d_{x}F(v)=\left.\frac{d}{dt}\right|_{t=0}F(x+tv)=\left.\frac{d}{dt}\right|_{t=0}(x+tv)\cdot(Ax+tAv)=x\cdot Av+v\cdot Ax $$ In particular, for $v=x$ we have $$ d_{x}F(x)=2x\cdot Ax= 2F(x)=2k\neq 0. $$ So the map $d_{x}F:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is not identically zero, and therefore it has rank $1$.