For a positive definite quadratic form $f: R^n \rightarrow R$, $f^{-1}(x)$, for any $x>0$, is diffeomorphic to $S^{n-1 }$

Question

How to show for a positive definite quadratic form $f: R^n \rightarrow R$, there exists $f^{-1}(x)$, for any $x>0$, is diffeomorphic to $S^{n-1 }$?

Answer 1

Let $B$ is positive definite symmetric matrix Then we have a quadratic form : $$ Q(v):= \langle Bv,v\rangle $$

If we diagonalize $B$ with ${\rm diag} (c_1,\cdots , c_n),\ c_i > 0$ then $$ Q(v)=\sum c_i v_i^2 $$

Hence $$Q^{-1} (x)=\{ v| \sum c_i v_i^2=x \} $$ Define $F(v)=(\sqrt{c_1} v_1,\cdots,\sqrt{c_i} v_n ) $. Since $F$ is a diffeomorphism, so $F^{-1} (S^{n-1}(\sqrt{x}) )=Q^{-1} (x)$ is a sphere.