When are these quadratic forms surjective from $R^n$ to $R^n$?
问题内容
Consider functions from $R^n$ to a subset of $R^n$.
So $f(x_1,x_2,...,x_n) = (y_1,y_2,...,y_n)$.
where $x_i,y_i$ are all real.
More specific consider
$$f(x_1,x_2,...,x_n) = (Q_1(x_1,x_2,...,x_n),Q_2(x_1,x_2,...,x_n),...,Q_n(x_1,x_2,...,x_n))$$
where the $Q_i$ are all Quadratic forms.
Even more specific :
The $Q_i$ are distinct and of the type
$$ Q_i(x_1,x_2,...,x_n) = \sum_{u=1}^{u=n} \sum_{v=1}^{v=n} a_{(u,v,i)} x_u x_v $$
and
$a_{(u,v,i)}$ belongs to the set ${-1,0,1}$.
(not to be confused with ternary quadratic forms what usually means using $3$ variables)
We want to know how to solve the equation $f(x_1,x_2,...,x_n) = (y_1,y_2,...,y_n)$ for a given set of reals $y_i$.
This is essentially the functional inverse or equivalently solving a given systems of quadratic forms over the reals.
Now this might be a difficult problem so we end up with the more general question, is there always a solution for a given $f$ and a set of reals $y_i$ ?
Or in other words :
When is for a given $f$ like above
$$R^n \to (f) \to R^n $$
a surjective function ?
回答 (1)
Let $F(x)=(Q_1(x),\dots,Q_n(x))$, where the $Q_i$ are homogeneous quadratic forms on ${\bf R}^n$.
The main observation is that $F(tx)=t^2F(x)$ for every real $t$. So if $x\ne 0$ and $x=ru$, with $r=|x|$ and $u\in S^{n-1}$, then $F(x)=r^2F(u)$.
Hence the whole image of $F$ is determined by what happens on the unit sphere. More precisely, $F({\bf R}^n)$ is obtained from $F(S^{n-1})$ by multiplying by all nonnegative scalars.
Therefore $F$ is surjective onto ${\bf R}^n$ exactly when $F(S^{n-1})$ meets every ray starting at the origin. Equivalently, the set of directions $F(u)/|F(u)|$, for $u\in S^{n-1}$ and $F(u)\ne 0$, has to be the whole sphere $S^{n-1}$.
[Edit] This is the natural criterion for surjectivity in this setting. A minor clarification: $F(-x)=F(x)$ only shows that the map is not injective in general. It does not rule out surjectivity.