Derivative of a function from $\mathbb{R}^n$ to $\mathbb{R}^m$ Suppose that the mapping $F:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is continuously differentiable and that there is a fixed $m \times n$ matrix $A$ so that $DF(x)=A$ for every $x$ in $\mathbb{R}^n$
Prove that there is some $c$ in $\mathbb{R}^m$ so that $F(x)=Ax+c$ for every $x$ in $\mathbb{R}^n$
Ok, so this makes sense intuitively to me when I consider $n=m=1$ because then it's just simple integration of a constant function. But, I'm not sure how to go about showing this in these higher dimensions. 
 A: Note that the Jacobian matrix (the matrix of partial derivatives) is constant, meaning that all partial derivatives $\frac{\partial f_i}{\partial x_j} $ are constants. 
A: One would like to apply the mean value theorem, but there is no such thing for vector-valued functions (see here). So one could apply it for functions like $t\longmapsto f_j(tx)$, using the nullity of the partial derivatives. 
But it is easier to write it all at once as follows, which is standard technique in calculus of functions wich take values in normed vector spaces. For instance, one uses such techniques to show that $C^1$ functions are locally $M$-Lipschitz with $M$ a local bound for $\|dF\|$. Then one uses this fact to prove that a function with zero derivative is locally constant! The proof below works more generally if $F:X\longrightarrow Y$ is differentiable with constant derivative on $X$, where $X$ and $Y$ are both normed vector spaces.
Fix $x$ and consider the function
$$
g(t)=F(tx) \qquad \forall t\in\mathbb{R}.
$$
By composition, this function is continuously differentiable on $\mathbb{R}$ and by the chain rule
$$
g'(t)=dF_{tx}(x).
$$
Since you assume $y\longmapsto dF_y$ is constant equal to $A$, this yields
$$
g'(t)=Ax\qquad \forall t\in\mathbb{R}.
$$
Integrating the latter over $[0,1]$ and applying the fundamental theorem of calculus (componentwise), we get
$$
F(x)-F(0)=g(1)-g(0)=\int_0^1g'(t)dt=\int_0^1Axdt=Ax.
$$
So 
$$
F(x)=Ax+F(0)\qquad\forall x\in\mathbb{R}^n.
$$
Remark: the assumption $F$ continuously differentiable with constant derivative is redundant, since a constant function is automatically continuous. So the assumption should be: $F$ is differentiable with constant derivative $A$.
