How can you prove that a function $\mathbf{F}:\mathbb{R}^n\longrightarrow \mathbb{R}^m$ $\leftrightarrow$ $\mathbf{F}(x_1,...,x_n)=\mathbf{0}$ The question comes from the textbook "Multivariable calculus with applications" and it is the following:
question 2.4 p.73
Show that a function $\mathbf{F}:\mathbb{R}^n\longrightarrow \mathbb{R}^m$ is linear if and only if:
$$
\mathbf{F}(x_1,\dots,x_n)=\mathbf{0}
$$
I reckon that the function $\mathbf{F}$ should have zeroes instead of the vector $(x_1,\dots,x_n)$, because the function $\mathbf{F}(x_1,\dots,x_n)=(1,\dots,1,1)$ is obviously not equal to the zero vector. Anyways, if the transformation $\mathbf{F}:\mathbb{R}^n\longrightarrow \mathbb{R}^m$ is linear then the following should be true:
$$
\mathbf{F}(X)=
\begin{pmatrix}
c_{11}&\dots&c_{1n}\\
&\dots&&\\
c_{m1}&\dots&c_{mn}\\
\end{pmatrix}
\begin{pmatrix}
x_{1}\\
\dots\\
x_{n}\\
\end{pmatrix}
=
\begin{pmatrix}
c_{11}x_{1}+\dots +c_{1n}x_{n}\\
\dots\\
c_{m1}x_{1}+\dots +c_{mn}x_{n}\\
\end{pmatrix}
$$
and if we set the resulting matrix to equal zero then it follows that $\exists x_1,\dots, x_n$ that are unique and dependent on the choice of $x$ which again disproves the statement because I can choose $c$ to be a non-zero number and by that and with the assumption that $x$ is not zero as well we can confirm that smth is wrong with the statement.
 A: There seems to be a misprint in your edition. The true statement should be:

Show that a constant function $\mathbf{F}:\mathbb{R}^n\longrightarrow \mathbb{R}^m$ is linear if and only if
$$\mathbf{F}(x_1,\dots,x_n)=\mathbf{0}.$$

Can you prove that result now?
A: (the author) I produced something like proof for the theorem. However, we have to deal with the responsibility to restate the theorem. As @Keplerto mentioned the correct formulation of the statement should be as follows:
Theorem. Every constant function $\mathbf{F}:\mathbb{R}^n\longrightarrow \mathbb{R}^m$ is linear if and only if:
$$
\mathbf{F}=0
$$
If $n>m$ we define a function $F(x_1,\dots,x_n)=(c_1,\dots,c_{n-1})$ where $c_1,\dots,c_{n-1}\in \mathbb{R}$,  which fulfills the conditions from the theorem. However, if defined in such way the first property of linear functions is not acquired:
$$ 
\alpha F(x_1,\dots,x_n)=\alpha (c_1,\dots, c_{n-1})=(\alpha c_1,\dots, \alpha c_{n-1})=F(\alpha x_1,\dots, \alpha x_{n-1}, x_n)
$$
If n<m the first condition for linearity holds. Although, the first condition for linearity is acquired the second one does not hold. Take the vectors $\vec{V},\vec{S}\in \mathbb{R}^n$, from here the following condition $F(\vec{V})+F(\vec{S})=F(\vec{V}+\vec{S})$ cannot be verified which is obvious from the following:
$$
F(\vec{V})+F(\vec{S})=2F(\vec{V})
$$
If we want to force the function to fulfill the condition for linearity the following system should have at least one solution:
$$
\begin{cases}
F(\vec{V})+F(\vec{S})=2F(\vec{V})\\
F(\vec{V})+F(\vec{S})=F(\vec{V}+\vec{S})
\end{cases}
\Longrightarrow 2F(\vec{V})=F(\vec{V}+\vec{S}) \Longrightarrow  2\vec{V}=\vec{V}+\vec{S} \Longrightarrow \vec{V}=\vec{S}
$$
A possible solution to the system is $F(\vec{V})=0$, then one can easily see that $F(\vec{V})+F(\vec{S})=2F(\vec{V})=0$ and $F(\vec{V})+F(\vec{S})=0=F(\vec{V}+\vec{S})$.
Honorable mentions There is a possible way to prove the theorem using dimensions. Since the linear function is always continuous and every continuous function is bijective and since the transformation $\mathbb{R}^n\longrightarrow \mathbb{R}^m$ is such that $\dim{R}^n\neq \dim{R}^m$ the condition cannot hold.
