If $T$ is linear, show that for any $\epsilon>0$, the behavior of $T$ on $B(\textbf{0}_n,\epsilon)$ determines the behavior of $T$ everywhere. Let $T:\mathbb{R}^n \mapsto \mathbb{R}^m$ be a linear mapping. Show that for any $\epsilon>0$, the behavior of $T$ on $B(\textbf{0}_n,\epsilon)$ determines the behavior of $T$ everywhere.
I think the linear transformation property $T(rv)=rT(v)$ for $r$ a scalar. Also, I am having trouble trying to understand the epsilon ball definition.
 A: Let $\epsilon>0$ and $\vec{x} \in \mathbb{R}^n$ be given. There exists $r>0$ such that $\frac{1}{r} \vec{x} \in B_\epsilon (\vec{0})$, i.e., such that $\frac{1}{r} \Vert x \Vert < \epsilon$. Then
$$T(\vec{x})= r\cdot \frac{1}{r} T(\vec{x}) = r \cdot T\left(\frac{1}{r} \vec{x}\right).$$
Notice that this last term only requires us to evaluate $T$ on $B_\epsilon (\vec{0})$ and then scale our result back up. Because $T$ is a linear transformation, we in fact have equalities. 
Contrast this with what would happen if $T$ weren't a linear transformation. For example, let $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$. We can still scale any input so that it is $B_\epsilon(0) = (-\epsilon, ~ \epsilon)$, but the calculations won't agree. Say, let $x=2$, $\epsilon=1/2$ and pick $r=10$. Then
$$4= 2^2 = f(2) = 10 \cdot \frac{1}{10} f(2) \neq 10 \cdot f(2/10) = 10 \cdot \left(\frac{2}{10}\right)^2 = \frac{4}{10}$$
A: Eric's answer is nice, and correct. (I upvoted it.) 
But what he doesn't say is that a linear transformation is fully determined by values on much "thinner" sets than sets like $B(\mathbf{0};\epsilon)$. Indeed, it follows immediately from the definition of linearity that a linear transformation is determined solely by its values on a basis of the domain. But a basis of the domain is a set with finite cardinality (and measure zero), in contrast to the ball $B(\mathbf{0};\epsilon)$, which has cardinality of the continuum (and nonzero measure). In other words, to determine $T$ it is sufficient but far from necessary to know how $T$ acts on an open ball.
So: another way to prove your claim is simply to observe that we can choose a basis of $\mathbb{R}^n$ that lies entirely in $B(\mathbf{0};\epsilon)$. (For concreteness: take the standard basis and scale each to have length $<\epsilon$.)
