Linear Transformations-Linear Algebra Give an example where T(u-v)=T(u)-T(v) but T is not a linear transformation.
Appreciate anyone who can help me with this. Really cannot think of any.
 A: As @abel pointed out, we do get q$\cdot T(u)=T(q\cdot u)$ if $q$ is a rational number. It is also true that 
$$T(u+v)=T(u--v)=T(u)-T(-v)=T(u)--T(v)=T(u)+T(v)$$
So, if my first identity could be extended to all real numbers (or to all fields), we would know that $T$ is a linear transformation.
Fortunately, we cannot extend that identity. We can show that if we accept the Axiom of Choice for sets. Here is a non-rigorous summary: if you understand the Axiom of Choice you will know how to make it rigorous.
We let $e_0=1$. We also order the real numbers in a well-order, starting with $0$ and $1$. In other words,
$$\mathbb R=\{0,1,r_1,r_2,\ldots\}$$
We want to find more $e_i$'s by looking at the $r_i$'s in order. If we can write an equation
$$q_0\cdot e_0+q_1\cdot e_1+\cdots =r_i$$
for finitely many $e$'s and $q$'s, and all the $q$'s are rational, then we do not add $r_i$ as an $e$; otherwise, we do. This defines an infinite set of $e$'s for us.
This choosing process was set up so that any real number $r$ can be written as a unique finite linear combination of $1$ and the $e$'s with rational coefficients.
We then define a transformation $T:\mathbb R\to \mathbb R$ such that:
$$T(1)=1$$
$$T(e_j)=0$$
$$T(q_0\cdot 1+q_1\cdot e_1+\cdots+q_n\cdot e_n)=q_0\cdot T(1)+q_1\cdot T(e_1)+\cdots q_n\cdot T(e_n)$$
for rational $q$'s. This defines a transformation that satisfies the given equation but is not linear. It is not linear because multiplying $v$ by an irrational number does not multiply $T(v)$ by that number.
Here is another way to think of all this. Our selection of the $e$'s basically partitions every real number into two unique parts $r=q+i$, where $q$ is a rational number and $i$ is either an irrational number or zero. Our construction makes $T(q+i)=q$. This makes $T$ linear in rational coefficients but not for all real coefficients. It meets the identity in the OP and even more, but does not meet the identity $T(b\cdot u)=b\cdot T(u)$ for $b$ irrational.
A: To elaborate on my hint, and to give a straightforward example:
Consider $\mathbb{C}$ as a vector space over itself, and let $T : \mathbb{C} \to \mathbb{C}$ be the conjugation map, i.e., $T(z) = \overline{z}$.  Then $T(w-z) = \overline{w - z} = \overline{w} - \overline{z} = T(w) - T(z)$ for all $w,z \in \mathbb{C}$, but given $\alpha, z \in \mathbb{C}$,
$$
T(\alpha z) = \overline{\alpha z} = \overline{\alpha} \, \overline{z} \neq \alpha \overline{z} = \alpha T(z)
$$
unless $\alpha \in \mathbb{R}$.  Thus $T$ respects subtraction, but is not $\mathbb{C}$-linear.
A: let me see what the consequences of $T(u-v) = T(u) - T(v)$ are:
(a) by putting $u = v = 0,$ we get $T(0) = 0$
(b) setting $u = 0,$ gives $T(-v)= -T(v)$
(c) setting $v = -u,$ and using (b) gets you $T(2u) = 2T(u)$ and we also have $T(u+v) = T(u) + T(v).$
(d) by induction $T(ku) = T(k-1)u + u) = T((k-1)u) + T(u) = kT(u)$ for all integers.
(e) you can also show $T(\frac{1}{k}u) = \frac{1}{k} T(u).$
(f) combining (d) and (e) we get  $T(\frac{m}{n}u) = \frac{m}{n} T(u).$
