Fact that affine transformations carries convex sets to convex sets My question is: it is usually told us that affine transformations have that property but cuuouldnt this be generalized for all linear transformations? While proving that affine transformations have images of set convex sets convex we do the folliwing;
$$T(ap+(1-a)p)=aT(p)+(1-a)T(p)$$
If $o<a<1$ then image is also convex. Didn't I use the property of linear transformations while proving? Why it is stated that affine transformations carry convex sets to convex sets instead of saying linear transformations?
 A: That linear transformations preserve convexity is not a generalization of the fact that affine transformations do.  It's really the other way around.  You do use the property that linear transformations map convex sets to convex sets, and then combine this with the fact that an affine transformation is a just a linear transformation plus a translation.  
First things first I'm a little confused about your notation; I think you're using $p$ as a double meaning and it causes your expression to simplify to an arithmetic tautology $T(p) = T(p)$ independent of any properties of $T$. 
To be precise in why a linear transformation $T$ preserves convexity, I would argue as follows: pick two points $u,v$, in the image of $T$, and write them as $T(x), T(y)$ for $x, y$ in the domain of $T$.  Then convexity of the domain implies that $ax + (1-a)y \in Domain$ for all $a \in [0,1]$, and linearity implies the image of the line $ax + (1-a)y$ under $T$ is a line, $au + (1-a)v$, connecting $u$ and $v$ in the image. 
Now an affine transformation is a slight generalization of a linear transformation.  It has to map linear objects to linear objects (lines go to lines, planes to planes...), like a linear transformation, but it doesn't have to respect subspaces (in particular we don't require $A(0) = 0$ for an affine transformation $A$).  So the important quality $T(ax + by) = aT(x) + bT(Y)$ doesn't hold for most affine transformations.  Indeed the simplest affine transformations are pure translation, $A(x) = c$, for which you can see linearity very much fails. But since you know that linear transformation preserve convexity and it's intuitive that translation should preserve convexity, you'd expect an affine map to still also preserve convexity.  
When you replace $T$ with $A(x) = T(x) + c$ in the above, nothing much changes, but you will notice that there's formally a little bit more going on (some cancellation happens).  The points $u,v$ that we choose in the image have preimages under $A$ of $x,y$, i.e. $T(x) + c = u$, $T(y) + c = v$.  Then 
$$A(ax + (1-a)y) = T(ax + (1-a)y) + c $$
$$= aT(x) + (1-a)T(y) + c = a(u-c) + (1-a)(v-c) + c$$
$$=au + (1-a)v$$.  
So again, since the affine transformation $A$ maps lines to lines we are able to find a line between any two points of the image of $A$.  
