Let $T$ be a map on $\Bbb{R}^3$ defined by $T(x,y,z)=(x-y+2z,2x+y,-x-2y+2z)$. What is the geometric interpretation behind this map?
I have proved that this transformation is linear, but I want to actually visualize this transformation graphically. I want to explore how this transformation changes the position or arrangement of the coordinate axes and other vectors in space. I also want to know how could I predict that this map is linear without formally solving and just by intuition or visualization.
The last question is easiest:
an expression of the form $$ H(x, y, z) = ax + by + cz $$ is linear when $a$, $b$, and $c$ are constants. So that lets you identify linear maps from, say $\Bbb R^3$ to $\Bbb R$. More generally, a map from $\Bbb R^n $ to $\Bbb R$ is linear exactly if it's a homogenous degree-one polynomial in the arguments (or is the constant zero map).
A map $\Bbb R^n \to \Bbb R^k$, say $$ T(x_1, \ldots, x_n) = (T_1(x_1, \ldots, x_n), \ldots, T_k(x_1, \ldots, x_n)) $$ is linear exactly if each of the maps $T_i : \Bbb R^n \to \Bbb R$ is linear, as described in the previous paragraph.
As for "understanding a map geometrically," that's probably what the rest of your linear algebra course will be (at least partly) about. The easiest maps to understand geometrically are the "diagonal" ones, things like
$$ T(x, y, z) = (2x, 3y, -z) $$ because they act on each axis independently. This example, for instance, streches by a factor of $2$ in $x$, by a factor of $3$ in $y$, and flips the $z$-axis.
But what if your transformation stretched the direction $(1,1,1)$ by two, the direction $(-1, 1, 0)$ by $3$, and "flipped" the direction $(1, 1, -2)$? In many ways that'd be just as simple as the diagonal one, if only you were using a different coordinate system. Finding such a new coordinate system, in which a particular transformation is "easy to understand," is part of what you'll do when you study eigenvalues and eigenvectors. But before you get to that, you'll need to work up to notions of what it means for a set of three vectors (like the ones in this paragraph) to "define a coordinate system", and other such topics; that's more than I can express in just one MSE answer -- whole books (i.e., linear algebra texts!) have been written about it.
In the case of your particular map, it turns out that there's no coordinate system in which it looks like a "different stretch along each of three axes." As it happens, it takes the vector $(-2, 4, 3)$ to $(0,0,0)$, so in that sense, it takes an "axis" in the $(-2, 4, 3)$ direction and "flattens it" (like projecting from $xyz$-space to $xy$-space by forgetting the $z$-coordinate). But in the plane perpendicular to that $(-2, 4, 3)$ direction, the function both "stretches" the plane uniformly, and rotates it a bit.
How did I learn those facts about your map? I computed its eigenvalues and eigenvectors, which you'll learn to do by hand, but personally, I just used Matlab to do it, because I've already done enough eigenstructures by hand for one lifetime, and I wasn't going to learn anything by doing another. :) Doing it by hand would have taken me about 30 minutes, so it's not something simple that you're missing!