Proving $(x, y, z) \mapsto (2x + y, 2y − z)$ is linear This is a problem with a sample solution I got from Linear Algebra class. The lectures are like trying to learn math from Wikipedia: it only really works when the listener/reader already understand much of the content. In my case, I don't.

 The question 
Prove that following map is a linear transformation.
$T : \mathbb R^3 → \mathbb R^2$
$(x, y, z) → (2x + y, 2y − z)$
 The given proof 
(1) 
$T(u + v)= T(x + x', y + y', z + z')$
$= (2(x + x') + (y + y'), 2(y + y') − (z + z')$
$= (2x + y, 2y − z) + (2x' + y', 2y' − z')$
$= T(u) + T(v)$
(2) 
$T(\alpha u)=T(\alpha x, \alpha y, \alpha z)$
$= (2αx + αy, 2αy − αz)$
$= α(2x + y, 2y − z)$
$= αT(u)$
By the definition of linear map, T is linear.

 My problem 
This feels like it skipped some pretty important steps and provides no actual explanation...
Here's my understanding of the given solution (after #s are my comments):
(1) 
$T(u + v)$ # (1a)
$= T(x + x', y + y', z + z')$ # (1b) adding vectors = adding components
$= (2(x + x') + (y + y'), 2(y + y') − (z + z')$ # (1c) Substitution $u$ and $v$'s components into the function $T(x)$
$= (2x + y, 2y − z) + (2x' + y', 2y' − z')$ # (1d) how???
$= T(u) + T(v)$ # (1e) long functions --> T(x) form
(2) 
$T(αu) $ (2a)
$= T(αx, αy, αz)$ # (2b)
$= (2αx + αy, 2αy − αz)$ # (2c) Substituting $u$'s components (with scalar) into the function $T(x)$
$= α(2x + y, 2y − z)$ # (2d) just take out the scalar $α$
$= αT(u)$ # (2e) long function --> short form $T(x)$
By the definition of linear map, $T$ is linear. # I understand this definition, as it is "trivial".
What I don't quite understand:


*

*How did (1c) = (1d) happen? Staring at them is just making me confused but I think it's similar to how $ab + ac = a( b + c )$ works.


(After >30 min of brainstorming, I got it down to just 1)
I'd like an explanation on "(1c) = (1d)". It looks simple but also confusing. If possible, I'd like my understanding checked with a quick walkthrough. 
 A: Maybe you’re being too picky/pedantic.
You know that for the reals (in fact, every ring), $a(b+c) = ab + ac$. Expand everything:  it becomes
$$(2x + 2x^\prime + y + y^\prime, 2y + 2y^\prime - z - z^\prime).$$
Since summation is commutative and associative, write this as 
$$((2x+y) + (2x^\prime + y^\prime), (2y-z) + (2y^\prime - z^\prime)).$$
Call $2x+y = a, 2y-z = b$ and the same for $a^\prime, b^\prime$. What you have left is $(a+a^\prime, b+b^\prime)$. But since you’re working over $\mathbb R^n$ with (presumably) componentwise sum, this is just $(a,b)+(a^\prime, b^\prime)$. Can you continue it from here?
Also, try to prove this more general fact: if $\mathbf{a_1}, \ldots, \mathbf{a_n}$ are vectors, then the mapping $\mathbf x \mapsto (\mathbf{a_1\cdot x}, \ldots, \mathbf{a_n\cdot x})$ is linear. This generalizes your problem with pretty much the same argument and, conversely, every linear transformation over a finite-dimensional vector space looks like this (given a basis). In fact, the dot products are row multiplications of the matrix $\begin{bmatrix}\mathbf{a_1}^\mathsf{T} \\ \vdots \\ \mathbf{a_n}^\mathsf{T} \end{bmatrix}$ that represents the linear transformation in this basis.
A: A different proof is that $T$ is equivalent to multiplication by the matrix $\begin{pmatrix}2&1&0\\0&2&-1\end{pmatrix}$.
