Proving $(x, y, z) \mapsto (2x + y, 2y − z)$ is linear

Question

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.

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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.

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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:

1. 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.

Answer 1

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.

Answer 2

A different proof is that $T$ is equivalent to multiplication by the matrix $\begin{pmatrix}2&1&0\\0&2&-1\end{pmatrix}$.