Evgeny.Makarov said:
There are two definitions of a linear transformation.
- $f$ is a linear transformation if $f(cx+dy)=cf(x)+df(y)$ for all vectors $x$ and $y$ and all scalars $c$ and $d$.
- $f$ is a linear transformation if two things hold: (a) $f(x+y)=f(x)+f(y)$ for all vectors $x$ and $y$, and (b) $f(cx)=cf(x)$ for all vectors $x$ and scalars $c$.
In fact, these definitions are equivalent. It is obvious that 1 implies 2 because both $f(x+y)=f(x)+f(y)$ and $f(cx)=cf(x)$ are special cases of $f(cx+dy)=cf(x)+df(y)$. (For the first equality take $c=d=1$ and for the second equality take $y=0$. Well, in fact it is not so obvious because you first have to prove that $f(0)=0$.) But 2 also implies 1:
\begin{align*}
f(cx+dy)&=f(cx)+f(dy)&&\text{using }f(x'+y')=f(x')+f(y')\text{ for }x'=cx,y'=dy\\
&=cf(x)+df(y)&&\text{using }f(ax')=af(x')\text{ for }a=c,x'=x\text{ and }a=d,x'=y
\end{align*}
Your textbook or instructor uses definition 1, while evinda used definition 2. In any case, they substitute the definition of this particular $f$ into the property to be checked (the definition of a linear transformation) and see if it is true.
Always be sure you know your definitions! Otherwise, no piece of math will make sense to you.
1. With respect, the solution I provided came from the text. This is self study for me, which is more difficult than than taking a course. Students who take a Linear Algebra course have access to an instructor who can answer their questions.
2. The solution I posted was derived from the definition of Linear Transformation, which from the text is stated as the following:
A transformation (or mapping) $T$ is $\textbf{linear}$ if:
(i) $T(\textbf{u} + \textbf{v}) = T(\textbf{u}) + T(\textbf{v})$ for all $\textbf{u,v}$ in the domain of $T$.
(ii) $T(c\textbf{u}) = cT(\textbf{u})$ for all scalars $c$ and all $\textbf{u}$ in the domain of $T$.
3. For: $f:\mathbb{R} \rightarrow \mathbb{R}$ by $f(x) = mx + b$, shouldn't $x$ and $y$ be described as coordinates rather than vectors? If not, then why not?
4. The only logical approach to this I can think of, I will post below. (There are still some things about this that remain unclear):
I suppose we have to assume $x$ and $y$ are vectors in $\mathbb{R}$ rather than coordinates, otherwise we won't be able to use Algebraic Properties of $\mathbb{R}^n$. I suppose using $f(y)$ in the manner presented below stems from the given assumption that $x$ and $y$ are vectors and the given function. I suppose whatever $x$ can do $y$ can also do? Other than this confusion, I believe I can successfully complete what is required using Algebraic Properties of $\mathbb{R}^n$ and definition of Linear Transformation.
Given:
$\textbf{Transformation}$:
$f:\mathbb{R} \rightarrow \mathbb{R}$
$\textbf{Function}$
$f(x) = mx + b$
Assumptions:
$b \ne 0$.
$x,y$ are vectors in $\mathbb{R}$.
$c$ and $d$ are scalars.
Statements:
$f(x) = mx$
$f(y) = my$
\begin{align} f(x) + f(y) &= mx + my \\&= m(x + y) \\&= f(x + y) \end{align}
\begin{align} c \dot\ f(x) &= c(mx) \\&= m(cx) \\&=f(cx)\end{align}
\begin{align} d \dot\ f(y) &= d(my) \\&= m(dx) \\&=f(dy) \end{align}
Conclusion:
$f$ is linear.
There were several things that contributed to my confusions and frustrations about this intially. I'm not even sure how to articulate my confusion regarding this problem clearly enough but I'll try:
1. The use of $x$ and $y$ as vectors rather than coordinates. I was under the assumption that $f(x) = mx + b$ is a linear function and $x$ and $y$ are coordinates. However, in this case, I suppose we assume $x$ and $y$ are vectors in order to use the properties of $\mathbb{R}^n$.
2. The use of the $y$ variable. I assume that the use of this variable is an essential necessity for the showing $f$ is linear and without the extra variable, it is not possible.
3. The use of $x$ and $y$ in place of $x$. I suppose because we announced $x$ and $y$ as vectors in $\mathbb{R}$ and both together are necessary for showing $f$ is linear.
4. The given textbook solution completely threw me off. I was not familiar enough with the definition to be given the solution in that manner. It did not help me understand at all.
5. With respect, it is not either of @evinda or @Evgeny.Makarov. fault. Nevertheless the only way I can explain my confusion at this particular point in time is that basically, when I received these explanations, I felt like the equivalent of a an algebra student who was already confused about algebra, but was given a differential equations problem to solve.
6. I feel that I have wasted a lot of time and that I can only benefit from the help of others if I ask specific questions about what I am confused on and if I receive specific, pointed, targeted responses to my specific question without giving any further information. And for each subsequent question I ask, the same approach is applied. Otherwise, if more information than I ask for is supplied, then the extra information will only add to my confusion.
If anyone is willing to provide further clarification on my solution and my confusions, feel free to do so.