MHB Is $f(x) = mx + b$ a Linear Transformation?

[bwpbruce]

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by $f(x) = mx + b$.
$\textbf{a.}$ Show that $f$ is a linear transformation when $b = 0$.
$\textbf{b.}$ Find a property of linear transformation that is violated when $b = 0$
$\textbf{c.}$ Why is $f$ called a linear function?

 

[evinda]

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by $f(x) = mx + b$.
$\textbf{a.}$ Show that $f$ is a linear transformation when $b = 0$.

A function $f$ from $R^n$ to $R^m$ is a linear transformation if and only if it satisfies the following two properties:

• For every two vectors $A$ and $B$ in $R^n$
  $f(A+B)=f(A)+f(B)$
  
• For every vector $A$ in $R^n$ and every number $k$
  $f(kA)=kf(A)$.

Can you use this theorem for $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f(x) = mx$ ?

 

[bwpbruce]

I believe so, however, I am trying to understand the reasoning behind how to construct the actual proof. I've already seen the proof but there is no demonstration of methodology behind it. The initial process doesn't seem very intuitive at all.

 

[Evgeny.Makarov]

Why don't you start checking the properties evinda wrote? Suppose that $f(x)=mx$. Is it true that $f(A+B)=f(A)+f(B)$?

 

[bwpbruce]

What you're asking is exactly what I'm trying to understand. What is the process of verifying this property for the given function?

 

[Evgeny.Makarov]

The process consists of substituting the definition of $f$ into the equation you want to verify and then actually checking it.

How would you check that $x=5$ satisfies the equation $x^2-3x-10=0$? You would replace $x$ with 5 in the equation and then perform the calculation.

 

[bwpbruce]

The proof that I have isn't written up that way. They don't just use $x$. Maybe I should just write it up the way they did it, then you can explain how it relates to what you're suggesting.

 

[Evgeny.Makarov]

Maybe I should just write it up the way they did it, then you can explain how it relates to what you're suggesting.

That makes sense.

 

[bwpbruce]

So here's the solution they provided for part $\textbf{a.}$:

When $b = 0, f(x) = mx$. In this case, for all $x,y$ in $\mathbb{R}$ and all scalars, $c$ and $d$, $f(cx + dy) = m(cx + dy) = mcx + mdy = c(mx) + d(my) = c \dot\ f(x) + d\dot\ f(y)$. This shows that $f$ is linear.

Perhaps someone can explain the above or present a different approach that is more intuitive.

 

[Evgeny.Makarov]

There are two definitions of a linear transformation.

1. $f$ is a linear transformation if $f(cx+dy)=cf(x)+df(y)$ for all vectors $x$ and $y$ and all scalars $c$ and $d$.
2. $f$ 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.

 

[bwpbruce]

There are two definitions of a linear transformation.

1. $f$ is a linear transformation if $f(cx+dy)=cf(x)+df(y)$ for all vectors $x$ and $y$ and all scalars $c$ and $d$.
2. $f$ 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.

 

[Evgeny.Makarov]

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 $T$.

OK, so this definition is definition 2 from https://driven2services.com/staging/mh/index.php?posts/66920/, while they used definition 1 from the same post in the solution you provided. So they used the equivalence of the two definitions implicitly. In fact, this equivalence is almost trivial, which you will likely see in a couple of weeks if you continue studying the subject. Authors can sometimes be guilty of not explaining some facts they remember as trivial. For example, I realized that one needs to use $f(0)=0$ in proving the equivalence only when I started explaining it (though one may avoid it as well).

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?

The problem (post #1) says: "Show that $f$ is a linear transformation". By definition, a linear transformation is a mapping between two vector spaces. So its argument as well as return value are vectors, even if in this particular case the set of vectors and the set of scalars are the same.

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?

I don't understand your difficulty with $x$ vs $y$. They are just variables with different names, and none of them has a special role: whenever $x$ is used, one can use $y$ and vice versa. It's like a journalist saying, "The person asked not to be named, so I'll call him Brian". There is nothing that makes Brian better than Eric, the journalist could have equally chosen that name. When we say that $f(x)=mx+b$, it does not make $x$ special; this definition means exactly the same as $f(y)=my+b$. In fact, the only role $x$ fulfills is showing where the argument is used in the body of the definition. The definition really ways, "Whatever is given as an argument, multiply it by $m$ and add $b$".

$\textbf{Function}$
$f(x) = mx + b$

Assumptions:
$b \ne 0$.

But then you say $f(x)=mx$, so you assume $b=0$. In fact, in part (a) of the problem, you should.

Your solution that follows is correct. It is not necessary to show $d\cdot f(y)=f(dy)$ once you showed $c\cdot f(x)=f(cx)$: this is the same statement.

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.

Know your definitions! (In this case, the definition of "linear transformation".)

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.

Don't be afraid of seeing something new, such as a new way to solve a problem. This happens all the time. The only problem is if you encounter a genuinely new concept; then you need to look it up or ask about it.