Question re: Linear maps

Having started reading a textbook titled Advanced Calculus, I'm stumped by an excercise. There are a few chapters of vector space revision which threw up this question:

Show that the most general linear map from R (the real number system) to R is multiplication by a constant.

Now showing that it is a linear map is easy, showing that T(x) = ax + b is not a linear map is easy, and showing that whole other classes of functions aren't linear maps is easy. What I can't seem to do is prove in full generality that multiplication by a constant is the most general.

Any ideas?

Edit: What I have is that a linear map T from R to R should satisfy
T(xy) = xT(y), T(x + y) = T(x) + T(y)
I can see that multiplication by a constant does satifsy this, but I don't see why no other kind of map should be linear.

Michael Collins

Hmm kind of a weird one. How about this:

All linear mappings [latex] f [/latex] from [latex]{\bf R^{n} \to R^{m} [/latex] may be expressed as

[latex] \displaystyle f(v) = {\bf A}v, \; \forall v \in {\bf R^{n}} [/latex]

where [latex]{\bf A} [/latex] is an m by n matrix.

Now if we think about the n = m = 1 case...we get the required result.

Fremen

MC, that sounds dangerously circular.

OP, I think the first step is probably as follows:

Let T be a linear map from R -> R
Without loss of generality, assume T(1) = c

Now what?

sponsoredwalk

Deleted User wrote: »

Having started reading a textbook titled Advanced Calculus,

Just curious, what book?

Deleted User wrote: »

Show that the most general linear map from R (the real number system) to R is multiplication by a constant.
...
What I can't seem to do is prove in full generality that multiplication by a constant is the most general.

Deleted User wrote: »

but I don't see why no other kind of map should be linear.

I think this stems from the same problem I had on this forum as
regards partial fractions ages ago You have already proven that
the most general form of a linear transformation of the form T : ℝ → ℝ
is a linear map by the argument T(xy) = xT(y), T(x + y) = T(x) + T(y)
but I don't think you understand why the general form of a linear equation
is relevant & how that answers your question. Hopefully these questions
will help you see:

What is the general form of a linear equation?
What is the least general form of a linear equation?
What other forms of linear equation are there?
Can you build a heirarchy of linear equations from least general to most
general?
What is a quadratic equation?
What is the most general form of a quadratic equation?
What is the least general form of a quadratic equation?
Why isn't a quadratic equation usable in a linear map?

Fremen

sponsoredwalk wrote: »

You have already proven that
the most general form of a linear transformation of the form T : ℝ → ℝ
is a linear map by the argument T(xy) = xT(y), T(x + y) = T(x) + T(y)
but I don't think you understand why the general form of a linear equation
is relevant & how that answers your question.

But that's the definition of a linear map. It's not a proof that a one dimensional map is always of the form T(x) = ax. Granted, the proof is two lines, but just saying "it follows from the definition" wouldn't satisfy an examiner.

The proof seemed much simpler once Fremen gave me the first line of two. :pac:

Assume WLOG that T(1) = c, c a real number.
By the fact that T is linear, T(x) = T(1.x) = x.T(1) = cx, for all real numbers x.

Micheal Collins' approach isn't circular, I don't think. It does seem to be using a sledgehammer to crack a nut but I like it!

And sponsoredwalk, the book is 'Advanced Calculus' by Lynn H. Loomis and Shlomo Sternberg, published by Jones and Bartlett publishers although I obtained mine through the free maths ebooks reddit.

Fremen

Well, if it's not circular, you certainly need to do a lot of work to prove that all linear maps are given by matrix multiplication.

Fremen

Fremen wrote: »

Well, if it's not circular, you certainly need to do a lot of work to prove that all linear maps are given by matrix multiplication.

I don't think so.
The way I first saw to derive that result (except restricted to square matrices) is to prove that a linear map is completely determined by how it acts on a basis, and then to construct a bijection between the set of NxN matrices and the set of linear maps.
To do this, all we need to do is to set the column vectors in an NxN matrix to be the image of each basis vector under the linear map.
We then show that given this matrix, we can recover the linear map, which gives us the bijection.

I will grant you that the result that a linear map is completely determined by its action on a basis is why considering T(1) allows us to construct the rest of the linear map in the proof you hinted at, but MC's isn't circular, it just uses more general results.

(Of course, since we're setting m=n=1, it's fine that the proof I sketched only works for square matrices).

sponsoredwalk

Fremen wrote: »

But that's the definition of a linear map. It's not a proof that a one dimensional map is always of the form T(x) = ax. Granted, the proof is two lines, but just saying "it follows from the definition" wouldn't satisfy an examiner.

Well yes, I was just more focused on the issue about understanding what
the most general form of a linear equation possible here is. It seems your
proof, which I assume was the one given, was different from the way I
was viewing this. Because λx is the most general linear equation possible,
(which can be shown by contradiction by taking λx + b), we just take
T(λx + y) = λT(x) + T(y) but because we're on ℝ this instantly implies
that T(y) = μT(x) so T(λx + y) = λT(x) + T(y) = λT(x) + μT(x) = (λ + μ)T(x).
I don't think that's assuming anything (other than the provable fact that T(λx + y)
also satisfies the linearity property which was just timesaving even though writing this equates
to about the same time if not longer but too late now :(:(), is it?

---

Conor, what a scary book! It's free on Sternberg's website. I only read
the first chapter on logic like last week & it cleared up some stupid
thing I'd read in another book. It's on the list of advanced calculus books
I am working on but it's probably the most neglected of them all at the
moment. This might get me started on it again :cool:

Eliot Rosewater

sponsoredwalk wrote: »

Because λx is the most general linear equation possible,

Well, all you've proved is that ax+b isn't a linear map: you haven't proved that ax is the most general. To do that you'd also have to do a contradiction proof on every other function around - x^n, e^x, sin(x), etc - and prove that every single one of those isn't linear either.

sponsoredwalk

Eliot Rosewater wrote: »

Well, all you've proved is that ax+b isn't a linear map: you haven't proved that ax is the most general. To do that you'd also have to do a contradiction proof on every other function around - x^n, e^x, sin(x), etc - and prove that every single one of those isn't linear either.

But none of those functions satisfies the definition of what a linear
equation is, none of those are "linear". All of those are defined as
nonlinear:

Since terms of linear equations cannot contain products of distinct or
equal variables, nor any power (other than 1) or other function of a
variable, equations involving terms such as xy, x2, y1/3, and sin(x)
are nonlinear.
link

That's why I emphasized an understanding of what a linear equation is
in my first post, it just falls out from this. I didn't even bother writing
anything further, such as T(x) = ax + b because, as conor already pointed
out, this can instantly be shown not to hold.

Eliot Rosewater

sponsoredwalk wrote: »

But none of those functions satisfies the definition of what a linear
equation is, none of those are "linear".

Well, yes, that's what we have to prove. The paragraph you quote merely states the fact without any justification.

EDIT: didn't mean to sound ratty here by the way!

sponsoredwalk

sponsoredwalk wrote: »

But none of those functions satisfies the definition of what a linear
equation is, none of those are "linear". All of those are defined as
nonlinear:
That's why I emphasized an understanding of what a linear equation is
in my first post, it just falls out from this. I didn't even bother writing
anything further, such as T(x) = ax + b because, as conor already pointed
out, this can instantly be shown not to hold.

I think you're working with the wrong definition of linearity, sponsoredwalk. Eliot Rosewater is right.
You see, the definition of a linear map T on a vector space V over a field F is one such that T(x + y) = T(x) + T(y) for x, y in V
and such that
T(ax) = aT(x) for x in V, a in F.

So in the case where T:R->R,
T(x + y) = T(x) + T(y) and
T(xy) = xT(y) for x and y real numbers.

We have to show that T(x) = ax is the only function that works, so you see, we can't just do this by contradiction, as we'd need to contradict every possible T(x), and there are infinitely many of these.

sponsoredwalk

Yeah you guys are right, honestly have it stuck in my head that those
functions are defined as nonlinear because I keep reading it stated as
fact. I'm actually over the moon because of this, thank you! I honestly
don't know how I could read sentences defining those functions as nonlinear
without asking why that's justifiable!

But think about it, if these functions were actually defined as nonlinear
as I was unfortunately led to believe then I'd be right (right?)

sponsoredwalk

sponsoredwalk wrote: »

Yeah you guys are right, honestly have it stuck in my head that those
functions are defined as nonlinear because I keep reading it stated as
fact. I'm actually over the moon because of this, thank you! I honestly
don't know how I could read sentences defining those functions as nonlinear
without asking why that's justifiable!

But think about it, if these functions were actually defined as nonlinear
as I was unfortunately led to believe then I'd be right (right?)

Yes, you would be right. The reason people chose the definition we gave over the definition you gave is that it holds for a general vector space rather than just for real numbers.

Basically, the question turned out to really be "show that our definition of linear in terms of vector spaces actually gives us back our definition of linear in terms of real numbers if we let the two vector spaces be the set of real numbers".