Is p(x) + p(2) a Linear Transformation in P_3?

[HMPARTICLE]

Homework Statement

t:P_3 -----> P_3
p(x) |---> p(x) + p(2)

Determine whether or not this function is linear transformation or not.

Homework Equations

For a function to be a linear transformation then t(0) = 0 , there are other axioms that must be satisfied, but that is not the problem.

The Attempt at a Solution

I don't see how this is a linear transformation, since the zero element of P_3 is where p(x) = 0, and t(p(x)) = t(0) = 0 + p (2) ?? surely?

could someone explain why i am totally incorrect please.

 

[HallsofIvy]

Homework Statement

t:p_3 -----> P_3

You have a "special character" here. Please say, in words, what this means and what "P_3" means- projective space or polynomials?

p(x) |---> p(x) + p(2)

Determine whether or not this function is linear transformation or not.

Homework Equations

For a function to be a linear transformation then t(0) = 0 , there are other axioms that must be satisfied, but that is not the problem.

The Attempt at a Solution

I don't see how this is a linear transformation, since the zero element of P_3 is where p(x) = 0, and t(p(x)) = t(0) = 0 + p (2) ?? surely?

could someone explain why i am totally incorrect please.

Unless you are using some special spaces, you are NOT "totally incorrect". What exactly was your answer and what makes you thing you are wrong?

 

[Fredrik]

It's not t(p(x)), it's t(p)(x), because t takes a function to a function, and that function takes a number as input. $t(p)(x)=(p+p(2))(x)=p(x)+p(2)$. Other than that, you seem to have the right idea about the problem. If we denote the zero polynomial by z, is t(z)=z?

 

[HMPARTICLE]

Hey, guys sorry about not replying sooner.
I understand what you are saying, the problem i have with the question is that, for a linear transformation to be a linear transformation then the transformation has to satisfy t(z) = z where z is the zero vector as Fredrik points out.

so it is, t(p)(x) and if the zero vector in P_3 is z, such that p(x)= z

then t(p)(x) = t(z) = z.

I'm going to have a go at trying to explain why. for the function t to map a polynomial p(x) to p(x) + p(2), it must be a polynomial? for example if we had t(p)(0) = p(0) + p(2).

 

[PeroK]

Hey, guys sorry about not replying sooner.
I understand what you are saying, the problem i have with the question is that, for a linear transformation to be a linear transformation then the transformation has to satisfy t(z) = z where z is the zero vector as Fredrik points out.

so it is, t(p)(x) and if the zero vector in P_3 is z, such that p(x)= z

then t(p)(x) = t(z) = z.

I'm going to have a go at trying to explain why. for the function t to map a polynomial p(x) to p(x) + p(2), it must be a polynomial? for example if we had t(p)(0) = p(0) + p(2).

I would take this opportunity to denote a polynomial by a single letter (e.g. "p"), rather than "p(x). I think this will help clarify your thinking and make your statements more precise. So, use p(x) only when you mean the polynomial value at x; and, use p to denote the polynomial.

Note that p(2) can then be a number (p evaluated at 2) or the constant polynomial with value p(2). You should make this distinction explicit in your work as well.

 

[Fredrik]

so it is, t(p)(x) and if the zero vector in P_3 is z, such that p(x)= z

I'm guessing that you think of p(x) as a function. Don't. It's a "function of x" in the sense that its value is determined by the value of x, but it's not a function. p is the function. p(x) is a number in its range.

Even if I assume that you intended for p(x) to denote a function, I still don't understand what you were trying to say.

then t(p)(x) = t(z) = z.

I don't understand why you wrote down these equalities. The first one says that a number is equal to a polynomial, so it can't be right. (Also, it says that t(p)(x) is equal to something that's independent of p. That should make you suspicious of your result). Did you do use the definition of t to conclude that the second one holds?

 

[Fredrik]

Here's a suggestion for a notation for constant functions. For each $r\in\mathbb R$, let $c_r$ be the function defined by $c_r(x)=r$ for all $x\in\mathbb R$. Now the zero polynomial can be denoted by $c_0$. What does the definition of t tell you about $t(p)$, where p is an arbitrary polynomial? What does it tell you about $t(c_0)$? Is it equal to $c_0$?

Also, I need to make sure that you understand an essential fact about functions. Do you know what it means to say that two functions are equal? If f and g are functions and I say that f=g, what does that mean?

 

[HMPARTICLE]

Ahh i see, i knew that p(x) is a number and that p is a function, my silly excuse of notation.

I should have done this in the first place, this is the question, the first part of the answer is the second image. that is what i do not understand... or didn't understand

The solution adds that the zero element of P_3 is p(x) = 0.

my confusion comes from that image on the right. It seems so simple but i just can't get my head around it! I sort of understand it, but I'm not the student that likes to 50-90 percent understand things :(.

what my solution to this was;

t(0) = 0 + p(2)

This is no the zero polynomial, so t is not a linear transformation.

I still don't 100% know why this is not true.

What does the definition of t tell you about t(p)

That t is a function that maps polynomials to polynomials. specifically that t maps polynomials of greatest power 2 to themselves plus p(2).

Also, I need to make sure that you understand an essential fact about functions. Do you know what it means to say that two functions are equal? If f and g are functions and I say that f=g, what does that mean?

suppose that f and g are functions such that f:X -> Y and g:X -> Y, then f = g if for each x in X and y in Y, f(x) = g(x) = y.

That sounds about right... i think. so if f and g are equal, each element in the domain of f and g get mapped to the same element in the co-domain by f and g.

 

[Fredrik]

suppose that f and g are functions such that f:X -> Y and g:X -> Y, then f = g if for each x in X and y in Y, f(x) = g(x) = y.

That sounds about right... i think. so if f and g are equal, each element in the domain of f and g get mapped to the same element in the co-domain by f and g.

There's no need to involve a second variable (the y in your statement) but you seem to have understand f=g correctly. It means that f and g have the same domain, and that f(x)=g(x) for all x in that domain. In the problem you're working out, all functions have domain $\mathbb R$, so if you want to check that two polynomials f and g are equal, you have to check if f(x)=g(x) for all real numbers x.

So how do you check if $t(c_0)=c_0$? If you prefer to use the book's notation, $\mathbf 0$ instead of $c_0$, that's fine. So how do you check if $t(\mathbf 0)=\mathbf 0$?

Ahh i see, i knew that p(x) is a number and that p is a function, my silly excuse of notation.

I think that a lot of problems are very hard to work out when you don't distinguish between f and f(x), but I don't think you could have been expected to pick the best notation from the start. The difference between f and f(x), and the difference between "is a function" and "is a function of", are both not explained well in math books or by math teachers. Even the actual problem statement fails to make the distinction between f and f(x). It says that t takes p(x) to p(x)+p(2), but what they mean is that we have $t(p)(x)=p(x)+p(2)$ for all $p\in P_3$ and all $x\in\mathbb R$. The right-hand side is equal to $(p+c_{p(2)})(x)$, so t is actually taking each p to $p+c_{p(2)}$. (Note that I'm using the definition of addition of functions, which says that for all X and all $f,g:X\to\mathbb R$, (f+g)(x)=f(x)+g(x) for all $x\in X$).

It's not necessary to use the notation for constant functions that I suggested in my previous post. If it doesn't confuse you, you can just write $r$ instead of $c_r$. This is standard. In this notation, we have $t(p)(x)=p(x)+p(2)=p(x)+p(2)(x)=(p+p(2))(x)$ for all p and all x, and therefore $t(p)=p+p(2)$ for all p. However, I think it would be a bit confusing to denote the zero polynomial by 0, so I would prefer to use a notation like z for it. The book uses $\mathbf 0$. That's OK too. Just don't use a plain (not bold) $0$.

Ahh i see, i knew that p(x) is a number and that p is a function, my silly excuse of notation.
The solution adds that the zero element of P_3 is p(x) = 0.

If I didn't already know what the zero element of $P_3$ is, I wouldn't have been able to make sense of that sentence. The zero element of $P_3$ is a polynomial, but you wrote down an equality between two numbers. What you should be saying is that the zero element in $P_3$ is the $p\in P_3$ such that $p(x)=0$ for all $x\in\mathbb R$. If we denote it by z (just because it's easier to type than $\mathbf 0$), then we have $p+z=z+p=p$ for all $p\in P_3$. Since these are equalities between functions with domain $\mathbb R$, this means that $(p+z)(x)=(z+p)(x)=p(x)$ for all $x\in\mathbb R$. By definition of the addition operation on $P_3$ (the sum of two functions), this means that $p(x)+z(x)=z(x)+p(x)=p(x)$ for all $x\in\mathbb R$. We know that these equalities hold because $z(x)=0$ for all $x\in\mathbb R$.

what my solution to this was;

t(0) = 0 + p(2)

Why is there a p on the right-hand side?

That t is a function that maps polynomials to polynomials. specifically that t maps polynomials of greatest power 2 to themselves plus p(2).

That's correct, but I was hoping that you would say that this means that $t(p)=p+p(2)$ for all $p\in P_3$, and that this means that $t(p)(x)=(p+p(2))(x)=p(x)+p(2)$ for all $p\in P_3$ and all $x\in\mathbb R$. Then you can use this to evaluate things like $t(\mathbf 0)$.

 

[HMPARTICLE]

So how do you check if t(c0)=c0t(c_0)=c_0? If you prefer to use the book's notation, 0\mathbf 0 instead of c0c_0, that's fine. So how do you check if t(0)=0t(\mathbf 0)=\mathbf 0?

using $t(p)(x)= ( p+p(2))(x)=p(x)+p(2)$ this makes a lot more sense.
since the zero element of $P_3$ is $p = \textbf{0}$ and using $t(p)(x)=(p+c_{p(2)})(x)$ then $t( \textbf{0})=( \textbf{0}+c_{ \textbf{0}}) = \textbf{0}$ Would that be correct? close ?

or
$t( \textbf{0})(x) =( \textbf{0}+c_{ \textbf{0}})(x) = \textbf{0}$ ??I see where you are coming from with books not explaining the difference between p and p(x), this thread has been a valuable lesson!

the reason for the p on the right hand side of this is that i was taking p(0) as the zero element or something, i can't quite remember;
$t( \textbf{0} ) = \textbf{0} + p(2)$ I hope i am going in the right direction!

 

[Fredrik]

using $t(p)(x)= ( p+p(2))(x)=p(x)+p(2)$ this makes a lot more sense.
since the zero element of $P_3$ is $p = \textbf{0}$

No need to say $p=\mathbf 0$. Just say that you're denoting the zero element by $\mathbf 0$. If you want to also mention what element of $P_3$ is the zero element, you can add that it's the $p\in P_3$ such that $p(x)=0$ for all $x\in\mathbb R$.

and using $t(p)(x)=(p+c_{p(2)})(x)$ then $t( \textbf{0})=( \textbf{0}+c_{ \textbf{0}}) = \textbf{0}$ Would that be correct? close ?

It's close. If we use the notation I suggested for constant functions, we have $t(p)=p+c_{p(2)}$ for all $p\in P_3$, and in particular $t(c_0)=c_0+c_{c_0(2)} =c_0+c_0=c_0$. The last equality holds because for all $x\in\mathbb R$, we have $(c_0+c_0)(x)=c_0(x)+c_0(x) =0+0=0=c_0(x)$. It's probably easier to follow what we're doing if we include an x from the start. For all $x\in\mathbb R$, we have
$$t(c_0)(x)=(c_0+c_{c_0(2)})(x) = (c_0+c_0)(x)=c_0(x)+c_0(x)=0+0=c_0(x).$$ This implies that $t(c_0)=c_0$.

When we write $t(p)=p+p(2)$ instead of using the notation above, the sum on the right should be interpreted as $p+p(2)c_1$. So for all $x\in\mathbb R$, we have
$$t(\mathbf 0)(x)=(\mathbf 0+\mathbf 0(2))(x) =(\mathbf 0+\mathbf 0(2)c_1)(x) =(\mathbf 0+0c_1)(x) =\mathbf 0(x)+(0c_1)(x) =\mathbf 0(x)+0c_1(x) =0+0\cdot 1=0 =\mathbf 0(x).$$ I realize now that it's not always useful to type $\mathbf 0$ instead of $0$, because when it's clear from the context that we're talking about a function, $0$ will be interpreted as $0 c_1$, which is equal to $\mathbf 0$.

Note that these conventions allow us to write $t(0)=0+0(2)=0+0=0$.

 

[Fredrik]

Note that if we had found that $t(\mathbf 0)\neq \mathbf 0$, we could have immediately concluded that t isn't linear. But the result $t(\mathbf 0)=\mathbf 0$ tells us nothing about whether t is linear. So you still have work to do.

 

[HMPARTICLE]

ah yes! i know.
Thats where it went all wrong with my initial solution!

I've done well with most of the questions up until this one.
To prove that t is a linear transformation;
I have to prove that for all p(x) and q(x) in $P_3$ we have $t( p(x) + q (x) ) = t (p(x)) + t(q(x))$ which is satisfied.
and also that $at(p(x)) = t(ap(x))$ for all p(x) in $P_3$ and all real numbers a.
This is also satisfied.

so vector addition under t and scalar multiplication.

 

[PeroK]

ah yes! i know.
Thats where it went all wrong with my initial solution!

I've done well with most of the questions up until this one.
To prove that t is a linear transformation;
I have to prove that for all p(x) and q(x) in $P_3$ we have $t( p(x) + q (x) ) = t (p(x)) + t(q(x))$ which is satisfied.
and also that $at(p(x)) = t(ap(x))$ for all p(x) in $P_3$ and all real numbers a.
This is also satisfied.

so vector addition under t and scalar multiplication.

No, no, no! For all p and q in $P_3$ we have t(p + q) = t(p) + t(q) etc.

t does not operate on polynomial values!

 

[HMPARTICLE]

Thanks!
The textbook I'm using uses this t(p(x)+q(x))=t(p(x))+t(q(x)) ??
But i can see what its implying is incorrect, t operates on polynomials not polynomial values.

 

[PeroK]

Thanks!
The textbook I'm using uses this t(p(x)+q(x))=t(p(x))+t(q(x)) ??
But i can see what its implying is incorrect, t operates on polynomials not polynomial values.

I'd say your textbook is making it difficult for you to take the step of working with operators of functions. This extra level of abstraction is not easy when you first encounter it. I think you see this now. Whoever wrote your book is being a bit lazy, IMHO.

 

[Fredrik]

So do you see what $t(p+q)(x)$ is? Or $t(ap)(x)$, where $a\in\mathbb R$? I have to go to bed, but perhaps PeroK will be awake a bit longer.

 

[HMPARTICLE]

using my textbooks notation i would say;

$t(p(x) + q (x)) = p(x) + q (x) + p (2) + q(2) = p(x) + p(2) + q(x) + q(2)$
but using the notation we have been using throughout this thread(and now actually prefer);

$t(p+q)(x) = ( p + q + c_{p(2)} + c_{q(2)})(x) = ( p + q)(x) + (c_{p(2)} + c_{q(2)})(x) = p(x) + q(x) + c_{p(2)}(x)+c_{q(2)}(x)$

then $t (ap)(x) = (ap + ac_{p(2)})(x)$, this notation is new to me, so i also want to say
$t (ap)(x) = (ap + c_{ap(2)})(x) = ap(x) + c_{ap(2)}(x)$. The second one looks like its the correct one.

 

[PeroK]

using my textbooks notation i would say;

$t(p(x) + q (x)) = p(x) + q (x) + p (2) + q(2) = p(x) + p(2) + q(x) + q(2)$
but using the notation we have been using throughout this thread(and now actually prefer);

$t(p+q)(x) = ( p + q + c_{p(2)} + c_{q(2)})(x) = ( p + q)(x) + (c_{p(2)} + c_{q(2)})(x) = p(x) + q(x) + c_{p(2)}(x)+c_{q(2)}(x)$

then $t (ap)(x) = (ap + ac_{p(2)})(x)$, this notation is new to me, so i also want to say
$t (ap)(x) = (ap + c_{ap(2)})(x) = ap(x) + c_{ap(2)}(x)$. The second one looks like its the correct one.

You could have used the definition of t more directly:

$t(p+q)(x) = (p+q)(x) + (p+q)(2)$

That's from the definition of t.

Also, you need to end up showing that this is equal to $t(p)(x) + t(q)(x)$

You also need to be able to see (ap) as a polynomial. And note the difference between:

$(ap)(x)$ and $ap(x)$

In one case a is a scalar acting on a vector space of polynomials. In the the other a is multiplying another real number. From the definition of t we have:

$t(ap)(x) = (ap)(x) + (ap)(2)$

And, you need to show this is equal to at(p)(x).