How to Show F is an Isomorphism for Polynomial Vectors?

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In summary: I come in. :)In summary, F is a nonsingular linear operator, and F^-1 exists and is the inverse of F.
  • #1
simpledude
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Homework Statement


Let V = P2(R) be the vector space of all polynomials P : R −> R that have order less
than 2. We consider the mapping F : V −> V defined for all P belonging to V , by

F(P(x)) = P'(x)+P(x) where
P'(x) denotes the first derivative of the polynomial P.

Question is: Show that F is an isomorphism from V into V

The Attempt at a Solution



So first I showed that F is a linear operator. Now I have to show Ker F={0}
However, when I start to solve the equation, I get lost at solving
P'(x) + P(x) = 0

I know this is a basic first order linear equation, can anyone point me in the right direction?

Thanks!
 
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  • #2
simpledude said:

Homework Statement


Let V = P2(R) be the vector space of all polynomials P : R −> R that have order less
than 2.
Do you mean degree? I don't know what you mean by the order of a polynomial. If you mean degree, then P2 is just the functions of the form p(x)= ax+ b.

We consider the mapping F : V −> V defined for all P belonging to V , by

F(P(x)) = P'(x)+P(x) where
P'(x) denotes the first derivative of the polynomial P.

Question is: Show that F is an isomorphism from V into V

The Attempt at a Solution



So first I showed that F is a linear operator. Now I have to show Ker F={0}
However, when I start to solve the equation, I get lost at solving
P'(x) + P(x) = 0

I know this is a basic first order linear equation, can anyone point me in the right direction?

Thanks!
Not only is it a basic equation, it is even more basic applied to P2!
F(ax+ b)= (ax+b)'+ (ax+b)= ax+ (a+b). If F(ax+b)= ax+ (a+b)= 0 for all x, then a= 0 and a+b= 0. What does that tell you?
 
  • #3
first, i think mean to say degree and not order
second, are you sure you don't mean, P_2(R) is the vector space of all polynomials of degree at most 2?

Let's assume this is what is meant.

so you have,

F(P(x)) = P'(x) + P(x).

so suppose F(P(x)) = 0, so
P'(x) + P(x) = 0, and this where you are stuck.{1, x, x^2} is a basis for P_2(R), so we can write P(x) = a + bx + cx^2 for some a, b, c in R.Now plug P(x) into P'(x) + P(x) = 0, and "equate the coefficients", you should get that a = b = c = 0, so P(x) = 0

edit: woops halls posted just as I did, i'll leave mine though just incase hehe
 
  • #4
Yes I did mean degree, sorry that's how we refer to it in our class :)

Why can't I just solve P'(x) + P(x) = 0 for p(x) explicitly?
Then get something of the form p(x) = A e^-x ?
 
  • #5
simpledude said:
Yes I did mean degree, sorry that's how we refer to it in our class :)

Why can't I just solve P'(x) + P(x) = 0 for p(x) explicitly?
Then get something of the form p(x) = A e^-x ?

P is a polynomial remember.
 
  • #6
Ah! Thanks Dan, working it out now :)
 
  • #7
Ok so I got a0 = a1 = a2 = 0
(I used that instead of a,b,c)

So can I say, since the only way to obtain {0} with this equation is
by a0=a1=a2=0, we see that F is indeed nonsingular?
 
  • #8
simpledude said:
Ok so I got a0 = a1 = a2 = 0
(I used that instead of a,b,c)

So can I say, since the only way to obtain {0} with this equation is
by a0=a1=a2=0, we see that F is indeed nonsingular?

a_0 = a_1 = a_2 = 0, means P(x) = 0, so kerF = {0}, so F is injective, then it surjective because ...(i'll let you fill this in)...


Also, i guess P_2(R) was suppose to be all polys with degree <= 2 and not < 2? Make sure to check this because if it is < 2 as you say, then do as halls did.
 
  • #9
Yeah, I just cleared it up, it is degree < 2.
The approach is the same, so thanks Halls/Dan.
I am having some issues with another part of the question, which asks to find an inverse mapping F^-1

I understand that since F is nonsingular we can find F^-1 : V-->V
Furthermore, since the dimensions of V and V are the same, F^-1 does exist.
So that's F^-1 ( P'(x) + P(x) ) = P(x)However how can I find the actual mapping?
EDIT: What I got so far is that F^-1 ( P'(x) + P(x) ) = P(x) can be written as
F^-1(a0 + a1 + a1x) = a0+a1x
 
Last edited:
  • #10
Ok so you need to find F^-1, let's call it G.

So when you get a problem like this and you really can't figure out how to solve for the inverse, always remember this fact.
G is the inverse of F means GF = i_d and FG = i_d , where i_d is the identity map, ie, i_d:V->V is the linear transformation given by i_d(ax + b) = ax + b.

So let's work backwards by assuming we already have found G = F^-1. So we have that,

G(F(ax + b)) = ax + b, and
F(G(ax + b)) = ax + b, let's work with this one, so
we need G such that

F(G(ax + b)) = ax + b. We want to find G, assume G(ax + b) = cx + d. Now write out what this means F(G(ax + b)) = ax + b and solve for c and d in terms of a and b and you have G.

Then check that
G(F(ax + b)) = ax + b
F(G(ax + b)) = ax + b

with the G you have found to make sure your answer is correct.
 
Last edited:
  • #11
Hi Dan, thanks for the last post.

So I worked on the problem this morning, and when I get to the part
F(G(a+bx)) = a+bx

Assuming G(a+bx)=c+dx

We can write this as F(c+dx) = a + bx

So now is where I get lost, since if I solve for cx + d = a + bx
I just get c = b and d=a ...

How can I write c and d in terms of a and b?
 

Related to How to Show F is an Isomorphism for Polynomial Vectors?

1. What does it mean for F to be an isomorphism?

An isomorphism is a mathematical concept that describes a one-to-one correspondence between two mathematical structures. In the context of showing F is an isomorphism, it means that the function F preserves the structure and properties of the two structures being compared.

2. How do you prove that F is an isomorphism?

To prove that F is an isomorphism, you must show that it is both injective (one-to-one) and surjective (onto). This means that each element in the first structure has a unique match in the second structure, and vice versa. You must also show that F preserves the operations and properties of the two structures being compared.

3. What are some common examples of isomorphisms?

Some common examples of isomorphisms include the isomorphism between addition and multiplication in the real numbers, the isomorphism between rotations and reflections in geometry, and the isomorphism between a vector space and its dual space in linear algebra.

4. Can two structures be isomorphic if they have different elements?

No, two structures cannot be isomorphic if they have different elements. Isomorphism requires a one-to-one correspondence between the elements of the two structures being compared. If the elements are different, they cannot be matched in a one-to-one manner.

5. How is the concept of isomorphism used in science?

The concept of isomorphism is used in various fields of science, including mathematics, physics, chemistry, and biology. In mathematics, isomorphisms are used to compare and classify different mathematical structures. In physics, isomorphisms are used to describe the relationship between different physical systems. In chemistry and biology, isomorphisms are used to understand the similarities and differences between different molecules and organisms.

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