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Showing F is an isomorphism

  • Thread starter simpledude
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  • #1
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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!
 

Answers and Replies

  • #2
HallsofIvy
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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.
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
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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
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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
229
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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
21
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Ah! Thanks Dan, working it out now :)
 
  • #7
21
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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
229
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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
21
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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
229
0
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
21
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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?
 

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