How to Show F is an Isomorphism for Polynomial Vectors?

[simpledude]

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!

 

[HallsofIvy]

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?

 

[ircdan]

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

 

[simpledude]

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 ?

 

[ircdan]

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.

 

[simpledude]

Ah! Thanks Dan, working it out now :)

 

[simpledude]

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?

 

[ircdan]

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.

 

[simpledude]

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

 

[ircdan]

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.

 

[simpledude]

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?