Proving the Non-Isomorphism of D(p(x))

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In summary: B.Yes it is like that before that I'm asked to find D with respect to the base B = { p1=1, p2=x, p3=x^2 and p^4 =x^3 }I already did that, then I'm asked to prove that D is not an isomorphism.
  • #1
Jimmy84
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Homework Statement



I have D(p(x)) = (the second derivative of p with respect to x ) - (2 derivative of p with respect to x) + p

Proof that D(p(x)) is not an isomorphism

Homework Equations





The Attempt at a Solution



Just by watching the problem it seems I can assign one unique value to D(p(x)) and receive an unique image. It seems by inspection that it is one to one. It also seems onto by inspection.
 
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  • #2
What are the domain and codomain??
Isomorphism of what?? Vector spaces, groups, rings, modules?
 
  • #3
micromass said:
What are the domain and codomain??
Isomorphism of what?? Vector spaces, groups, rings, modules?

Sorry, It is a linear Operator D from P3 to P3 where P3 is the polynomial space of degree equal or less than 3.
 
  • #4
Did you check whether it is injective or surjective??
 
  • #5
can I say that it is not onto because it is not defined for degrees greater than 3?
 
  • #6
Jimmy84 said:
can I say that it is not onto because it is not defined for degrees greater than 3?

No, degrees greater than 3 don't matter because they are not in the domain or codomain.
 
  • #7
micromass said:
Did you check whether it is injective or surjective??

How can I start doing that?
 
  • #8
Jimmy84 said:
How can I start doing that?

Do you know about the matrix form of a linear operator?? That might help you.
 
  • #9
micromass said:
Do you know about the matrix form of a linear operator?? That might help you.

No I haven't, how can I do that, and where can I find some more information about it? that might come useful for nexts week test. I would appreciate it
 
  • #10
Jimmy84 said:
No I haven't, how can I do that, and where can I find some more information about it? that might come useful for nexts week test. I would appreciate it

Oh well, if you haven't seen it yet, then you probably aren't allowed to use it. You'll see it when the time comes!

Anyway, can you calculate the kernel of the linear map?? Find out which p are such that D(p)=0??
 
  • #11
micromass said:
Oh well, if you haven't seen it yet, then you probably aren't allowed to use it. You'll see it when the time comes!

Anyway, can you calculate the kernel of the linear map?? Find out which p are such that D(p)=0??

well I guess that would be just P = 0 because there is a + p , that's the reason why I thought it would be injective.
 
  • #12
Hmm, yes, it does turn out to be an isomorphism... Weird...
 
  • #13
micromass said:
Hmm, yes, it does turn out to be an isomorphism... Weird...

Are you sure? because the problem states that it is not :P
 
  • #14
how did you find out it was onto just by inspection or is there a another way?
 
  • #15
Something is an isomorphism if there exists a linear bijective transformation T such that :

T(T-1) = Id

Where Id is the identity transformation ( The do nothing transformation ).

So your question is abit vague, but you have a transformation :

D(p(x)) = p''(x) - 2p'(x) + p ?

You phrased it terribly so I am guessing that's what you meant? Also is p a real number?
 
  • #16
Zondrina said:
Something is an isomorphism if there exists a linear bijective transformation T such that :

T(T-1) = Id

Where Id is the identity transformation ( The do nothing transformation ).

So your question is abit vague, but you have a transformation :

D(p(x)) = p''(x) - 2p'(x) + p ?

You phrased it terribly so I am guessing that's what you meant? Also is p a real number?

Yes it is like that before that I'm asked to find D with respect to the base B = { p1=1, p2=x, p3=x^2 and p^4 =x^3 }

I already did that, then I'm asked to prove that D is not an isomorphism I guess p can be a real number in P3 which is the space of polynomials less or equal to 3.
 
  • #17
Jimmy84 said:
Yes it is like that before that I'm asked to find D with respect to the base B = { p1=1, p2=x, p3=x^2 and p^4 =x^3 }

What was your answer to this question?
 
  • #18
Zondrina said:
So your question is abit vague, but you have a transformation :

D(p(x)) = p''(x) - 2p'(x) + p ?

You phrased it terribly so I am guessing that's what you meant? Also is p a real number?

I believe p is an element of P3, i.e. it is a polynomial of degree <= 3. This means p could be a real number (which is a polynomial of degree 0), but it doesn't have to be.
 
  • #19
Jimmy84 said:
Are you sure? because the problem states that it is not :P
I get that it's an isomorphism too.

Jimmy84 said:
Yes it is like that before that I'm asked to find D with respect to the base B = { p1=1, p2=x, p3=x^2 and p^4 =x^3 }
What did you mean by this? It sounds like you found the matrix representing D relative to B, but you told micromass you didn't know how to find the matrix representation of a linear operator.
 
  • #20
vela said:
I get that it's an isomorphism too.What did you mean by this? It sounds like you found the matrix representing D relative to B, but you told micromass you didn't know how to find the matrix representation of a linear operator.

Well I'm a bit confused because I'm taking this subject for the first time but I know how to find a linear transformation with respect to a given base.

For D with respect to the basis B = { p1=1, p2=x, p3=x^2 and p^4 =x^3 } the result was D relative to B =
1 -2 2 0
0 1 -4 6
0 0 1 6
0 0 0 1
 
  • #21
Okay, that matrix is what micromass was referring to earlier. If its determinant isn't 0, it's invertible, which tells you D is bijective.

Could just be a typo, but I got a -6 in the third row.
 
  • #22
vela said:
Okay, that matrix is what micromass was referring to earlier. If its determinant isn't 0, it's invertible, which tells you D is bijective.

Could just be a typo, but I got a -6 in the third row.

Yes it is -6 thanks a lot for your time.
 

1. What is an isomorphism?

An isomorphism is a mathematical concept that describes a one-to-one correspondence between two mathematical structures, such as groups, rings, and vector spaces. This means that the structures have the same number of elements and the same operations, and they can be mapped onto each other in a way that preserves their structure and properties.

2. How do you prove an isomorphism?

To prove that two structures are isomorphic, you need to show that there exists a bijective function (a function that is both injective and surjective) between them. This means that every element in one structure has a unique match in the other structure, and vice versa. Additionally, you need to demonstrate that this function preserves the structure and operations of the structures.

3. What are some common techniques for proving an isomorphism?

One common technique is to show that the structures have the same cardinality (number of elements) and then construct a bijective function between them. Another approach is to demonstrate that the structures have the same algebraic properties, such as the same number of subgroups or the same dimension. You can also use specific properties of the structures, such as the existence of an identity element, to prove an isomorphism.

4. Can two structures be isomorphic but look different?

Yes, two structures can be isomorphic even if they look different. For example, the group of integers modulo 3 (Z3) and the group of rotations of an equilateral triangle (D3) are isomorphic, but they have different elements and operations. This is because the structures have the same algebraic properties and can be mapped onto each other in a way that preserves these properties.

5. Are there any limitations to proving an isomorphism?

One limitation is that two structures can only be proven to be isomorphic if they have the same cardinality. This means that if one structure has more elements than the other, they cannot be isomorphic. Additionally, proving an isomorphism can be a complex and time-consuming task, especially for more complicated structures, so it may not always be feasible to prove an isomorphism.

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