micromass said:What are the domain and codomain??
Isomorphism of what?? Vector spaces, groups, rings, modules?
Jimmy84 said:can I say that it is not onto because it is not defined for degrees greater than 3?
micromass said:Did you check whether it is injective or surjective??
Jimmy84 said:How can I start doing that?
micromass said:Do you know about the matrix form of a linear operator?? That might help you.
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
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??
micromass said:Hmm, yes, it does turn out to be an isomorphism... Weird...
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?
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 }
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 get that it's an isomorphism too.Jimmy84 said:Are you sure? because the problem states that it is not :P
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.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 }
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.
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.
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.
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.
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.
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.
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.