The discussion revolves around proving that the linear operator D(p(x)) defined as the second derivative of p with respect to x minus twice the first derivative of p plus p itself is not an isomorphism. The context is within the polynomial space P3, which includes polynomials of degree less than or equal to 3.
The discussion is ongoing, with participants examining various aspects of the operator D. Some have suggested calculating the kernel of the linear map, while others have expressed confusion about the implications of their findings. There is no explicit consensus on whether D is an isomorphism, as differing opinions are present.
Participants note that the operator D is defined for polynomials of degree up to 3, and there is a mention of a matrix representation that may indicate whether D is bijective based on its determinant. There is also a reference to homework constraints that may limit the use of certain concepts.
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.