basis of kernel of derivative LT

i am a little stuck becuase there is so many concepts at once. i know what each of things are by the definitions, but to put them together, i'm not sure of the idea. here is the problem "let p(n) be the real vector space denoting the polynomial functions up to degree n. consider the linear transformation D(f) = d/dx (f). give a basis for the kernel of D. also, what is the rank of the linear transformation D?". now, I know kernel means, once you apply the LT, you get zero. so its the functions where the derivative is zero. but does this mean, for example, all the cooeficcients are zero? or that the whole function goes? so only the constant term, so the basis would just be 1?. i have mostly done vector spaces of matrixes, not really polynomials so i dont know if the rules apply"

Re: basis of kernel of derivative LT

if anyone can help it is great, it is quite time sensitive so please as soon as possible

Re: basis of kernel of derivative LT

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Originally Posted by

**learning** i am a little stuck becuase there is so many concepts at once. i know what each of things are by the definitions, but to put them together, i'm not sure of the idea. here is the problem "let p(n) be the real vector space denoting the polynomial functions up to degree n.

So any polynomial in p(n) is of the form $\displaystyle a_0+ a_1x+ a_2x^2+ \cdot\cdot\cdot+ a_nx^n$.

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consider the linear transformation D(f) = d/dx (f).

Okay, what is the derivative of $\displaystyle a_0, a_x+ a_2x^3+ \cdot\cdot\cdot+ a_nx^n$?

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give a basis for the kernel of D". now, I know kernel means, once you apply the LT, you get zero. so its the functions where the derivative is zero. but does this mean, for example, all the cooeficcients are zero? or that the whole function goes?

If there is a difference between the two? If all the coefficients are 0 then the polynomial value is 0 for any x isn't it? And if the polynomial is 0 for all x, the coefficients must be 0.

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so only the constant term, so the basis would just be 1?

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Yes that's the point. The derivative of $\displaystyle a_0+ a_1x+a_2x^2+ \cdot\cdot\cdot+ a_nx^n$ is $\displaystyle a_1+ 2a_2x^2+ \cdot\cdot\cdot\+ na_nx^{n-1}$. The "0" **function** is the function that has value 0 for all x. For polynomials that is true if and only if all coefficients are 0. So you correct. In order to be "0" all coefficients must be 0: $\displaystyle a_1= a_2= \cdot\cdot\cdot= a_n= 0$. P(x) is in the kernel if and only if [itex]P(x)= a_0[/itex] which is, as you say, a constant times "1".

(Strictly speaking, **any** single, non-zero, number will do as a basis.)

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i have mostly done vector spaces of matrixes, not really polynomials so i dont know if the rules apply"

Re: basis of kernel of derivative LT

what i meant is, would i have to consider all examples where, i plug in values of x= something , and it comes out to zero, there are infinite combinations for this surely. i.e x^3 - 4x when x = 2 etc.

or is x NOT treated as a variable? is the only "zero" when there are no terms left, or solutions to equations = 0 as well?

Re: basis of kernel of derivative LT

does it make sense to write that the kernel = 1? i know any constant is the same since you can just multiple by any constant so any constant spans the kennel. i felt better writing 1 since its analagous to the 'standard' base i.e. for vectors, (1,0,0), (0,1,0), (0,0,1). would it have to be written like {1} since its technically a set. (with dimension 1? and rank(D)=n?) just want to check i have the concepts right since linear algebra isn't my thing at all

Re: basis of kernel of derivative LT

ok i also have to show that the eigenvalue of D is zero,

i know how to find eigenvalue of LT when there is a matrix but for polynomial differentiation LT how do you find the eigenvalue?