What is the dim(N(T))?

  • Thread starter Mr.Tibbs
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Define ne a linear operator T : Pn [itex]\rightarrow[/itex] Pn by
T(p(x)) = [itex]\frac{d^{k}x}{dx^{n}}[/itex]p(x)
where 1 [itex]\leq[/itex]k [itex]\leq[/itex]n. Show that dim(N(T)) = k.

i. The null space of a linear operator is all vectors v [itex]\in[/itex] V such that T(v) = 0

ii. The dimension of a vector space V, is the number of vectors in any basis of V.

My logical understanding is that the null space of T is any k derivative that gives 0 i.e. k is bigger than n. Thus the dimension has k number of possible vectors. My problem is I don't know how to write the proof. . . I'm assuming that I just have to write the null space as a summation of derivatives from 1 to k thus the dimension of the null is k (the largest possible number).
 

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LCKurtz
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Define ne a linear operator T : Pn [itex]\rightarrow[/itex] Pn by
T(p(x)) = [itex]\frac{d^{k}x}{dx^{n}}[/itex]p(x)
where 1 [itex]\leq[/itex]k [itex]\leq[/itex]n. Show that dim(N(T)) = k.

I assume you mean ##\frac {d^k x}{dx^k}p(x)##. Or for easy display, ##D^kp(x)##.

i. The null space of a linear operator is all vectors v [itex]\in[/itex] V such that T(v) = 0

ii. The dimension of a vector space V, is the number of vectors in any basis of V.

My logical understanding is that the null space of T is any k derivative that gives 0

No. The null space isn't "any k derivative". The null space is a subspace of your nth degree polynomials.

i.e. k is bigger than n. Thus the dimension has k number of possible vectors. My problem is I don't know how to write the proof. . . I'm assuming that I just have to write the null space as a summation of derivatives from 1 to k thus the dimension of the null is k (the largest possible number).

No again, the null space is not a "summation of derivatives". For example if ##n=5## so you are considering polynomials of degree ##5##, and you operate on them with ##D^2##, which ones would map to the zero polynomial?
 
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Awesome! So I've been sitting here and had an epiphany. . . I think. . . so here is what I came up with.

Let B = {1,x,x[itex]^{2}[/itex],...,x[itex]^{n}[/itex]} be the standard basis for P[itex]_{n}[/itex]. Since p(x) is in the N(T) iff it's degree is < k, then the null space is all the polynomials with degree k-1 or less, thus [1,x,x[itex]^{2}[/itex],...,x[itex]^{k-1}[/itex]] is a basis for N(T),

and the dim(N(T)) = k


:) I think I have it what do you think?
 
  • #4
LCKurtz
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Awesome! So I've been sitting here and had an epiphany. . . I think. . . so here is what I came up with.

Let B = {1,x,x[itex]^{2}[/itex],...,x[itex]^{n}[/itex]} be the standard basis for P[itex]_{n}[/itex]. Since p(x) is in the N(T) iff it's degree is < k, then the null space is all the polynomials with degree k-1 or less, thus [1,x,x[itex]^{2}[/itex],...,x[itex]^{k-1}[/itex]] is a basis for N(T),

and the dim(N(T)) = k


:) I think I have it what do you think?

I think that's much better. :smile:
 

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