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Insights about $Tv_j=w_j$, the linear maps and basis of domain.

I have a question about the theorem mentioned in the link above. I understand what the theorem is saying, but a little uncertain. it states that the linear map is unique if given a set of basis, but I think there is a counter-example.

Suppose 1 is a basis in a polynomial vector space, then D(1) = 0, D^2(1) = 0, where D is the differential operator. But D and D^2 are not the same linear map, therefore the linear map is not unique.

Can anyone tell me if I misinterpret the theorem? Thanks in advance.

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Harry
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1 Answers1

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"Suppose that $1$ is a basis in a polynomial vector space". That would necessarily mean that the vector space you're looking at is the space of constant functions.

What is it that the differential operator does on the space of constant functions? It maps everything to zero. In that sense, I would say that $D$ and $D^2$ are indeed the same map over the space of constant functions.

If there is any one thing that you have misunderstood, I would say it's the definition of a "basis".

Ben Grossmann
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  • To make it clear, if I choose another basis (1,x,x^2) with the same differential operator, then the unique part is still true since x and x^2 are mapped into different values, despite that the constant term is mapped into 0 by both operators? – Harry Jul 18 '16 at 22:43
  • Yes, that's one way to put it. I'd say that rather than "choosing a new basis", you're "choosing a new space". – Ben Grossmann Jul 18 '16 at 22:44