Consider the vector space, $V$ of doubly infinite sequences of complex numbers over the complex field. Let $e_i$ denote the sequence which is 1 at the integer $i$ and $0$ elsewhere. I am looking for a linear transformation $T$ from $V$ to $V$ and a $ x = (x_i)$, such that $T(x) = \sum_{i} x_i T(e_i)$ is "invalid" (i.e., some coordinate value on the RHS does not converge to the corresponding value on the LHS).
counterexample on a infinite dimensional vector space
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linear-algebra
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1Let $x$ be the all-ones sequence, and let $T$ be any linear map which sends every $e_i$ to $x$ itself (there are many such maps, because the set $\{e_i\}$ does not span $V$) – 2012-04-01
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0Thanks a lot, is it possible to find such a $T$ explicitly. The only argument I can think of is that the $e_i$'s are contained in some basis. I would prefer an explicit map since I need to explain the problem with the infinite sum to someone with an engineering background. – 2012-04-01
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2You are not going to be able to explicit exhibit a basis of that vector space to an engineer :) – 2012-04-01
1 Answers
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Put $T(e_i):=e_0$ for all $i\in{\mathbb Z}$ and consider the all ones vector $u:=(\ldots,1,1,1,\ldots)$. If $T(u)$ is already defined, fine; otherwise put $T(u):=u$. In any case, the equation $$T(u)=\sum_i u_i\, T(e_i) =\sum_i 1\ e_0$$ makes no sense.
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0There is still the objection that you haven't given $T$, that you have only indicated that it exists. I understand the difficulties involved here, but I'm not sure I understand how to explain them to an engineer. – 2012-04-02