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This is related to a question I answered earlier which raised a question in my mind.

My question is the following,

Suppose we have a vector space $\mathbb{V}$ with real coefficients.

Let $\textbf{T}$ be an operator on this space which has the following two properties:

  1. $ \textbf{T}( \textit{u} + \textit{v})= \textbf{T}(\textit{u}) + \textbf{ T}( \textit{v}) \qquad (\textit{u},\textit{v} \in \mathbb{V})$
  2. $\textbf{T}(c v) = c \textbf{T}(\textit{v}) \qquad (\textit{v} \in \mathbb{V}, c \in \mathbb{Q})$

Can we prove that $\textbf{T}(c\textit{v}) = c\textbf{T}(\textit{v}) $ for $c \in \mathbb{R}$ ?

I suspect we can prove this for any particular $r\in \mathbb{R}$ by taking a sequence of rational points $\lbrace q_n \rbrace$ which converge to $r$ and writing the following equality.

$$\textbf{T}(q_n \textit{v}) = q_n \textbf{T}(\textit{v}) \qquad (\forall n)$$

$$\lim_{n\rightarrow \infty} \textbf{T}(q_n \textit{v}) = \lim_{n\rightarrow \infty} q_n \textbf{T}(\textit{v}) $$

$$\lim_{n\rightarrow \infty} \textbf{T}(q_n \textit{v}) = r \textbf{T}(\textit{v}) $$

The statement would be proven if we can pull the limit inside but I'm not enough of an analyst to be able to justify that at this stage.

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  • $\begingroup$ Without a continuity assumption, you can't. And there are $\mathbb{Q}$-linear functions that aren't $\mathbb{R}$-linear. $\endgroup$ – Daniel Fischer Oct 13 '13 at 1:45
  • $\begingroup$ Thanks for the input, would you mind sharing an example of such a function? $\endgroup$ – Spencer Oct 13 '13 at 1:50
  • $\begingroup$ Take a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ including $1$. Let $f(1) = 1$, and $f(v) = 0$ for every other basis element. Extend $\mathbb{Q}$-linearly. You can't have a nice formula for such functions, but they exist. $\endgroup$ – Daniel Fischer Oct 13 '13 at 2:03
  • $\begingroup$ Thanks Daniel, I like to have counter examples for things like this to keep me from falling for plausible "proofs". I you want to post your comments as an answer I'll accept it. $\endgroup$ – Spencer Oct 13 '13 at 2:06
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You can move the limit past the $T$ only if $T$ is continuous (strictly, if the restriction of $T$ to every one-dimensional subspace is continuous).

Every $\mathbb{R}$ vector space can be canonically considered also a $\mathbb{Q}$ vector space, and the additivity of $T$ gives you $\mathbb{Q}$-linearity, but not $\mathbb{R}$-linearity.

We can "construct" operators that are $\mathbb{Q}$-linear but not $\mathbb{R}$-linear if we extend an $\mathbb{R}$-basis of $\mathbb{V}$ to a $\mathbb{Q}$-basis, and define the operator to take some non-zero values on the basis elements belonging to the original $\mathbb{R}$-basis, and $0$ on the added basis elements. Since, however, one cannot explicitly construct a $\mathbb{Q}$-basis of an $\mathbb{R}$ vector space, one cannot obtain an explicit example in that way.

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