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I'm looking at a definition for linear functionals on a vector space. It says it's a scalar valued function and then states the linearity condition. I've just been looking at some exercises in a book and I was narrowing down which ones were functionals.

question: when it says 'scalar valued' does that mean that if $V$ is a vector space over a field $\cal F$ then a linear functional should be a map $y:V\rightarrow \cal F $. Or can it map to any field?

I'd assume not because you'd have to make sense of $\alpha\cdot y (x)$ where $\alpha \in \cal F$ and $x \in V$ otherwise.

In which case the map $y$ from $\Bbb C$ as a real vector space to $\Bbb C$ via complex conjugation would be ruled out.

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    $\begingroup$ My recent answer may be helpful: math.stackexchange.com/questions/886026/… $\endgroup$ Commented Aug 5, 2014 at 15:03
  • $\begingroup$ @andybenji Thanks for the link. How's does a 1x4 row matrix act on a 2x2 matrix in terms of the trace operation though? $\endgroup$
    – snulty
    Commented Aug 5, 2014 at 16:23
  • $\begingroup$ Once you choose the basis mentioned in the post, it behaves like any other 4 dimensional vector space. So in the context of $2 \times 2$ matrices, you encode the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ as the column vector $(a \ b \ c \ d )^T$, so that by doing normal matrix multiplication $$(1 \ 0 \ 1 \ 0) \cdot (a \ b \ c \ d)^T = (a + c)$$ (Here, $^T$ stands for the transpose, eg taking a row to a column and vice versa) $\endgroup$ Commented Aug 6, 2014 at 4:49
  • $\begingroup$ @Andybenji yep I see what you're saying now :) I forgot that's what happens when you choose a basis! Thanks again for the response and clarification! $\endgroup$
    – snulty
    Commented Aug 6, 2014 at 6:42

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Yes, a linear functional on a vector space over $\mathcal{F}$ by definition maps into $\mathcal{F}$. Otherwise, as you correctly say, the linearity condition would not make sense.

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