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.

  • 2
    $\begingroup$ My recent answer may be helpful: math.stackexchange.com/questions/886026/… $\endgroup$ – Andrew Maurer Aug 5 '14 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 Aug 5 '14 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) $\endgroup$ – Andrew Maurer Aug 6 '14 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 Aug 6 '14 at 6:42

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.