-1
$\begingroup$

Let $V$, a finite dimensional vector space, and $L$, a subspace of $V$. Let $T:V^*\rightarrow L^*$ defined as: $T(\varphi)(x)=\varphi(x)$ for all $\varphi \in V^*$. Prove $T$ is onto.

Well, I'm struggling with understanding what are $L^*, V^*$ with relation to $L, V$.

Do you have an explanation for that?

$\endgroup$
2
  • $\begingroup$ I assume you mean only finite-dimensional, not necessarily finite? $\endgroup$
    – Hagen von Eitzen
    Jun 29, 2014 at 9:31
  • $\begingroup$ Indeed, thank you. Corrected that.. $\endgroup$
    – Elimination
    Jun 29, 2014 at 9:32

1 Answer 1

Reset to default
1
$\begingroup$

$V^*$ is the dual space to $V$, that is the vector sapce of all linear maps $V\to k$ to the ground field (this is a vector space via pointwise addition and mutlpplication with scalars). So if $\phi\in V^*$ is a linear map $V\to k$, $v\mapsto \phi(v)$, then $T(\phi)$ is the linear map $L\to k$ given by $x\mapsto \phi(x)$, i.e. the restriction to $L$.

Remark: Unless you have issues with the Axiom of Choice, the condition that $V$ be finite dimensional can be dropped.

$\endgroup$
3
  • $\begingroup$ Oh, so it's a familiar notation I'm not aware of. Thanks! $\endgroup$
    – Elimination
    Jun 29, 2014 at 9:33
  • $\begingroup$ I bet it git introduced just a few paragraphs before the problem statement :) $\endgroup$
    – Hagen von Eitzen
    Jun 29, 2014 at 9:34
  • $\begingroup$ Well, actually it's not part of our syllabus. Some teachers do teach dual-spaces though. I bet I'll be introduced with that material at Linear Algebra II. $\endgroup$
    – Elimination
    Jun 29, 2014 at 9:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .