Let $V$ be a vector space over $F$, and $S$ a non-empty subset of $V$. We say that $S$ generates $V$ if every $v\in V$ can be written as finite $F$-linear combination of elements of $S$.

I want to express the above definition of generating set in terms of universal property. My question is whether the following is the universal property of $S$ (generating set) and how to prove that it is equivalent to above definition? Just hint is sufficient, I will try to write complete proof.

We say that $S$ generates $V$ if the following holds: any map $\varphi$ from $S$ into any vector space $W$ extends to a linear map $\varphi_1$ from $V$ into/onto $W$.

  • $\begingroup$ Do you wanna qualify that $\varphi$ be a homomorphism? $\endgroup$ – AJY Jul 10 '16 at 5:28
  • $\begingroup$ no, in vector space category, morphisms should be linear maps (it was initial typo from me, sorry for inconvenience, I corrected it) $\endgroup$ – p Groups Jul 10 '16 at 5:28

The universal property you wrote (leaving out any assertion about $\varphi_1$ being injective or surjective) is equivalent to $S$ being linearly independent, not generating $V$. You want to say instead that any $\varphi$ extends to at most one linear map $\varphi_1$ (again, with no condition that $\varphi_1$ is injective or surjective). The hard direction of the proof is that if $S$ has this property, then $S$ generates $V$. To prove this direction, you need to give an example of a $\varphi$ for which $\varphi_1$ is not unique, assuming $S$ does not generate $V$. To do this, let $U$ be the subspace generated by $S$, $W=V/U$, and $\varphi=0$.

  • $\begingroup$ if $\varphi_1$ is onto linear map, will it not imply that $S$ generates $V$? $\endgroup$ – p Groups Jul 10 '16 at 5:30
  • $\begingroup$ I don't know what you mean by that. What choice of $\varphi:S\to W$ are you making? The basic problem with your condition is that if $S$ is not linearly independent, then for most choices of $\varphi$ there will not exist any linear extension $\varphi_1$. $\endgroup$ – Eric Wofsey Jul 10 '16 at 5:32
  • $\begingroup$ OK, I will make it precise: let $S$ be a subset of $V$ satisfying following property: given any map from $S$ to a vector space $W$, it extends to a linear map from $V$ onto $W$ (i.e. $\varphi_1\colon V\rightarrow W$ is surjective linear map). Then is this (property) equivalent to say that $S$ generates $V$? $\endgroup$ – p Groups Jul 10 '16 at 5:34
  • $\begingroup$ That condition is not true for any subset of any vector space. You can always choose $W$ to have larger dimension than $V$, and then there can't exist any linear surjection $V\to W$. $\endgroup$ – Eric Wofsey Jul 10 '16 at 5:35
  • 1
    $\begingroup$ Yes, everything after my first sentence is about characterizing generating sets. $\endgroup$ – Eric Wofsey Jul 10 '16 at 5:41

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.