10

Let $V$ and $W$ be vector spaces over a field $K$. If a linear map $L:V \rightarrow W$ is surjective then its dual is injective. If $V$ and $W$ are finite dimensional then the converse holds, i.e. $L^*:W^* \rightarrow V^*$ injective implies $L$ surjective.

I have proved both statements but I don't see where I used the finite dimensional requirement for the second. Here is my proof:

Assume $L$ is not surjective, say the element $e_i$ of the basis of $W$ is not in the image of $L$. Take its corresponding dual $\alpha_i \in W^*$, then $L^*(\alpha_i)=\alpha_i \circ L =0$ so the kernel of $L^*$ is not 0 and therefore $L^*$ is not injective.

inquisitor
  • 1,740
  • Your argument seems correct, but I might be missing something. Where did you see this statement? Do you know that every vector space has a basis (i.e. are you allowed to use the Axiom of Choice)? – Michael Albanese Aug 31 '13 at 20:35
  • 4
    You claimed that $e_i\not\in L(V)$ implies $\alpha_i\circ L=0$. I don't think this is true, even for finite-dimensional spaces. – Julian Rosen Aug 31 '13 at 20:40
  • I think you need to make sure to first choose a basis of the image of $L$, and then extend it to a basis of $W$. If you don't do that, there is no guarantee that your dual basis element acts properly on the image of $L$. – Carl Aug 31 '13 at 21:31
  • It is problem 10.5 of Tu's book "An introduction to Manifolds". Yes the axiom of choice is assumed so every vector space has a basis. @Pink Elephants the only elements which are not mapped to $0$ by $\alpha_i \circ L$ are those whose image by $L$ is a multiple of $e_i$. – inquisitor Sep 01 '13 at 00:36
  • @inquisitor Let $e_1,\ldots,e_n\in W$ be a basis, $\alpha_1,\ldots,\alpha_n\in W^$ the dual basis. It is not true that the only elements of $W$ not mapped to 0 by $\alpha_i$ are scalar multiples of $e_i$. For example, $\alpha_1(e_1+e_2)=1$. So if $e_1$ is not in the image of $L$ but $e_1+e_2$ is, then $\alpha_1$ is not in the kernel of $L^$. – Julian Rosen Sep 01 '13 at 00:55
  • @Pink Elephants you are right! – inquisitor Sep 01 '13 at 01:06
  • @PinkElephants Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Sep 11 '13 at 15:55
  • Have you notice that you are assuming some finite element of basis of W are not in image L, What if there are infinite such elements? The idea given by @JulianRosen is valid in case of finite dim. only. – Sry Jan 01 '14 at 04:48

2 Answers2

1

My first argument has to be changed as the comments of @Julian Rosen show. Finite dimension is not needed:

Take a basis $B'$ of the image of $L$ in $W$ and complete it to a basis $B$ of $W$ (by assumption $B \setminus B'$ is not empty). Define the linear functional $\alpha$ by $\alpha(e)=1$ where $e\in B\setminus B'$ and $\alpha(v)=0$ for $v \in B'$ (and extend linearly). Then $L^*(\alpha)=\alpha \circ L = 0$, and $\alpha$ is not the $0$ linear functional, hence $L^*$ is not injective.

For the proof of $L^*$ surjective iff $L$ injective see the last answer of: Are injectivity and surjectivity dual?

Community
  • 1
inquisitor
  • 1,740
1

I think Pete L. Clark's answer is intuitive and I don't mean to obfuscate the problem, but could I add some abstract nonsense (which obviously one could ignore)?

In the category of vector spaces, we can easily show every mono (injective) and epi (surjective) splits. The dual construction is faithfully functorial (another straightforward exercise), hence it both preserves and reflects split monos and split epis, i.e. monos iff the dual is epi and epi iff dual is mono.

Thank you for the indulgence.

Community
  • 1
Rachmaninoff
  • 2,650