# Is the finite dimension required in this proof?

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.

• 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)? Aug 31, 2013 at 20:35
• 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. Aug 31, 2013 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, 2013 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 $e_i$. Sep 1, 2013 at 0:36
• @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 $L^*$. Sep 1, 2013 at 0:55

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?