Proof of injectivity of dual map using rank-nullity theorem

Question

I am trying to solve the following using just rank/kernel properties, just wanted to check if my solution works. Thanks

(a) Suppose $V$ and $W$ are vector spaces of possibly infinite dimension over a field $K$. Show that if a linear map $L:V \rightarrow W$ is surjective, then its dual $L^v : W^v \rightarrow V^v$ is injective.

(b) Suppose $V$ and $W$ are vector spaces of finite dimension over a field $K$. Show that if $L^v : W^v \rightarrow V^v$ is injective then the dual $L:V \rightarrow W$ is surjective.

We define $L^v=\alpha L$ where $ \alpha \in W^v$.

For the (a) part, suppose there is $w \in W$ such that $\alpha w \neq 0$; if $L$ is surjective there is $v \in V$ such that $L v = w$ and then $\alpha L v \neq 0$.

This means that if $L$ is surjective $$\neg(\alpha w = 0 \ \forall w \in W) \implies \neg(\alpha L v = 0 \ \forall v \in V)$$ hence $$ \alpha L v = 0 \ \forall v \in V\implies \alpha w = 0 \ \forall w \in W $$ hence $L^v$ is injective as its kernel has to be $0$. $$ \alpha L = 0 \implies \alpha = 0 $$

(Here I use $0$ with some freedom meaning both the null element of $K$ as well as the "null" element of $V^*$ and $W^*$ depending on the context).

In this first part I have not used reference to the dimensions of $V$ and $W$ so I believe this should hold regardless of the dimension (i.e. also for infinite dimension spaces).

For the (b) case I have tried to apply the rank-nullity theorem, that only holds for finite dimension vector spaces, so the proof should hold only for a finite number of dimensions. This is how it goes. If $L^v$ is injective, then its nullity is $0$, and the rank of $L^v$ is $dim(W^v)=dim(W)$ (equality holds as dual spaces have the same dimension). Say $n = dim(W)$.

This means that if we take $n$ linearly independent covectors $\alpha_1, ..., \alpha_n$ in $W$ we can produce a set of $n$ linearly independent covectors in V that we can write as $$\alpha_1L, ..., \alpha_nL$$ We can take than $n$ dual vectors $e_1,...e_n$ such that $\alpha_iLe_j=\delta_i^j$ and generate $n$ vectors in W $Le_i$ that are linearly independent. In fact, if I take $$t_1Le_i+t_2Le_j = 0 $$ for any couple of vectors $Le_i$ and $Le_j$ with $i,j \leq n$, if I apply in turn $\alpha_i$ and $\alpha_j$ I get that both $t_1$ and $t_2$ must be $0$, as $$\alpha_i(t_1Le_i+t_2Le_j) = t_1 = 0$$ $$\alpha_j(t_1Le_i+t_2Le_j) = t_2 = 0$$ this means that $L$ can generate an image on $W$ whose dimension is $n=dim(W)$ hence $L$ is surjective. Would this work? Thanks!

Answer 1

I'm not sure if this result holds without the hypothesis of finite dimension. I know a proof that only requires $W$ to be finite-dimensional. First note that

$$ \ker L^{v} = (\operatorname{im} L )^{0}, $$

where $(\operatorname{im} L)^{0}$ denotes the annihilator of $\operatorname{im} L$. This holds even if $W$ is infinite-dimensional. Then

\begin{equation*} \begin{split} L \text{ is surjective} & \iff \operatorname{im} L = W \\ & \iff (\operatorname{im} L)^{0} = \{ 0 \} \\ & \iff \ker L^{v} = \{ 0 \} \\ & \iff L^{v} \text{ is injective}, \\ \end{split} \end{equation*}

and we can conclude that if $W$ is finite-dimensional, $L$ is surjective if and only if $L^{v}$ is injective.

The proof of the equivalence $$ \operatorname{im }L = W \iff (\operatorname{im} L)^{0} = \{0\}$$ that I know uses the fact that $W$ is finite-dimensional in the following way: \begin{equation*} \begin{split} \operatorname{im} L = W & \iff \dim \operatorname{im} L = \dim W \\ & \iff \dim (\operatorname{im }L)^{0} = \dim W - \dim \operatorname{im }L = 0 \\ & \iff (\operatorname{im }L)^{0} = \{ 0 \}. \end{split} \end{equation*}

Answer 2

Suppose $L\colon V\to W$ is surjective; take $\alpha\in\ker L^v$; this means that, for every $v\in V$, $\alpha L(v)=0$.

Since $L$ is surjective, we get that $\alpha (w)=0$, for every $w\in W$. Hence $\alpha=0$ and the proof that $L^v$ is injective is complete.

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Suppose $V,W$ are finite dimensional and that $L\colon V\to W$ is injective. Take $\beta\in V^v$: you want to find $\alpha\in W^v$ such that $\alpha L=\beta$.

Fix a basis $\{v_1,v_2,\dots,v_n\}$ of $V$. You can extend $\{w_1=L(v_1),\dots,w_n=L(v_n)\}$ to a basis of $W$, say with the vectors $\{w_{n+1},\dots,w_m\}$. Now you can define $\alpha\in W^v$ on the basis by setting \begin{align} \alpha (w_i)&=\beta(v_i), &&(i=1,\dots,n)\\ \alpha (w_i)&=0 &&(i=n+1,\dots,m) \end{align} Now $\alpha L=\beta$ and you're done.

The key is that $L$ injective implies $\{L(v_1),\dots,L(v_n)\}$ is linearly independent.

You could use the rank-nullity theorem, but the above proof is much simpler.