# Thread: dual liner map isomorphism

1. ## dual liner map isomorphism

Hi!

We have two finite dimensional vector spaces V,W and a linear map f between them. We know that the dual map f' is an isomorphism. What can we say about f? Is it also an isomorphism? I think i miss something quite elementary here.

Banach

2. ok, first of all i'm not sure what you mean by the dual map but i can guess and it looks to me that it should be what you meant:

the dual $\displaystyle f': W^* \longrightarrow V^*$ of a linear map $\displaystyle f: V \longrightarrow W$ is defined by $\displaystyle f'(a)=af$ for all $\displaystyle a \in W^*.$

if that's the definition, then the answer to your question is "yes". to prove this, we only need to show that $\displaystyle f$ is onto because, since $\displaystyle f'$ is an isomorphism, we have

$\displaystyle \dim V = \dim V^*=\dim W^*=\dim W.$

so suppose it is not. then there exists a subspace $\displaystyle (0) \neq W_0$ of $\displaystyle W$ such that

$\displaystyle W=f(V) \oplus W_0.$

now let $\displaystyle 0 \neq a_0 \in W_0^*$ and define $\displaystyle a \in W^*$ by $\displaystyle a(w)=a_0(w_0), \ w \in W,$ where $\displaystyle w=f(v)+w_0.$ note that $\displaystyle a \neq 0$ and $\displaystyle a \in W^*$ is well-defined because the sum of $\displaystyle f(V)$ and $\displaystyle W_0$ is direct. clearly $\displaystyle a(f(v))=0$ for all $\displaystyle v \in V,$ i.e. $\displaystyle f'(a)=0,$ contradicting our hypothesis that $\displaystyle f'$ is an isomorphism.

the coverse is also true, i.e. if $\displaystyle f$ is an isomorphism, then $\displaystyle f'$ is an isomorphism too.

3. Originally Posted by NonCommAlg
ok, first of all i'm not sure what you mean by the dual map but i can guess and it looks to me that it should be what you meant:

the dual $\displaystyle f': W^* \longrightarrow V^*$ of a linear map $\displaystyle f: V \longrightarrow W$ is defined by $\displaystyle f'(a)=af$ for all $\displaystyle a \in W^*.$

if that's the definition, then the answer to your question is "yes". to prove this, we only need to show that $\displaystyle f$ is onto because, since $\displaystyle f'$ is an isomorphism, we have

$\displaystyle \dim V = \dim V^*=\dim W^*=\dim W.$

so suppose it is not. then there exists a subspace $\displaystyle (0) \neq W_0$ of $\displaystyle W$ such that

$\displaystyle W=f(V) \oplus W_0.$

now let $\displaystyle 0 \neq a_0 \in W_0^*$ and define $\displaystyle a \in W^*$ by $\displaystyle a(w)=a_0(w_0), \ w \in W,$ where $\displaystyle w=f(v)+w_0.$ note that $\displaystyle a \neq 0$ and $\displaystyle a \in W^*$ is well-defined because the sum of $\displaystyle f(V)$ and $\displaystyle W_0$ is direct. clearly $\displaystyle a(f(v))=0$ for all $\displaystyle v \in V,$ i.e. $\displaystyle f'(a)=0,$ contradicting our hypothesis that $\displaystyle f'$ is an isomorphism.

the coverse is also true, i.e. if $\displaystyle f$ is an isomorphism, then $\displaystyle f'$ is an isomorphism too.
Alternatively, if $\displaystyle V=W$ (the proof shouldn't be too hard if one just does the natural identification of $\displaystyle \text{Hom}\left(V,F)$ with $\displaystyle \text{Hom}\left(W,F\right)$, but I leave that to you). In particular, let $\displaystyle \mathcal{B}=(x_1,\cdots,x_n)$ be an ordered base for $\displaystyle V$. Lift $\displaystyle \mathcal{B}$ to its associated dual bases$\displaystyle \mathcal{D}$. It's fairly easy to prove then that $\displaystyle \displaystyle \left[T\right]_{\mathcal{B}}=\left[T^*\right]_{\mathcal{D}}^{\top}$ from where the result should be fairly clear from $\displaystyle \displaystyle \det\left(M\right)=\det\left(M^{\top}\right)$ for any matrix $\displaystyle M$.

4. Thats exactly what i mean by the dual map. What is the correct english term for it?

Thank you both very much for taking your time to answer. Good news for my situation that the answer is yes.
I tried to work with the diagram that is associated with the maps f: V -> W, f': W' -> V' and the (non-canonical) isomorphisms between V and V', W and W', respectively but got confused there.