# An example of a linear transformation that is not 1-1 or onto but has bijective dual

This is a question from some ring theory problems in Linear Algebra.

I am not sure if this question is simple enough to come up with a linear map on a finite dimensional vector space but here it goes.

Give an example of a linear map between two vectors spaces such that the dual is bijective but the original map is neither 1-1 or onto.

If the vector spaces in question have finite dimension we would require the matrix representing the transpose to have determinant zero but the transpose to be singular so that is why I am thinking there must be an infinite dimensional example.

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I know what the transpose of a matrix is. I don't know what the transpose of a linear map is, especially on an infinite-dimensional vector space. Do you perhaps mean dual, adjoint,...? –  Pete L. Clark Sep 20 '11 at 6:58
Hi @Pete: I think it's more or less standard to call transpose what you probably call dual. It's Bourbaki's terminology. I personally use it. (Of course, it's more important to know what the OP means...) –  Pierre-Yves Gaillard Sep 20 '11 at 7:11
No such example exists: If $f$ is not injective, then $f^*$ is not surjective. If $f$ is not surjective, then $f^*$ is not injective. –  Pierre-Yves Gaillard Sep 20 '11 at 7:36
Thanks is there still no examples if we relax the hypothesis to R-module homomorphisms between two modules where R is a commutative ring with identity? –  user7980 Sep 20 '11 at 8:03
Here is an example: Let $R$ be $\mathbb Z$ and $f$ the zero endomorphism of $\mathbb Z/2\mathbb Z$. –  Pierre-Yves Gaillard Sep 20 '11 at 8:20
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I'll follow Wikipedia's terminology: If $f:V\to W$ is a linear map, then the transpose (or dual) $f^*:W^*\to V^*$ is defined by $$f^*(\varphi)=\varphi\circ f.$$

The first question is: Give an example of a linear map between two vectors spaces such that the transpose is bijective but the original map is neither 1-1 nor onto.

There are no such examples. Indeed, in the above notation, if $f:V\to W$ is not injective, then $f^*:W^*\to V^*$ is not surjective, because the image of $f^*$ consists of those linear forms on $V$ which vanish on $\ker f$. Similarly, if $f$ is not surjective, then $f^*$ is not injective, because the kernel of $f^*$ consists of those linear forms on $W$ which vanish on $f(V)$.

Another question was added in the comments: Are there still no examples if, instead of vector spaces, we consider modules over a commutative ring $R$ with identity?

Such examples exist. A cheap one is given by setting $R=\mathbb Z$, $V=W=\mathbb Z/2\mathbb Z$, $f=0$.

EDIT. In fact we have canonical isomorphisms

(1) $\text{Ker}(f^*)=\text{Coker}(f)^*$ and $\text{Coker}(f^*)=\text{Ker}(f)^*$.

This can bee seen as follows:

(2) If
$$0\to A\overset{i}{\to}V\overset{f}{\to}W\overset{p}{\to}B\to0$$ and $$0\to A'\overset{i'}{\to}V\overset{f}{\to}W\overset{p'}{\to}B'\to0$$ are exact sequences of $K$-vector spaces, then there is a unique linear map $a:A\to A'$, and a unique linear map $b:B\to B'$ such that $i'\circ a=i$ and $b\circ p=p'$. Moreover $a$ and $b$ are bijective. The proof is easy.

(3) We have the exact sequences $$\begin{matrix} 0\to&\text{Ker}(f)&\overset{i}{\to}V\overset{f}{\to}W\overset{p}{\to}&\text{Coker}(f)&\to0,\\ \\ 0\to&\text{Coker}(f)^*&\overset{p^*}{\to}W^*\overset{f^*}{\to}V^*\overset{i^*}{\to}&\text{Ker}(f)^*&\to0,\\ \\ 0\to&\text{Ker}(f^*)&\to W^*\overset{f^*}{\to}V^*\to&\text{Coker}(f^*)&\to0. \end{matrix}$$ Then (1) follows from (2) and (3).

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Thanks for the answer - I am just going through checking the details and cannot remember if the following isomorphisms are correct: $Hom_\mathbb{Z}(\mathbb{Z}_2, \mathbb{Z}) \cong \mathbb{Z}_2$ and $Hom_\mathbb{Z}(\mathbb{Z}_2, \mathbb{Z}_2) \cong \mathbb{Z}_2$ –  user7980 Sep 21 '11 at 12:38
Dear @user7980: You're welcome. Put $A:=\mathbb Z/2\mathbb Z$. We have $$H:=\text{Hom}_\mathbb Z(A,\mathbb Z)=0.$$ Indeed, an $f\in H$ is just a morphism of abelian groups, and for any $a$ in $A$, we have $$f(a)+f(a)=f(a+a)=f(0)=0,$$ and thus $f(a)=0$. The iso $\text{End}_\mathbb Z (A)\cong A$ is correct. Here is a possible generalization: if $C$ is a cyclic group with generator $c$, then $\text{End}_\mathbb Z (C)\cong C$, the iso being the evaluation at $c$. –  Pierre-Yves Gaillard Sep 21 '11 at 13:34