Let $\alpha: \mathbb{Z}^a \to \mathbb{Z}^b$ be a homomorphism with matrix representation $M$. I wish to show that the dual map, $\alpha^*: \operatorname{Hom}(\mathbb{Z}^b,\mathbb{Z}) \to \operatorname{Hom}(\mathbb{Z}^a,\mathbb{Z})$, has a matrix representation $M^T$, the transpose of $M$.

Let $(e_1,e_2,......,e_a)$ be a bases for $\mathbb{Z}^a$ and $(f_1,f_2,......,f_b)$ be a basis for $\mathbb{Z}^b$.

Then, lets define $\alpha$ by $\alpha(e_j) = \sum_{i=1}^b m_{ij} f_i$ for each $j=1,\dots,a$.

Then, the matrix $M=(m_{ij})\in \mathbb Z^{b \times a}$ describes $\alpha$ in the sense that $[\alpha(v)]_f = M[v]_e$.

I'm stuck here... How do I show $\alpha^*$ has a matrix representation of $M^T$?

  • $\begingroup$ Pick a basis for $Hom(\mathbb{Z}^a, \mathbb{Z})$ and $Hom(\mathbb{Z}^b, \mathbb{Z})$ first, and think about how $\alpha^*$ is represented in it. Choose your basis carefully.. $\endgroup$ – B. Mehta Apr 19 '18 at 0:33
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    $\begingroup$ $\alpha^*$ should be contravariant $\endgroup$ – Andres Mejia Apr 19 '18 at 0:39
  • $\begingroup$ The answers here and here only deal with vector spaces, but the proofs carry over to this case. $\endgroup$ – André 3000 Apr 24 '18 at 4:14


Use the dual bases $(e_1^*, \dots, e_a^*)$ and $(f_1^*, \dots, f_b^*)$ and compute the images of the compositions $f_i^*\circ\alpha $ in basis $(e_1^*, \dots, e_a^*)$.

  • $\begingroup$ Do you mean something like this? $f^*_k (\alpha(e_j)) = f^*_k(\Sigma^b_1m_{ij}f_i)=\Sigma^b_1m_{ij}f^*_k(f_i)=\Sigma^b_1m_{ij}\delta_{ik}$ I don't know how i'm closer to answering the question, where does the transpose come into play? $\endgroup$ – Math is hard Apr 20 '18 at 16:09
  • $\begingroup$ Yes, that's what I mean. There remains to express the coordinates of the images of each $f_k^*$ in basis $(e_i^*)$. $\endgroup$ – Bernard Apr 20 '18 at 17:04
  • $\begingroup$ Hmm, can you help me a bit with that? $\endgroup$ – Math is hard Apr 24 '18 at 0:13
  • $\begingroup$ I'll have some time to help more tomorrow. Just the starting idea:the linear form $f_k*$ is just the $k$-th coordinate map. With this you should find the coordinates of $f_k^*\circ \alpha$ in basis (e_i^*)$. $\endgroup$ – Bernard Apr 24 '18 at 0:32
  • $\begingroup$ so, above, did i find $f_k^*(\alpha(e_j))$ with respect to basis $(f_i^*)$, which ended up with the term involve delta kronecker function? And now I should find $f_k^*(\alpha(e_j))$ with respect to basis $(e_i^*)$? Or rather is the string of equalities that I wrote unfinished? Where the last term has the delta Kronecker function, can I still go somewhere from there? Honestly I am a bit confused, my linear algebra is shaky at best. $\endgroup$ – Math is hard Apr 24 '18 at 2:05

Hint: for $\mathrm{Hom}(\mathbb Z^a,\mathbb Z)$ let $e^1, \dots e^a$ be functions that are characteristic on $e_1, \dots e_a$, and likewise for $\mathrm{Hom}(\mathbb Z^b,\mathbb Z)$.

Both of these constitute a basis for each vector space.


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