# Matrix representation of the dual homomorphism of $\alpha: \mathbb{Z}^a \to \mathbb{Z}^b$

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 $M^T$?

• 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.. – B. Mehta Apr 19 '18 at 0:33
• $\alpha^*$ should be contravariant – Andres Mejia Apr 19 '18 at 0:39
• The answers here and here only deal with vector spaces, but the proofs carry over to this case. – 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 $(e_1^*, \dots, e_a^*)$.
• 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? – Math is hard Apr 20 '18 at 16:09
• Yes, that's what I mean. There remains to express the coordinates of the images of each $f_k^*$ in basis $(e_i^*)$. – Bernard Apr 20 '18 at 17:04
• 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^*)$. – Bernard Apr 24 '18 at 0:32 • so, above, did i find$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. – 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 $$\mathrm{Hom}(\mathbb Z^b,\mathbb Z)$$.