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

Question

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 $M^T$?

Answer 1

Hint:

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 $(e_1^*, \dots, e_a^*)$.

Answer 2

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.