For $A \in \operatorname{Mat}_{n,m}(K)$ and $B \in \operatorname{Mat}_{m,n}(K)$, let $\tau_A(B) := \operatorname{trace}(AB)$. Show that the map $\operatorname{Mat}_{n,m}(K) \to \operatorname{Mat}_{m,n}(K)^*, A \mapsto \tau_A$ is an isomorphism.
I already showed that $\tau_A: \operatorname{Mat}_{m,n}(K) \rightarrow K$ is linear. Of course the asterix refers to the dual space.
For the linearity of the map $A\mapsto \tau_A$ just work through the definitions. It is similar to the linearity of $\tau_A$.
Let us show that the given map is injective. So let $A\in K^{n\times m}$ with $\tau_A=0$, i.e. $\operatorname{Tr}(AB)=0$ for all $B\in K^{m\times n}$. We want to show that $A=0$. For $i=1,\dots,m,j=1,\dots,n$ define the matrix $E_{ij}\in K^{m\times n}$ so that $E_{ij}$ has only zero entries except in the $i$-$j$-index where it has a $1$. Verify that $$\operatorname{Tr}(AE_{ij})$$ is just the entry of $A$ at the position $j,i$ (notice the indices are swapped). Conclude that $A=0$.
So we proved the injectivity of the map. Now argue via a dimension argument why this map is then already an isomorphism.