Here is an exercise from Artin:
Let $V$ be an $F[t]$-module, and let $\mathbf B=(v_1,\dots,v_n)$ be a basis for $V$ as $F$-vector space. Let $B$ be the matrix of $T$ with respect to this basis. Prove that $A=tI-B$ is a presentation matrix for the module.
(Note that $T$denotes the linear operator $V\to V$ defined by $T(v)=tv$.)
Artin defines a presentation matrix as follows:
Left multiplication by an $m \times n$ matrix defines a homomorphism of $R$-modules $A: R^n \rightarrow R^m$. Its image consists of all linear combinations of the columns of $A$ with coefficients in the ring, and we may denote the image by $AR^n$. We say that the quotient module $V=R^m/AR^n$ is presented by the matrix $A$. More generally, we call any isomorphism $\sigma: R^m/AR^n \rightarrow V$ a presentation of a module $V$, and we say that the matrix $A$ is a presentation matrix for $V$ if there is such an isomorphism.
Back to the problem, I'm not even sure what I'm supposed to show. Presentation matrices are defined for quotient modules. Why is an arbitrary $F[t]$-module $V$ a quotient module? If $V$ were finitely generated, this would certainly be true since there would be a surjective $F[t]$-module homomorphism $(F[t])^n\to V$, and we could apply the First Isomorphism Theorem. So, why is a presentation matrix defined for $V$ in question, what exactly do I need to show, and at least how to get started?
Edit: alright, indeed, as @Mohan pointed out in the comments, since $V$ is generated by $\mathbf B$ as an $F$-vector space, then it is also generated by $\mathbf B$ as an $F[t]$-module. So there is a surjective module homomorphism $\phi: F[t]^n\to V$ given by $X\mapsto \mathbf B X$, where the RHS is the product of the row $\mathbf B$ by the column $X$. So $V$ is isomorphic to $F[t]^n/\ker \phi$. If we show that $\ker \phi= AF[t]^n$ where $AF[t]^n$ stands for the image of the map $A: F[t]^n\to F[t]^n$ given by $X\mapsto AX$ (here I denote the map and the matrix by the same letter $A$), then we will be done.