Show that $(F^n)^A$ is cyclic as an $F[x]$-module Problem statement:
Let $A$ be an $n \times n$ matrix over the field $F$ and let $A$ have $n$ distinct eigenvalues.  Show that $(F^n)^A$ is cyclic as $F[x]$-module.
I'm not sure I understand the notation $(F^n)^A$.  What does this refer to?
Thanks.
 A: I'm fairly sure that the notation means the following. We can turn the vector space $F^n$ into a module over the polynomial ring $F[x]$ by declaring that
$$
x\cdot v=Av
$$
for all vectors $v\in F^n$ (viewed as column vectors so that the matrix multiplication makes sense). Consequently we have to define $x^k\cdot v=A^kv$, and
if $p(x)=\sum_{i=0}^k p_ix^i,$ $p_i\in F$ for all $i=1,2,\ldots, k$, is an arbitrary polynomial, we are forced to define
$$p(x)\cdot v=\sum_{i=0}^k p_iA^iv$$
for all $v\in F^n$. If you have never done this, it is a useful exercise to convince yourself of the fact that this gives, indeed, the space $F^n$ a structure of an $F[x]$-module. This is sometimes stated by saying that $F[x]$ is
a free $F$-algebra. Basically meaning that we can let the monomial $x$ act any which way it wants, and simply extend the action in the obvious way to its powers and then $F$-linearly to all polynomials. The crux is that the powers of $x$ do not satisfy any hidden relations in the ring $F[x]$ (they are linearly independent).
Notice that above we were free to choose the matrix $A$ any which way we wanted.
Using a different matrix in place of $A$ gives a different gives a different module structure. This makes it necessary to tell the reader, which matrix was used. Hence the superscript. 
But most likely the notation was explained somewhere. It may have been at an unexpected place.
