Question about Noncommutative Rings by I. N. Herstein Here's an example that I do not quite understand it fully, it's on page 6 of the book. And here's what it says:

Let $F$ be a field, and $F_n$ be a ring of all $n\times n$ matrices over $F$. We consider $F_n$ as the ring of all linear transformations on the vector space $V$ of $n-$tuples of elements of $F$. If $A$ is a subset of $F_n$, let $\overline{A}$ be the sub-algebra generated by $A$ over $F$. Clearly, $V$ is a faithful $F_n-$module, and so, $\overline{A}$-module. $V$ is in addition, both a unitary, and irreducible $F_n-$module.
We say the set of matrices $A \subset F_n$ is irreducible if $V$ is an irreducible $\overline{A}-$module. In matrix terms, it merely says that there is no invertible matrix $S$ in $F_n$ so that
$$S^{-1}aS = \left( \begin{array}{c|c} a_1 & 0 \\ \hline * & a_2  \end{array} \right)$$ for all $a \in A$.

The bolded parts are what I don't understand

*

*Firstly, why must $V$ be a faithfull $F_n-$module in order to be an $\overline{A}-$module? What will happen if $V$ is not faithful?


*Secondly, I don't understand the bolded part start with "in matrix terms...", say I have $V$ is an irreducible $\overline{A}-$module. How can I deduce the last part? And I don't get the star in the expression $S^{-1}aS$ as well, what does it stand for, is it a wild-card? I think this has to do with some matrix knowledge that I'm missing, so if the explanation is pretty long, please just give me a reference on what I should read.
Thank you guys very much,
And have a good day,
 A: *

*I don't see that you need faithful in order to have an $\overline{A}$-module. Whenever you have an algebra homomorphism $i: \overline{A}\to F_n$ (such as inclusion in your case) then any $F_n$-module is also an $\overline{A}$-module by restriction, i.e. $a\cdot m:= i(a)m$. The author probably means that $V$ is also a faithful $\overline{A}$-module, which is quite clear since $i$ is the inclusion (although that $V$ is faithful is not used in the paragraph you provided). 

*For the last part, you are right $*$ stands for a wildcard. The explanation is not really long, the intuition you should have is that block matrices correspond to direct sum decompositions and block matrices with a wild card instead of a zero correspond to submodules. This is how you proceed. Suppose $V$ is not irreducible. Then it has a submodule $U$. This means that $a\cdot U\subset U$ for all $a\in A$. Now doing a base change from $U$ to the standard basis this means that $S^{-1}aS V'\subset V'$ where $V'$ denotes the vector space with the first $r=n-\dim U$ entries equal to zero (and the others arbitrary). And if you right this down in matrix language, this means that $S^{-1}aS$ has the claimed form with $a_1$ a matrix of size $r$. For the reverse direction view $S$ as such a base change matrix, then you have your submodule (in the same way as above) by changing back $V'$. I'll leave the details for you. Feel free to ask if I should explain something in more detail.

