precise definition of "irreducible representation" (of associative algebras with unit) Let $K$ be a field and $A$ an associative $K$-Algebra with unit. By a representation of $A$ I mean a homomorphism of $K$-Algebras with unit $f\colon V\rightarrow{End}_K(V)$ where $V$ is a finite dimensional $K$-vector space.

Definition 1: A representation $f$ is called irreducible, if it has no invariant subspaces besides $0$ and $V$.

From this definition it follows that
$f\colon A\rightarrow End_K(0)$ is irreducible.
Then the following theorem is false for $n\neq 0$:

Theorem: The matrix algebra $\mathbb{C}^{n\times n}$ has, up to equivalence, only one irreducible complex representation which is just the natural representation of $\mathbb{C}^{n\times n}$ on $\mathbb{C}^n$. This implies in particular, that every irreducible complexe representation of $\mathbb{C}^{n\times n}$ has dimension $n$.

(It's false since $f\colon \mathbb{C}^{n\times n}\rightarrow End_\mathbb{C}(0)$ is irreducible and has dimension $0$.)
So it seems to me the precise definition of "irreducible" should be

Definition 2: A representation $f\colon A\rightarrow{End}_K(V)$ is called irreducible, if $V\neq  0$ and if it has no invariant subspaces besides $0$ and $V$.

Am I right? (I struggle with that because every book I looked up uses definition 1 and no one uses definition 2.)
edit: corrected typo, the domain of $f$ should be $A$.
 A: You're right. Many books give definitions that incorrectly specialize to the empty set and related objects; see too simple to be simple on the nLab for a discussion.  
My preferred phrasing of the correct definition is that a simple module $V$ is a module that has exactly two submodules (which must therefore be $0$ and $V$), so the zero module is never simple because it has exactly one submodule. Hence the zero ring has no simple modules. 
A: The same issue arises when we ask whether $1$ is a prime number.
This just came up in a discussion of topological dimension.  Typically, the empty set is not considered to be an irreducible topological space.  Similarly, $0$ is not a field or an integral domain, and $A$ is not a prime ideal of a ring $A$.
It's not very surprising that a lot of texts would make a mistake when dealing with the trivial case.  You are quite correct that the zero representation should not be considered to be irreducible.  A decomposable representation should have a unique decomposition into irreducible components, and $0=0\oplus 0$ would contradict this philosophy.
