Show that if $V$ is an irreducible finite dim. representation of $A$, then $z \in Z(A)$ acts in $V$ by multiplication by some scalar $\chi_V(v)$. 
Let $A$ be an algebra over a field $k$. The center $Z(A)$ of $A$ is
  the set of all elements $z \in A$ which commute with all elements of
  $A$. For example, if $A$ is commutative, then $Z(A)=A$. 
Show that if $V$ is an irreducible finite dimensional representation
  of $A$, then any element $z \in Z(A)$ acts in $V$ by multiplication by
  some scalar $\chi_V(v)$. Show that  $\chi_V : Z(A) \to k$ is a homomorphism. It is called the
  central character of $V$.

This is my try:
As $V$ is representation of $A$, there is some action such that $a.v\in V$ for $a \in A$ and $v\in V$. We need to show that if $a \in Z(V)$, then $a.v = k_1 v$ for some $k_1 \in k$ for all $v\in V$. 
So we need to show that there exists a $k_1$ such that $a.v=k_1.v$. 
Or equivalently, $(a-k_1).v=0$, for all $v \in V$. 
So we need to show that there exists  a $k_1$ such that $\rho(a)=k_1I$.
In other words, we need to show that $\rho(a)$ is a scalar matrix, for any $a \in Z(A)$. 
We define $\rho(a)=X$, and we know that $XB=BX$ for every $B=\rho(b)$ where $b\in A$. 
Okay, so this is where I'm stuck. My intuition says that if $\rho$ is surjective, then this must be true, but I'm not sure why $\rho$ should be surjective. 
 A: First, I think the theorem is false is $k$ is not algebraically closed.  To see this, let $A = \mathbb{C}$ as an algebra over $\mathbb{R}$, and consider the action of $\mathbb{C}$ itself ($=\mathbb{R}^2$) given by left multiplication.
That is, for $z\in \mathbb{C}$, $(x,y)\in\mathbb{R}^2$, $z(x,y) = (\text{Re}(z(x+iy)), \text{Im}(z(x+iy))$.
Then this action is irreducible as the orbit through any point is all of $\mathbb{R}^2$.  Nonetheless, not every element of $\mathbb{C}=Z(\mathbb{C})$ acts as a scalar multiple of the identity.  For example, $i(x,y) = (-y,x)$, so the matrix $\rho(i)$ for $i$ is $\begin{bmatrix} 0 & -1\\ 1&0\end{bmatrix}$, which is not a multiple of the identity.
Now, as far as your proof goes, you need to use irreducibility somewhere, otherwise the statement isn't true.  Also, we'll assume $V\neq \{0\}$ because other wise the theorem is trivially true.
Pick an $a\in Z(A)$ and let $X = \rho(a)$.  As you noted, $X$ commutes with $\rho(b)$ for any $b\in A$.  We show that $X$ is a multiple of the identity.  To do this, we need two key facts.
Key fact 1:  $X:V\rightarrow V$ is a map of representations.
Proof:  We need to show that for any $b\in A$, any $v\in V$, that $\rho(b) (Xv) = X(\rho(b)v)$.  But we just noted that $X$ commutes with $\rho(b)$ for any $b\in A$. $\square$
Key fact 2: (Schur's Lemma)  Over an algebraically closed field, a map $X:V\rightarrow V$ from an irreducible finite dimensional representation to itself is a multiple of the identity.
Proof: The assumptions on $V$ imply that $X$ has an eigenvalue $\lambda$.  Then note that $X-\lambda I$ is a map of representations because $X$ and $I$ are.
The kernel of $X-\lambda I$ is therefore a subrepresentation of $V$.  Since $V$ is irreducible, this subrepresntation must be either $\{0\}$ or $V$.  But since $\lambda$ is an eigenvalue of $X$, there is a corresponding eigenvector, and this eigenvector will be in the kernel of $X-\lambda I$.  This implies the kernel is all of $V$, so $X-\lambda I = 0$, i.e., $X = \lambda I$.$\square$
Finally, the part about $\chi$ being a homomorphism comes directly from the defintion of representation.  We need to show that if $a,a'\in Z(A)$, then $\chi_V(aa') = \chi_V(a)\chi_V(a')$.
But by definition, $$\chi_V(aa')I = \rho(aa') = \rho(a)\rho(a') = \chi_V(a)I \chi_V(a')I = \chi_V(a)\chi_V(a')I. $$
