Characters of (distinct) irreducible finite-dimensional representations of $A$ I need help to understand the proof of this theorem. The theorem can be found in the book Introduction to representation theory by Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina. Page 27

Characters of (distinct) irreducible finite-dimensional representations of $A$ are
  linearly independent.

Proof: If $V_1,...,V_r$ are nonisomorphic irreducible finite-dimensional representations of $A$, then $\rho_{V_1}\oplus...\oplus\rho_{V_r}:A\rightarrow \text{End}V_1\oplus...\oplus\text{End}V_r$ is surjective by the density theorem, so $\chi_{V_1},...,\chi_{V_r}$ are linearly independent. 
I can't understand this part that suggest for a more detailed proof:
(Indeed, if $\Sigma_{i=1}^{r} \lambda_i\chi_{V_i}(a)=0$ for all $a \in A$, then $\Sigma_{i=1}^{r}\lambda_i\text{Tr}(M_i)=0$ for all $M_i \in \text{End}_k V_i$. But each $Tr(M_i)$ can range independently over $k$, so it must be that $\lambda_1,...,\lambda_r=0$)
How I can see this part? But each $Tr(M_i)$ can range independently over $k$, so it must be that $\lambda_1,...,\lambda_r=0$)
Any help would be greatly appreciated.
Thanks.

Each $\mathrm{Tr}(M_i)$ can range independently over $k$ because $k$ is algebraically closed so if $\dim V_i = n$ then for any $a \in k$ we may take $M_i = \sqrt[n]{a}I$.  Then $\mathrm{Tr}(M_i) = a$.
