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Let $A$ be a $\mathbb{C}$-algebra.

For finite dimensional representation $\rho:A \rightarrow \rm{End}(V)$, the trace $\chi_\rho$ of $\rho$ is defined by $\chi_\rho(a):=\rm{Tr}(\rho(a))$ for $a \in A$.

It is true that if two finite dimensional representations $\rho,\sigma$ are isomorphic then $\chi_\rho =\chi_\sigma$? Now, if $\chi_\rho = \chi_\sigma$, are $\rho$ and $\sigma$ isomorphic?

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Not in general. For instance, let $A=\mathbb{C}[x]/(x^2)$. Consider two different representations of $A$ on $\mathbb{C}^2$: $\rho$, which sends $x$ to $\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$, and $\sigma$, which sends $x$ to $0$. For any $a\in A$, $\rho(a)$ and $\sigma(a)$ have the same diagonal entries, and hence the same trace. But $\rho$ and $\sigma$ are not isomorphic.

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  • $\begingroup$ Oh! it's simple counter example. Thank you. $\endgroup$ – H.Kakuhama Sep 27 '16 at 9:45
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That the answer to your question is „no” has been shown in Eric Wofsey's answer. But there is a little bit more that can be said.

Let $A$ be a finite dimensional algebra over a field $\mathbb{k}$ of characteristic zero, and $M_1, M_2$ finitely generated $A$-modules with corresponding characters $\chi_1, \chi_2$. Then $\chi_1 = \chi_2$ if and only if $M_1$ and $M_2$ have the same composition factors, counted with multiplicities. Alternatively, $[M_1] = [M_2]$ as elements of the Grothendieck group $G_0(A)$.

In the counterexample of Eric Wofsey, both modules have the composition factor $\mathbb{C}$ (on which $x$ acts trivially) with multiplicity 2, so their characters have to coincide.

Details can be found in chapter 3 of [Lam - a first course in noncommutative rings]. See in particular Theorem 7.19.

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