Trying to compute eigenvalues using traces. Let $A$ be an $n$ by $n$ matrix over $\mathbb{C}$. There are $n$ eigenvalues of $A$: $\lambda_1, \lambda_2, \ldots, \lambda_n$. How can we guess $\lambda_1, \lambda_2, \ldots, \lambda_n$ using traces? We have
\begin{align}
& tr(A) = \lambda_1 + \cdots + \lambda_n.
\end{align}
Let $D = diag(\lambda_1, \ldots, \lambda_n)$. Then there is an invertible matrix $P$ such that $D = PAP^{-1}$. Therefore $A^2 = P^{-1}D^2P$. We have 
\begin{align}
& tr(A^2) = tr(P^{-1}D^2P) = tr(D^2)= \lambda_1^2 + \cdots + \lambda_n^2.
\end{align}
Using these results, can we find $\lambda_1, \ldots, \lambda_n$? Thank you very much.
 A: Short answer: Yes using Newton's Identities.
A: In your argument, you have assumed that $A$ is diagonalizable and there's no need to do that. Anyway, if you know $c_1 := \operatorname{tr}(A), \dots, c_n := \operatorname{tr}(A^n)$, you can can use them together with Newton's identities to reconstruct the characteristic polynomial of $A$ and then by solving it (explicitly, if you can, or numerically), you can find the eigenvalues. Note that you don't get an explicit expression for the eigenvalues themselves, only for the characteristic polynomial of $A$ so if $n \geq 5$, you generically won't have an algebraic formula for the eigenvalues in terms of $c_n$ (which shouldn't be surprising).
For example, if $n = 2$ and you know $c_1 := \lambda_1 + \lambda_2$ and $c_2 := \lambda_1^2 + \lambda_2^2$ then the characteristic polynomial of $A$ is given by
$$ x^2 - (\lambda_1 + \lambda_2)x + (\lambda_1 \lambda_2) = x^2 - c_1 x + \frac{c_1^2 - c_2}{2}.$$
To solve for the eigenvalues, we need the roots of
$$ 2x^2 - 2c_1x + (c_1^2 - c_2) = 0 $$
which are given by
$$ \lambda_{1,2} = \frac{2c_1 \pm \sqrt{4c_1^2 - 8(c_1^2 - c_2)}}{4} = \frac{c_1 \pm \sqrt{2c_2 - c_1^2}}{2}. $$
