 Characteristic polynomial from traces of powers

For a complex $n\times n$ matrix $A$, $\mathrm{tr}(A^k)$ is the sum of the $k$th powers of its eigenvalues, counted with algebraic multiplicity. This does not require diagonalizability: a similar upper triangular matrix has the eigenvalues on its diagonal, and taking powers raises those diagonal entries to their corresponding powers. Newton's identities therefore recover its characteristic polynomial from the first $n$ such traces.

For the Jordan block $A=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$, $\mathrm{tr}A=4$ and $\mathrm{tr}(A^2)=8$. Hence $e_1=4$ and $e_2=(4^2-8)/2=4$, giving $\det (zI-A)=z^2-4z+4=(z-2{)}^2$. The repeated eigenvalue is recovered even though this matrix is not diagonalizable.^[ref]