The polynomial given by the determinant of:

\begin{equation} det(A-\lambda I) \end{equation}

for some Linear Map \(A\). Solutions for \(\lambda\) are the eigenvalues. This is because something is an eigenvalue IFF \((A-\lambda I)v = 0\) for some \(\lambda, v\), so we need \((A-\lambda I)\) to be singular.

Characteristic polynomial of a 2x2 matrix is given by \(\lambda^{2}-tr(A)\lambda + det(A)\).