Suppose
[tex]\mathbf A=\begin{bmatrix}a&b\\c&d\end{cases}[/tex]
with [tex]\mathrm{tr}\,\mathbf A=a+d=-4[/tex] and [tex]\det\mathbf A=ad-bc=0[/tex]. We can find the eigenvalues of [tex]\mathbf A[/tex] in the usual way, by taking the determinant of [tex]\mathbf A-\lambda\mathbf I[/tex] and setting it equal to 0.
[tex]\begin{vmatrix}a-\lambda&b\\c&d-\lambda\end{vmatrix}=(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)[/tex]
and we observe that the characteristic polynomial has linear coefficient [tex]-\mathrm{tr}\,\mathbf A[/tex] and constant coefficient [tex]\det\mathbf A[/tex]. Thus in this case we have the characteristic polynomial equation,
[tex]\lambda^2+4\lambda=\lambda(\lambda+4)=0\implies\lambda_1=0,\lambda_2=-4[/tex]
