Question

(1 point) suppose that the trace of a 2×22×2 matrix aa is tr(a)=−4tr(a)=−4, and the determinant is det(a)=0det(a)=0. find the eigenvalues of aa.

Answer 1

Suppose

$\mathbf A=\begin{bmatrix}a&b\\c&d\end{cases}$

with $\mathrm{tr}\,\mathbf A=a+d=-4$ and $\det\mathbf A=ad-bc=0$. We can find the eigenvalues of $\mathbf A$ in the usual way, by taking the determinant of $\mathbf A-\lambda\mathbf I$ and setting it equal to 0.

$\begin{vmatrix}a-\lambda&b\\c&d-\lambda\end{vmatrix}=(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)$

and we observe that the characteristic polynomial has linear coefficient $-\mathrm{tr}\,\mathbf A$ and constant coefficient $\det\mathbf A$. Thus in this case we have the characteristic polynomial equation,

$\lambda^2+4\lambda=\lambda(\lambda+4)=0\implies\lambda_1=0,\lambda_2=-4$