Can someone help me show that the characteristics equation for a $2\times 2$ matrix, say $A$, can be written as: $$ \lambda^2 - \mathrm{tr}(A)\lambda + \det(A) = 0,$$
where $\mathrm{tr}(A)$ is the sum of the diagonal entries of $A$?
I'm having trouble figuring out the question, we haven't been taught this yet. Please help.
$\endgroup$ 12 Answers
$\begingroup$Alternative proof: The the roots of the characteristic polynomial $\chi_A$ of $A$ are the eigenvalues of $A$, so if $\lambda_0,\lambda_1$ are eigenvalues of $A$, by Vieta's theorem we have
$$ \chi_A(\lambda) = \lambda^2 - \lambda (\lambda_0 + \lambda_1) + \lambda_0\lambda_1 $$ Since $\operatorname{tr} A = \lambda_0 + \lambda_1, \det A = \lambda_0\lambda_1$,
$$ \chi_A(\lambda) = \lambda^2 - \lambda \operatorname{tr} A + \det A $$
$\endgroup$ $\begingroup$$$A=\left(\begin{matrix} a & b\\ c & d\\ \end{matrix}\right).$$
The characteristic polynomial is $$\det (A-\lambda I)=(a-\lambda)(d-\lambda)-bc=ad-a\lambda-d\lambda+\lambda^2-bc.$$ Setting this polynomial equal to zero we have $$\lambda^2-(a+d)\lambda+ad-bc=0.$$
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