Prove: The characteristic equation of a $2\times2$ matrix can be expressed as $\lambda^2-tr(A)\lambda+\det(A)=0$.
Let $$A=\begin{pmatrix} a&b\\ c&d\\ \end{pmatrix}$$ We want to show that the characteristic equation derived from $\det(A-\lambda I)=0$ is equivalent to the one shown above. We will define $tr(A)$ and $\det(A)$ for later:
$tr(A)=a+d$
$\det(A)=ad-bc$
So, the characterstic equation is for $A$ is defined as:$$\det(A-\lambda I)=\begin{pmatrix} a-\lambda&b\\ c& d-\lambda\\ \end{pmatrix}=0$$ $$(a-\lambda)(d-\lambda)-bc=0$$ $$ad-a\lambda-d\lambda+\lambda^2-bc=0$$
Rearranging and grouping terms: $$\lambda^2-(a+d)\lambda+(ad-bc)=0$$
Substituting our earlier formulas for $tr(A)$ and $\det(A)$: $$\lambda^2-tr(A)\lambda+\det(A)=0$$
Q.E.D.
Is this proof correct and true in full generality? Is there room to be more formal anywhere? Pointers would be appreciated! Thanks in advance.
This equation comes from the following facts:
1) the trace is a sum of eigenvalues
2) the determinant is the product of eigenvalues
3) Vieta's formulas