characteristic equation

$\det(A-\lambda I) = 0$ — solved for $\lambda$ , its roots are $A$ 's eigenvalues.

Rearranging $A\mathbf v=\lambda\mathbf v$ gives $(A-\lambda I)\mathbf v=\mathbf 0$ , so $\mathbf v$ is an eigenvector exactly when it's a nonzero solution of this homogeneous system — which exists exactly when $A-\lambda I$ is singular, i.e. $\det(A-\lambda I)=0$ , the characteristic equation. Its unknown is $\lambda$ alone; every root is an eigenvalue of $A$ .