If we let \[ A = \begin {bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \] the characterstic equation can be found such that \[ (\lambda I - A)v =0 \] In our case, that means \[ (\lambda I - A)v = \begin {bmatrix} \lambda - 1 & 2 \\ 3 & \lambda - 4 \end{bmatrix} \] Then, we compute the determinant
\[ \det\begin {bmatrix} 1-\lambda & 2 \\ 3 & 4-\lambda \end{bmatrix} \] Whch is \[ (1-\lambda)(\lambda-4)-2(3) = 0\\ \lambda^2-5\lambda-2 = 0 \]
otherwise known as the characteristic polynomial. The reason it is called the characteristic equation is because the eigenvalues of a matrix are the linearly independent vectors that span the space of \(R^n\).