Once the characteristic equation hands you an eigenvalue \lambda, the matching eigenvectors are the non-zero solutions of

Subtract \lambda from the diagonal of A, then find the direction that gets sent to zero — a linear system you already know how to solve. Because A - \lambda I is singular by construction, there's a whole line of solutions: every eigenvector comes with all its scalar multiples, so we just report the direction.

Worked example, both directions

Take A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}, whose eigenvalues are \lambda = 3 and \lambda = 1. Pick one below: the diagram draws the eigenvector direction \vec{v} and confirms A\vec{v} = \lambda\vec{v} by landing A\vec{v} right on the same line, scaled by \lambda.

The recipe, start to finish

1. Solve \det(A - \lambda I) = 0 for the eigenvalues.
2. For each \lambda, solve (A - \lambda I)\vec{v} = \vec{0} for the eigenvector direction.

That pair of steps unlocks the payoff: a matrix with a full set of independent eigenvectors can be diagonalized — rewritten as a pure stretch along its own axes.