a generalized eigenvector
An ordinary eigenvector is a direction that the matrix merely stretches — A v = lambda v, with no change of direction. But when a repeated eigenvalue cannot supply enough such directions, you need stand-in vectors that are almost eigenvectors: they fail the equation by a little, and that little is itself a true eigenvector. These near-misses are generalized eigenvectors.
For an eigenvalue lambda, an ordinary eigenvector satisfies (A - lambda I) v = 0. A generalized eigenvector of rank 2 satisfies the looser condition (A - lambda I)^2 w = 0 while (A - lambda I) w is not zero. The key consequence is that (A - lambda I) w lands exactly on an ordinary eigenvector v — so w sits one step 'behind' v in a chain. More generally a rank-k generalized eigenvector is killed by (A - lambda I)^k but not by (A - lambda I)^(k-1), and applying (A - lambda I) repeatedly walks it down a chain w, ..., to a genuine eigenvector. To find one in the rank-2 case you simply solve the linear system (A - lambda I) w = v for w, taking v to be the eigenvector you already have.
Generalized eigenvectors exist precisely to repair the defective (repeated, eigenvector-short) case of the eigenvalue method, and they are exactly the vectors that produce the t-multiplied solutions e^(lambda t)(t v + w). They are also the backbone of the Jordan canonical form, where each chain becomes a Jordan block. In short: when ordinary eigenvectors run out, generalized ones finish the basis.
For A = [3, 1; 0, 3], lambda = 3 has eigenvector v = (1, 0). A generalized eigenvector solves (A - 3I) w = v: [0,1;0,0] w = (1, 0), so w = (0, 1). Check: (A - 3I)^2 w = 0 but (A - 3I) w = (1,0) is nonzero — rank 2, as required.
Solving (A - lambda I) w = v makes w the generalized eigenvector sitting one rung above the true eigenvector v.
A generalized eigenvector is not an eigenvector — it does not satisfy A w = lambda w. Forgetting this and treating w as an ordinary eigenvector gives a wrong second solution that does not actually solve the system.