the repeated eigenvalues case
The eigenvalue method needs n independent solutions, but sometimes a repeated eigenvalue is greedy — it claims two slots yet only supplies one eigenvector. You are left one solution short, and the clean formula e^(lambda t) v cannot fill the gap. This is the case that needs a genuinely new idea.
Suppose lambda is a double eigenvalue but A has only one eigenvector v for it. One solution is still e^(lambda t) v. By analogy with the second-order repeated-root case, where the second solution gains a factor of t, you try x(t) = t e^(lambda t) v + e^(lambda t) w. Substituting into x' = A x forces a condition on w: (A - lambda I) w = v. A vector w solving this is a generalized eigenvector — it is not an eigenvector, but applying (A - lambda I) to it lands you on the true eigenvector v. The second solution is then e^(lambda t) (t v + w), and together with e^(lambda t) v it completes the fundamental set. For higher multiplicities you build a longer chain w_1, w_2, ... each mapped by (A - lambda I) onto the previous, and solutions pick up higher powers of t.
The honest subtlety: a repeated eigenvalue does not always misbehave. If it happens to supply as many independent eigenvectors as its multiplicity (a non-defective, diagonalizable case), no t-factors are needed and you proceed as in the distinct case. The generalized-eigenvector machinery is required only when eigenvectors run short — the so-called defective case.
A = [2, 1; 0, 2] has the double eigenvalue 2 but only one eigenvector v = (1, 0). Solve (A - 2I) w = v: [0,1;0,0] w = (1,0) gives w = (0, 1). Solutions: e^(2t)(1,0) and e^(2t)(t(1,0) + (0,1)) = e^(2t)(t, 1).
The missing solution is rescued by a factor of t and a generalized eigenvector w with (A - lambda I) w = v.
A repeated eigenvalue is defective only when it has fewer independent eigenvectors than its multiplicity. The identity matrix's eigenvalue 1 is repeated but fully non-defective — it has every vector as an eigenvector and needs no t-terms.