function of a Jordan block

How do you apply a function f (like exp, sqrt, or a polynomial) to a matrix? On a single Jordan block the answer is beautifully explicit: f(J_k(lambda)) is upper-triangular, with f(lambda) on the diagonal and the successive derivatives f'(lambda), f''(lambda)/2!, ... marching up the superdiagonals — the Taylor coefficients of f at lambda.

Write J = lambda*I + N with N the nilpotent shift (N^k = 0). Then formally f(J) = sum over j>=0 of f^{(j)}(lambda)/j! * N^j, and since N^k = 0 the sum stops after k terms. Because N^j has 1s on the j-th superdiagonal, the (p,q) entry of f(J) is f^{(q-p)}(lambda)/(q-p)! for q >= p, and 0 below the diagonal. Only the values f(lambda), f'(lambda), ..., f^{(k-1)}(lambda) matter.

This is the engine of the matrix functional calculus. To compute f(A) for a general A: change to a Jordan basis so A = P J P^-1, apply f block by block using the Taylor formula above, then conjugate back: f(A) = P f(J) P^-1. It works whenever f is analytic near each eigenvalue (and the needed derivatives exist).

The caveat that makes Jordan form indispensable here: for a defective matrix you genuinely need the derivatives, not just the values f(lambda). A merely diagonalizable matrix would only use f(lambda) on each eigenvalue; the superdiagonal derivative terms are exactly the imprint of the off-diagonal 1s.