matrix powers via Jordan form

Computing A^n for large n directly is hopeless, but with Jordan form it is a one-block-at-a-time formula. Since A = P J P^-1 gives A^n = P J^n P^-1, and J is block-diagonal, you only need to raise each Jordan block to the n-th power — and for a single block that has a clean closed form.

Apply the function-of-a-block formula with f(x) = x^n. For a block J_k(lambda) = lambda*I + N, the binomial expansion (lambda*I + N)^n = sum_{j=0}^{k-1} C(n,j) lambda^{n-j} N^j stops at j = k-1 because N^k = 0. So the entries of J_k(lambda)^n are binomial-coefficient combinations C(n,j) lambda^{n-j} sitting on the j-th superdiagonal.

The shape of the answer is the headline: for a defective matrix, A^n grows like (polynomial in n) times lambda^n. A size-k block contributes a polynomial of degree up to k-1 in n. This is why a single eigenvalue of modulus 1 inside a nontrivial Jordan block makes powers grow polynomially rather than staying bounded.

This drives the asymptotics of discrete dynamical systems x_{n+1} = A x_n, Markov chains, Fibonacci-type recurrences, and PageRank-style iterations. The dominant eigenvalue's modulus sets exponential growth/decay, while its block size sets the polynomial correction — both read straight off the Jordan form.