Recipe

Compute a function of a Jordan block

Use when: you need f(A) for an operator A that isn't diagonalizable, given A's Jordan form, and f has enough derivatives at each eigenvalue

Goal: f(J) for each Jordan block, assembled and conjugated back into f(A)

Taught in Functions of a general matrix, via Jordan blocks

1. 1

   Identify J_k(\lambda)'s eigenvalue \lambda and its nilpotent shift part N (1's on the superdiagonal, 0 everywhere else).

   Example

   For J=\begin{pmatrix}2&1\\0&2\end{pmatrix}: \lambda=2 and N=\begin{pmatrix}0&1\\0&0\end{pmatrix}.

   Why: this split is what the Taylor-expansion formula behind this whole section is actually built around.

2. 2

   Differentiate f up to one fewer derivative than the block's size k, then evaluate every derivative at \lambda.

   Example

   For f(x)=x^3 and block size k=2: f(2)=8 and f'(x)=3x^2, so f'(2)=12; only one derivative is needed since k=2.

   Why: the Taylor terms beyond order k-1 would multiply N^k=\mathbf 0, so this section's own formula genuinely never needs them.

3. 3

   Put f(\lambda) on the main diagonal, f'(\lambda) on the superdiagonal, f''(\lambda)/2! two diagonals up, and so on, dividing each derivative by its own factorial.

   Example

   f(J)=\begin{pmatrix}f(2)&f'(2)\\0&f(2)\end{pmatrix}=\begin{pmatrix}8&12\\0&8\end{pmatrix}.

   • block size k=1 — only f(\lambda) is needed; no derivative terms exist at all, matching section 6's diagonalizable case exactly.
   • block size k\ge2 — every diagonal above the main one needs one more derivative, divided by its own factorial.

   ⚡ Shortcut

   For f=\exp, every derivative of f equals f itself, so every diagonal of f(J) uses the identical value e^\lambda; no separate derivative computation is needed.

   Why: this placement pattern is exactly what the Taylor expansion of f around \lambda, applied to N, produces term by term.

4. 4

   Repeat the steps above for every Jordan block, place the results block-diagonally in the Jordan form, then conjugate by the change-of-basis matrix used to reach it.

   Example

   Direct computation confirms J^3=\begin{pmatrix}8&12\\0&8\end{pmatrix}, matching f(J) from the previous step exactly.

   Why: this matches the same A=PJP^{-1} conjugation the diagonalizable case already used, just with a block-structured J in place of a purely diagonal D.