## Compute the Jordan canonical form of a given matrix

Let $A = \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix}$. Find the Jordan canonical form for $A$.

Clearly the characteristic polynomial of $A$ is $(x-1)^4$, which factors completely over any field. Certainly $A$ does not satisfy $x-1 = 0$, and a quick computation verifies that $A$ does not satisfy $(x-1)^2 = 0$ or $(x-1)^3 = 0$. So the minimal polynomial of $A$ is $(x-1)^4$, and thus the Jordan canonical form is $\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}$.