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}.

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