 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/2/a/iv/solution

F_{2}

, the matrix becomes

A = 000100001 .

On the first two coordinate vectors it is a nonzero nilpotent Jordan block of size two, and on the last it is the identity map. Consequently

m_{A} (x) = x^{2} (x - 1) over F_{2} .

The repeated zero root prevents diagonalizability. Directly, the only possible eigenvalues are zero and one, so diagonalizability would imply

A^{2} = A

, whereas the displayed off-diagonal entry disappears on squaring.

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