 Solution

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

F_{3}

(x - 1)^{3}

. For

B = A - I

, reduction of the entries gives

B = 201120012, B^{2} = 111111111 \neq = 0, B^{3} = 0.

Hence

m_{A} (x) = (x - 1)^{3} over F_{3} .

Its repeated linear polynomial prevents diagonalizability. Equivalently, a diagonalizable map with only the eigenvalue one would have to be

I

, and this matrix is not

I

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