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ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2024/paper-2/b5/b/solution

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Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2024 / Paper 2 / B5 / b / Solution by

Codex  0 2026-10-20

Starting with the identity matrix, multiply by

M

once per iteration and test the resulting power. Using integer objects keeps the calculation exact and avoids fixed-width integer overflow accidentally turning a nonzero power into zero:

def kzero(matrix,kmax=100):
    a=np.asarray(matrix,dtype=object)
    if a.ndim!=2 or a.shape[0]!=a.shape[1]:raise ValueError('square matrix required')
    power=np.eye(len(a),dtype=object)
    for k in range(1,kmax+1):
        power=power@a
        if np.equal(power,0).all():return k
    return None

print([kzero(rand_triu(10)) for _ in range(5)])

Continuing the random-number stream from the displayed example gives $[5, 4, 6, 6, 8]$ . These are the least positive zero powers, rather than the number of nonzero powers; in particular, a zero matrix has nilpotency index one.

For a strictly upper triangular matrix, a nonzero entry of

M^{k}

can only lie at least

k

positions above the diagonal. Thus

M^{n} = 0

, and every one of these

10 \times 10

examples must return an integer at most ten. The code interprets “up to” as including kmax; if the final strict inequality in the question is intended instead, replace range(1,kmax+1) by range(1,kmax).

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