 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/33c/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Adopt the boundary convention

a_{0} = a_{n} = 0

. Since

L

has nonzero entries only on its first off-diagonals and

B

only on its second off-diagonals,

B L

and

L B

can contribute to

[B, L]

only on the first and third off-diagonals. On the upper first off-diagonal,

(B L)_{j, j + 1} = a_{j} a_{j + 1}^{2}, (L B)_{j, j + 1} = a_{j - 1}^{2} a_{j},

[B, L]_{j, j + 1} = a_{j} (a_{j + 1}^{2} - a_{j - 1}^{2}) .

The lower first off-diagonal is equal to this by part (a). On the third off-diagonal the two products cancel:

(B L)_{j, j + 3} = a_{j} a_{j + 1} a_{j + 2} = (L B)_{j, j + 3} .

The diagonal and every remaining entry vanish. Therefore, if

c_{j} = a_{j} (a_{j + 1}^{2} - a_{j - 1}^{2}),

then

[B, L] = (0 c_{1} 0 c_{1} 0 c_{2} 0 c_{2} 0 ⋱) .

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