Putnam 1996/B4 solution

Putnam 1996

Problem B4

Let B be the 2 x 2 matrix (b_ij) with b₁₁ = b₂₂ = 1, b₁₂ = 1996, b₂₁ = 0. Can we find a 2 x 2 matrix A such that sin A = B? [We define sin A by the usual power series: A - A³/3! + A⁵/5! - ... .]

Solution

Answer: no.

Easy.

If sin A has the value given, then 1 - sin²A = (c_ij) with c₁₁ = c₂₁ = c₂₂ = 0, c₂₁ ≠ 0. But we should have cos²A = (c_ij).

This is not possible. Suppose cos A = (d_ij), then we need d₁₁d₁₂ + d₁₂d₂₂ ≠ 0, and hence (d₁₁ + d₂₂) ≠ 0. But we also need 0 = c₂₁ = d₂₁(d₁₁ + d₂₂) and so d₂₁ = 0. Now 0 = c₁₁ = d₁₁² + d₁₂d₂₁ = d₁₁², so d₁₁ = 0. Similarly, 0 = c₂₂ = d₂₂² + d₁₂d₂₁ = d₂₂², so d₂₂ = 0. So d₁₁ + d₂₂ = 0. Contradiction.

Hence sin A cannot have the value given.

The only problem with this argument is that it is too easy! One worries about justifying the various steps. The relation sin²A + cos²A = 1 is an easy calculation directly from the power series definitions, once one has established absolute convergence.

Putnam 1996