Let B be the 2 x 2 matrix (bij) 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! - ... .]
If sin A has the value given, then 1 - sin2A = (cij) with c11 = c21 = c22 = 0, c21 ≠ 0. But we should have cos2A = (cij).
This is not possible. Suppose cos A = (dij), 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.
© John Scholes
12 Dec 1998