Consider two square matrices $A_1$ and $A_2$. Consider the following matrix involving matrix trigonometric functions:
\begin{equation} M_1(t)=\begin{bmatrix} \cos(tA_1) & t\mathrm{sinc}(t A_1) \\ -A_1\sin(tA_1) & \cos(tA_1) \end{bmatrix} \end{equation} and similarly $M_2(t)$ defined by changing $A_1$ to $A_2$. Using the double-angle identities, it can be shown that
\begin{align} \delta &= M_1(2t_1)-M_2(2t_2) \\ &= 2\begin{bmatrix} t_1\mathrm{sinc}(t_1A_1) & -t_2\mathrm{sinc}(t_2A_2) \\ \cos (t_1 A_1) & -\cos(t_2A_2) \end{bmatrix} \begin{bmatrix}-A_1\sin(t_1A_1) & \cos(t_1 A_1) \\ -A_2\sin(t_2A_2) & \cos(t_2 A_2) \end{bmatrix} \end{align}
which provides a factorization of the difference $\delta$.
This equality can be checked in MMA using random values for $A_1,A_2,t_1,t_2$:
M1[t_] :=
ArrayFlatten[{{MatrixFunction[Cos, t*A1],
t*MatrixFunction[Sinc, t*A1]}, {-A1.MatrixFunction[Sin, t*A1],
MatrixFunction[Cos, t*A1]}}]
M2[t_] :=
ArrayFlatten[{{MatrixFunction[Cos, t*A2],
t*MatrixFunction[Sinc, t*A2]}, {-A2.MatrixFunction[Sin, t*A2],
MatrixFunction[Cos, t*A2]}}]
delta := M1[2 t1] - M2[2 t2]
zero := delta -
2 ArrayFlatten[{{t1*MatrixFunction[Sinc, t1*A1], -t2*
MatrixFunction[Sinc, t2*A2]}, {MatrixFunction[Cos,
t1*A1], -MatrixFunction[Cos,
t2*A2]}}].ArrayFlatten[{{-A1.MatrixFunction[Sin, t1*A1],
MatrixFunction[Cos, t1*A1]}, {-A2.MatrixFunction[Sin, t2*A2],
MatrixFunction[Cos, t2*A2]}}]
Block[{A1 = RandomReal[{-1, 1}, {2, 2}],
A2 = RandomReal[{-1, 1}, {2, 2}], t1 = RandomReal[10],
t2 = RandomReal[10]}, zero] // Chop
(* {{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}} *)
My question is, would it have been possible to find this factorization using MMA?
δ
. What is the criterion for selecting the one in the question? $\endgroup$ – bbgodfrey Nov 10 '17 at 21:38