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?

  • $\begingroup$ There is an infinity of matrix products that equal δ. What is the criterion for selecting the one in the question? $\endgroup$ – bbgodfrey Nov 10 '17 at 21:38
  • $\begingroup$ @bbgodfrey In the question I ask to find this precise factorization. The reason is that $\det \delta = 1$ iff at least one of the two matrices of the factorized form is singular. Having $\delta = I_n \delta$ would not be useful. Here, it provides a factorization of $\det\delta$. Hope that answers your question. $\endgroup$ – anderstood Nov 10 '17 at 22:13
  • $\begingroup$ @bbgodfrey I doubt your were asking for that level of detail, but this factorization is associated with a very nice physical interpretation (in short, symmetry of some motions of impact oscillators), see this article, near equation (3.15) --- unfortunately I can't think of a short explanation... It is used to find such motions. $\endgroup$ – anderstood Nov 10 '17 at 22:19
  • $\begingroup$ Are $ A_1 $ and $ A_2 $ simultaneouly diagonalizable? $\endgroup$ – Αλέξανδρος Ζεγγ Oct 7 '18 at 7:43
  • $\begingroup$ @ΑλέξανδροςZεγγ unfortunately not... $\endgroup$ – anderstood Oct 8 '18 at 12:49

I did the problem in reverse and showed the matrix product (which I called δ2) reduces to the difference of matrices with

δ2 // FunctionExpand // Simplify 

Only then I could reduce the Sinc[A] as Sin[A]/A. The result is that of a difference of two matrices.

  • 2
    $\begingroup$ OK but my question was on factorizing, not expanding :) $\endgroup$ – anderstood Nov 10 '17 at 20:20

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