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Consider two real square matrices $A_1$ and $A_2$ and $t_1,t_2\in\mathbb{R}$. $A_1$ and $A_2$ do not commute. 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 $M_2(t)$ defined similarly 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 $\Delta$.

Is there a similar factorization for $M_1(2t_1)M_2(2t_2) - M_2(2t_4)M_1(2t_3)$ as the product of two matrices (or more)? What about the factorization of the more general case

$$\prod_{i=1}^k M_{\epsilon(i)}(2t_i) - \prod_{i=1}^k M_{\epsilon(i+1)} (2t_{2k-i+1})$$ with $\epsilon(i)=1$ if $i$ is odd and $2$ if $i$ is even?

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  • $\begingroup$ Interesting question. But how sure are you that such a factorization should exist? The LHS does not contain a product of two sinc terms, at least not for k=2, but by symmetry considerations, I think that any suitable factorization should contain, once expanded, at least one such product. (I guess rather 2 or 4). Well, that would mean those'd have to cancel out, which would yield a powerful constraint. $\endgroup$
    – Wolfgang
    Commented Apr 10, 2017 at 14:24
  • $\begingroup$ @Wolfgang I am really not sure, but I guess there is a factorization with 2 product (my guess is based on observing on some numerical examples two distinct "behaviours" of the kernel of $\Delta$; by "behaviour" I mean something dubious which I can't explain clearly, yet). As per the $\mathrm{sinc}$ terms, I got rid of them by assuming $A_i$ nonsingular (then $t_i\mathrm{sinc}(t_iA_i)=A_i^{-1}\mathrm{sin}(t_iA_i)$), and by doing so I might have missed something. I will think about your argument. $\endgroup$
    – anderstood
    Commented Apr 10, 2017 at 15:09
  • $\begingroup$ My comment (talking about symmetry considerations) was based on the idea that you might look for 4 factors. $\endgroup$
    – Wolfgang
    Commented Apr 10, 2017 at 19:53

1 Answer 1

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Even if we replace the $A_i$ by $1\times1$ scalars $a_i$, I don't think that there is such a factorization, at least not for $k=2$ and a forteriori neither for bigger $k$. Putting $u_i:=2t_i$, we have, modulo sign errors,$$ M:=M_1(2t_1)M_2(2t_2) - M_2(2t_4)M_1(2t_3)=\begin{pmatrix} a_1^{-1}&0\\ 0&1\\ \end{pmatrix}\Bigl( (a_1-a_2) M_-+(a_1+a_2) M_+\Bigr) \begin{pmatrix} 1&0\\ 0&a_2^{-1}\\ \end{pmatrix},$$ where $ M_-=\begin{pmatrix} \cos(u_1-u_2) -\cos(u_3-u_4) & \sin(u_1-u_2) -\sin(u_3-u_4) \\ - \sin(u_1-u_2) +\sin(u_3-u_4) & -\cos(u_1-u_2) +\cos(u_3-u_4)\\ \end{pmatrix}$

and $ M_+=\begin{pmatrix} \cos(u_1+u_2) -\cos(u_3+u_4) & -\sin(u_1+u_2) +\sin(u_3+u_4) \\ -\sin(u_1+u_2) +\sin(u_3+u_4) & \cos(u_1+u_2) -\cos(u_3+u_4) \\ \end{pmatrix}. $

Using formulas like $\displaystyle \cos \alpha-\cos \beta=-2\sin {\frac {\alpha+\beta}{2}}\sin {\frac {\alpha-\beta}{2}}$, the entries of $M_-$ and $M_+$ can be written as products, which turn out to have common factors: $$ M_-=2\sin(-t_1+t_2+t_3-t_4)\begin{pmatrix} \sin(t_1-t_2+t_3-t_4) & -\cos(t_1+t_2+t_3+t_4) \\ \cos(t_1+t_2+t_3+t_4) &\sin(-t_1+t_2-t_3+t_4) \\ \end{pmatrix}\\ =:2\sin x\begin{pmatrix} \sin z & -\cos w \\ \cos w &-\sin z \\ \end{pmatrix}\\ M_+=2\sin(-t_1-t_2+t_3+t_4)\begin{pmatrix} \sin(t_1+t_2+t_3+t_4) & \cos(t_1+t_2+t_3+t_4) \\ \cos(t_1+t_2+t_3+t_4) &\sin(t_1+t_2+t_3+t_4) \\ \end{pmatrix}\\ =:2\sin y\begin{pmatrix} \sin w & \cos w \\ \cos w &\sin w \\ \end{pmatrix}.$$ The four linear combinations $x,y,z,w$ of $t_1,t_2,t_3,t_4$ are linearly independent, so by virtue of this form there is no chance that $(a_1-a_2) M_-+(a_1+a_2) M_+,$ or equivalently $M,$ can be factorized as a whole.

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