My question is that given an arbitrary $2\times2$ matrix with unit determinant, say $$
G_r =\left(\begin{array}{cc}\cos(g d) & -\frac{1}{n_0 g} \sin(g d) \\ n_0 g \sin(g d) & \cos(g d)\end{array}\right)$$ or $$G_r = \begin{pmatrix} \cosh(\hat g d) & -\frac{1}{n_0 \hat g} \sinh(\hat g d) \\ -n_0 \hat g \sinh(\hat g d) & \cosh(\hat g d) \end{pmatrix}$$ where $n_0, g, \hat g, d\in \mathbb{R}$,
is there a way to determine if the matrix can be decomposed into the multiplication of a series of shear matrices $R_a = \begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix}$
and $T_a = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ where $a\in \mathbb{R}$, i.e. $G_r = R_{a1}T_{b1}R_{a2}T_{b2}\dots$
My thoughts: It is very clear that those matrices all belong to $SL(2)$ group, which means $det = 1$. I have tried several unsuccessful way of decomposition:
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LU decomposition: this can decompose the matrix into upper and lower diagonal matrices, but with no guarantee that the diagonal entries are ones.
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Try to prove ${I, T_a, R_a, G_r}$ forms a subgroup, and $T_a$ and $R_a$ are group generators. I'm not entirely sure about how to prove group generators and hope someone can give me some advice.
Best Answer
Consider the following matrix decompositions into products of shears: \begin{eqnarray} R &=& \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix} &=& \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix} \\ S_c &=& \begin{bmatrix}c & 0\\0 & c^{-1}\end{bmatrix} &=& \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\c-1 & 1\end{bmatrix}\begin{bmatrix}1 & -c^{-1}\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\c(1-c) & 1\end{bmatrix} \end{eqnarray} Now, consider an arbitrary matrix 2x2 matrix $A$ of unit determinant. If the upper left entry of $A$ is $0$, use $RA$ instead. Find the LDU-decomposition of $A$ (or $RA$). $L$ and $U$ will be shears, since they are triangular matrices with unit diagonal. Since $D$ is a diagonal matrix with unit determinant, it be equal to $S_c$ for some $c$. Thus, either $A = LS_cU$ or $A = RLS_cU$. Since $L$ and $U$ are themselves shears and $R$ and $S_c$ can be decomposed into shears, $A$ can be decomposed into shears.
I think this generalizes to $n\times n$ matrices with unit determinant. All of the row swap-and-negate matrices $R$ can be analogously written as products of shears, the $L$ and $U$ of the LDU-composition are unit triangular matrices which obviously decompose into shears, and a unit determinant diagonal matrix should be factorable into products of $S_c$-like matrices, which themselves decompose into shears.
Follow-up: what's the minimum number of shears needed to represent any unit determinant matrix?