Let's say we have an endormorphism $\lambda: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, describing a volume-preserving transformation on a two-dimensional plane, so we have $\det \lambda = 1$.
I'm trying to prove that $\lambda$ can be written as composition of (at most) three shearing transformations, e.g. $\sigma_1 \circ \sigma_2 \circ \sigma_3 = \lambda$. Trivially, since shears are volume-preserving, they all have a determinant of 1, so this must also be true for their composition.
I started by looking at the transformation matrix of $\lambda$
$$ \Lambda := \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad a,b,c,d \in \mathbb{R}, $$
and trying to find three shear matrices, such that their product equals $\Lambda$. I came up with
\begin{equation*}
\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ y & 1 \end{pmatrix} \begin{pmatrix} 1 & z \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} xy+1 & x+z+xyz \\ y & yz+1 \end{pmatrix}, \\
x=\frac{a-1}{c}, \quad y=c, \quad z=\frac{d-1}{c},
\end{equation*}
for when $c \neq 0$, and
\begin{equation*}
\begin{pmatrix} 1 & 0 \\ x & 1 \end{pmatrix} \begin{pmatrix} 1 & y \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ z & 1 \end{pmatrix} = \begin{pmatrix} yz+1 & y \\ xyz+x+z & xy+1 \end{pmatrix}, \\
x=\frac{d-1}{b}, \quad y=b, \quad z=\frac{a-1}{b}.
\end{equation*}
when $b \neq 0$.
The issue is, that I can't seem to figure out a combination of three shear matrices in the event that $c = b = 0$. So I went back to the drawing board, and tried to apply the general definition of a shearing transformation
$$ \sigma = \textrm{id}_V + w\omega, \quad \textrm{where } \omega \in V^*, w \in \ker \omega, $$
but so far without success. If for example $c \neq 0$, we can pick $[\omega, w]$ as $ [e_2^*, e_1x], [e_1^*, e_2y], [e_2^*, e_3z] $, respectively.
Am I overlooking something here?
Best Answer
I have managed to find a solution involving three shears in the case where $c = 0 = b$. It's still possible - if we allow shears that are not constrained to an axis.
If we apply the general definition of a shear $$ \sigma = \textrm{id}_V + w\omega, \quad \textrm{where } \omega \in V^*, w \in \ker \omega, $$ and choose $$ \omega : V \rightarrow \mathbb{R}, v \mapsto v_1 + v_2, \quad w = \begin{pmatrix} -y \\ y \end{pmatrix} \in \ker \omega, $$ we get the transformation matrix $$ S = \begin{pmatrix} 1-y & -y \\ y & 1+y \end{pmatrix}. $$ We can also verify $\det S = 1$.
Using two other "basic" shears, we get $$ \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 - y & -y \\ y & 1 + y \end{pmatrix} \begin{pmatrix} 1 & 0 \\ z & 1 \end{pmatrix} = \begin{pmatrix} 1-y+xy+xz-yz+xyz & x-y+xy \\ y+z+zy & y+1 \end{pmatrix}, $$ which we can solve with respect to $\Lambda$ with $$ x = \frac{b+d-1}{d}, \quad y = d - 1, \quad z = \frac{c-d+1}{d}. $$
We know $d \neq 0$, since $$ \det \lambda = ad - bc = 1 \quad \textrm{and} \quad a = 0 = d \implies b \neq 0 \neq c. $$
As such, combined with the two other solutions stated in my initial solution, we have covered every case with a potential solution.
$$\tag*{$\blacksquare$}$$