Decomposition of shear matrix into rotation & scaling How can I decompose the affine transformation:
$$ \begin{bmatrix}1&\text{shear}_x\\\text{shear}_y&1\end{bmatrix}$$
into rotation and scaling primitives?
$$ \begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$$
\begin{bmatrix}\text{scale}_x&0\\0&\text{scale}_y\end{bmatrix}

You can only do this if $\textrm{shear}_x = -\textrm{shear}_y$, in which case
$$
\begin{align}
\theta &= \textrm{atan}(\textrm{shear}_y)\\
\textrm{scale}_y&=\sqrt{\textrm{shear}_y^2 + 1}\\
\textrm{scale}_x&=-\sqrt{\textrm{shear}_y^2 + 1}\\
\end{align}
$$

To see this we can identify both matrices:
$$
\begin{bmatrix}1& \text{shear}_x \\ \text{shear}_y&1\end{bmatrix} =
\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}
\begin{bmatrix}\text{scale}_x&0\\0&\text{scale}_y\end{bmatrix}
$$
$$
\begin{bmatrix}1& \text{shear}_x \\ \text{shear}_y&1\end{bmatrix} =
\begin{bmatrix}\text{scale}_x~\cos\theta&-\text{scale}_y~\sin\theta\\ \text{scale}_x~\sin\theta&\text{scale}_y~\cos\theta\end{bmatrix}
$$
Which yields the following equation system:
$$
\begin{eqnarray}
1 &= \text{scale}_x~\cos\theta\\
1 &= \text{scale}_y~\cos\theta\\
\text{shear}_x &= -\text{scale}_y~\sin\theta\\
\text{shear}_y &= \text{scale}_x~\sin\theta\\
\end{eqnarray}
$$
. This can only work if neither $\text{scale}_x$, $\text{scale}_y$, or $\cos\theta$ are zero. If that is the case, then we can immediately see that the scaling is isotropic ($\text{scale}_x=\text{scale}_y=\textrm{scale}$) and that
$$\textrm{scale} = \frac{1}{\cos\theta}$$
. We can use this to simplify the other two equations
$$
\begin{align}
\text{shear}_x &= - \frac{1}{\cos\theta}~\sin\theta = - \tan\theta\\
\text{shear}_y &=  \frac{1}{\cos\theta}~\sin\theta = \tan\theta\\
\end{align}
$$
. Both can only be true at the same time, if $\text{shear}_x = -\text{shear}_y$, yielding above constraint. From here, we can find values for both $\theta$ and $\textrm{scale}$:
$$
\begin{align}
\theta &= \textrm{atan}~\textrm{shear}_y\\
\textrm{scale} &= \frac{1}{\cos{\theta}} = \frac{1}{\cos(\textrm{atan}~\textrm{shear}_y)} = \sqrt{\textrm{shear}_y^2 + 1}
\end{align}
$$
