What's the method to finding the scale factor of enlargement and rotation of a 2D matrix? The matrix M is defined by:
\begin{bmatrix}
  -1 & -1 \\
  1 & -1 \\
\end{bmatrix}
Assuming the matrix represents an enlargement followed by a rotation
My idea here was to make an equation so you're left with simultaneous equations to solve.
$\begin{bmatrix}
  \cos \left(θ\right) & \sin \left(θ\right) \\
  -\sin \left(θ\right) & \cos \left(θ\right) \\
\end{bmatrix}
\begin{bmatrix}
  x & 0 \\
  0 & x \\
\end{bmatrix}
=
\begin{bmatrix}
  -1 & -1 \\
  1 & -1 \\
\end{bmatrix}$
$\begin{bmatrix}
  x\cos \left(θ\right) & x\sin \left(θ\right) \\
  -x\sin \left(θ\right) & x\cos \left(θ\right) \\
\end{bmatrix}
=
\begin{bmatrix}
  -1 & -1 \\
  1 & -1 \\
\end{bmatrix}$
This is where I get stuck. I don't think you can solve this problem like this but if you can, please answer. Regards
A couple more questions,
Does the type of enlargement and type of rotation alter this method? e.g. a scale factor more or less than 1 and a clockwise or counter clockwise rotation.
Also if there is an easier method to finding the matrices could someone please answer with working?
Regards Tom

If you suspect that this matrix is a scaling followed by a rotation, you can apply it to some basis vectors to get a clue.
For instance multiplying your matrix on $[1,0]^T$ yields $[-1, 1]$. And applying it to $[0,1]^T$ yields $[-1, -1]$.
The scale factor for $[1,0]^T$ is $\sqrt{2}$. The rotation can be figured out from the dot product $[1,0]^T \cdot [-1, 1]^T = -1 = \sqrt{2} \cos(\theta)$. Thus $\cos(\theta) = -1/\sqrt{2}$.
The scale factor for $[0,1]^T$ is again $\sqrt{2}$. The dot product gives us $[0,1]^T \cdot [-1,-1]^T = -1 = \sqrt{2} \cos(\theta)$. So again $\cos(\theta) = -1/\sqrt{2}$.
This tells us that $\theta = 3\pi/4$ or $\theta=-3\pi/4$.
Now we just test some matrices:
For $\theta = 3\pi/4$:
$$\left[\begin{array}{cc} -1/\sqrt{2} & 1/\sqrt{2}\\ -1/\sqrt{2} & -1/\sqrt{2} \end{array}\right] \left[ \begin{array}{cc} \sqrt{2} & 0\\ 0 & \sqrt{2} \end{array}\right] = \left[ \begin{array}{cc} -1 & 1\\ -1& -1\end{array}\right]$$
For $\theta = -3\pi/4$:
$$\left[\begin{array}{cc} -1/\sqrt{2} & -1/\sqrt{2}\\ 1/\sqrt{2} & -1/\sqrt{2} \end{array}\right] \left[ \begin{array}{cc} \sqrt{2} & 0\\ 0 & \sqrt{2} \end{array}\right] = \left[ \begin{array}{cc} -1 & -1\\ 1& -1\end{array}\right].$$
And this gives the decomposition you wanted.
