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
 A: Looking at determinants, we find $x^2=2$,  so if at all, we should have $x=\sqrt 2$.
A: Apply your matrix to the unit vectors 
$\begin{bmatrix}
  -1 & -1 \\
  1 & -1 \\
\end{bmatrix}
\begin{bmatrix}
  1 \\
  0 \\
\end{bmatrix}
=
\begin{bmatrix}
  -1 \\
  1 \\
\end{bmatrix}$
and
$\begin{bmatrix}
  -1 & -1 \\
  1 & -1 \\
\end{bmatrix}
\begin{bmatrix}
  0 \\
  1 \\
\end{bmatrix}
=
\begin{bmatrix}
  -1 \\
  -1 \\
\end{bmatrix}$
The enlargment is the ratio between the lenght of the vectors, that is $\sqrt{2}$ and the angle is the one between the unit vectors and the transformed vectors, that is $\dfrac{3\pi}{4}$.
A: 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.
