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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.

$\begin{bmatrix} cos(θ) & sin(θ) \\ -sin(θ) & cos(θ) \\ \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & x \\ \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 1 & -1 \\ \end{bmatrix}$

$\begin{bmatrix} xcos(θ) & xsin(θ) \\ -xsin(θ) & xcos(θ) \\ \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 1 & -1 \\ \end{bmatrix}$

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

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Looking at determinants, we find $x^2=2$, so if at all, we should have $x=\sqrt 2$.

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  • $\begingroup$ So x is equal to the square root of the determinant? Is this always the case? $\endgroup$ – Thomas Winkworth Feb 10 '15 at 16:17
  • $\begingroup$ The determinant of a product is the product of the determinants. The determinant of a rotation matrix, M, is 1. The determinant of a diagonal matrix is the product of the diagonals. Thus $$det(M(xI_2)) = det(M) det(xI_2) = 1 \cdot (x^2) = x^2.$$ $\endgroup$ – Joel Feb 10 '15 at 18:27
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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 $\dfrac{3\pi}{4}$.

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  • $\begingroup$ In principle this is right. However, there are two (non-coterminal) values of $\theta$ that result in $\cos(\theta) = -1/\sqrt{2}$. $\theta=3\pi/4$ does not produce the matrix in question. $\endgroup$ – Joel Feb 10 '15 at 16:09
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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.

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