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If I have points of a unit circle (centered at an origin)

$$ \left\{ \left. \begin{pmatrix} \cos(\varphi) \\ \sin(\varphi) \end{pmatrix} \right| \varphi \in [0;2\pi) \right\} $$

and I affect them using ANY linear transformation (if I understand correctly those transformations are isomorphic to $2 \times 2$ matrices) I should get

$$ \left\{ \left. \begin{pmatrix} A & B \\ C & D \end{pmatrix}\begin{pmatrix} \cos(\varphi) \\ \sin(\varphi) \end{pmatrix} \right| \varphi \in [0;2\pi);A,B,C,D \in \mathbb{R} \right\} $$

The question now is can my result be anything else than an ellipse (centered at an origin)?

I can't quite imagine anything else, but at the same time I realize that an ellipse is defined by two semi-axes and a degree of rotation - that's $3$ characteristics. Meanwhile we have $4$ characteristics in a $2 \times 2$ matrix.

Edit: Using semi-axes $P,Q$ and a degree of rotation $\alpha$ I should then be able to represent the same effect any $2 \times 2$ matrix has on a unit circle points hence: $$ \begin{pmatrix} A & B \\ C & D \end{pmatrix}\begin{pmatrix} \cos(\varphi) \\ \sin(\varphi) \end{pmatrix} = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix}\begin{pmatrix} P\cos(\varphi) \\ Q\sin(\varphi) \end{pmatrix} $$

I don't have much luck with it because apparently I cannot freely eliminate $\varphi$.

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If the matrix is singular, you get a closed bounded line segment. That could be considered a degenerate ellipse. It's not the sort of thing people usually picture when the word "ellipse" is used. –  Michael Hardy Feb 23 '13 at 14:45

2 Answers 2

up vote 1 down vote accepted

Yes, you only get ellipses. You have a fourth degree of freedom which is due to the fact that you can 'rotate' your figure without changing it. For example if you take the matrix $$ \left(\begin{array}{cc} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{array} \right) $$ your circle remains fixed.

addendum

In general the singular value decomposition asserts that your matrix can be written as $$ \left(\begin{array}{cc} \cos \alpha & -\sin \alpha\\ \sin \alpha & \cos \alpha \end{array} \right) \left(\begin{array}{cc} P & 0\\ 0 & Q \end{array} \right) \left(\begin{array}{cc} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{array} \right) $$ Here you see the four parameters. The rotation $\theta$ does not change the initial circle. The diagonal matrix $diag(P,Q)$ performs a scaling by the two factors which represents the semi-axes of the ellipse. Finally another rotation of $\alpha$ completes the transformation.

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How does this one degree of freedom manifest on an arbitrary A,B,C,D matrix? There should be a way to remove one variable then. Also please see my edited question. –  Ranas Feb 23 '13 at 9:00
    
you must notice that replacing $\phi$ with $\phi+k$ does not change your picture. –  Emanuele Paolini Feb 23 '13 at 12:23

Suppose $x^2+y^2=1$, then consider the action of an invertible matrix $$ \begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\tag{1} $$ Let $$ \begin{bmatrix}e&f\\g&h\end{bmatrix} =\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}\right)^T\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}\tag{2} $$ Then $$ \begin{align} 1 &=\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\ &=\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}e&f\\g&h\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix}\\ &=eu^2+(f+g)uv+hv^2\tag{3} \end{align} $$ Equation $(3)$ is the equation for an ellipse centered at the origin.

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