What can a linear transformation do in $\mathbb{R}^2$? 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 $\varphi$.
 A: 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.
A: 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.
