Geometrically explained, why do Linear Transformations Take a circle to an ellipse Short Version:
How can it be geometrically shown that non-singular 2D linear transformations take circles to ellipses? 
(Also, its probably important to state I'd prefer an explanation that doesn't use SVD, as I don't really understand it yet...although I see it everywhere)
Long Version:
Let's use the definition of an ellipse as being a circle stretched in two perpendicular directions. The two directions its stretched in will correspond to the two axes of the ellipse.  
We begin by defining the circle as the endpoints of all possible 2D unit vectors. The ellipse (or at least I"m TOLD it's an ellipse) is the shape drawn by the endpoints of all these vectors after they have all been transformed in the same way (aka multiplied by the same nonsingular matrix $A$). 


*

*For linear transformations represented by diagonal matrices, it's easy to see. We're just stretching the circle in the X and Y directions. 

*For linear transformations represented by symmetric matrices...its a little harder, but I can see the transformation because the eigenvectors of the symmetric matrix are perpendicular, and if we change to a basis where those eigenvectors are the basis vectors, the transformation can be represented by a diagonal matrix (as for WHY symmetric matrices can be decomposed this way I don't yet really understand - but for the purpose of this question I'm just accepting that they can; I'm accepting that the eigenvectors of symmetric matrices are perpendicular to one another). 
So, just like diagonal matrices, symmetric matrices also correspond to stretching a unit circle in perpendicular directions - but unless the symmetric matrix is diagonal, these are perpendicular directions different from the X and Y directions. 

*Buuut...what about for nonsymmetric matrices?
Thanks!

EDIT:
I've now learned of the polar decomposition of any real matrix, and that provides a beautiful explanation for why any real matrix takes a circle to an ellipse!
$A=QS$, where $Q$ is an orthogonal matrix (rotation) and $S$ is a symmetric matrix (stretching in the direction of the eigenvectors).
The symmetric matrix will definitely correspond to making an ellipse (since it scales in orthogonal directions, although perhaps not our regular $x$ and $y$ directions) and all the orthonormal matrix will do is rotate this ellipse. 
However, all the explanations I've seen so far that PROVE that polar decompositions of real matrices are always possible use an algebraic explanation instead of a geometric one...so they aren't really what I'm looking for. 
Thanks again!
 A: Every real square matrix has a polar decomposition into the product of an orthogonal matrix $U$ and a positive-semidefinite (symmetric) matrix $P$. If the original matrix is nonsingular, then $P$ is positive-definite. In 2-D, orthogonal matrices represent either rotations or reflections, which are both isometries, so they don’t affect the shape of the transformed circle. As you’ve mentioned, $P$ can be orthogonally diagonalized, so it represents a stretch in some set of perpendicular directions.  
The existence of this decomposition is equivalent to the existence of the SVD, but can be shown without relying on the latter. In a similar vein, the SVD decomposes the matrix into the product of a rotation or reflection, a scaling, and another rotation or reflection.  
You might also have a look at the Steiner generation of an ellipse. This uses intersecting line segments drawn between points on the sides of a parallelogram to generate ellipses, including circles. Affine transformations preserve incidence relationships (the image of the intersection of a pair of lines is the intersection of the lines’ images) and maps paralellograms to parallelograms, so the image of an ellipse under an affine transformation is another ellipse.
A: The equation of a circle is $x^2 + y^2 = r^2$, or in terms of vectors $(x,y) \pmatrix{x\cr y} = r^2$  An invertible linear transformation $T$ takes $\pmatrix{x\cr y}$ to $\pmatrix{X\cr Y} = T\pmatrix{x\cr y}$.  Thus $\pmatrix{x\cr y\cr} = T^{-1} \pmatrix{X\cr Y}$, and $(x,y) = (X, Y) (T^{-1})^\top$.  The equation becomes
$$(X, Y) (T^{-1})^\top T^{-1} \pmatrix{X\cr Y} = r^2 $$
Note that $(T^{-1})^\top T^{-1}$ is a real symmetric matrix, so it can be diagonalized, and its eigenvalues are positive.
A: The answers on this thread are quite insightful but I am attempting here a rather very geometric answer, as the OP demanded so. For this I am going to use another interesting geometric interpretation of linear transformation (which is easier to imagine)
An alternate geometric interpretation of (dimension-preserving) linear transformation is that it's a transformation such that any line in original space is always transformed into a line and origin is not shifted.
Now imagine a freshly made pizza base/crust. Even more, let's make grids on it.
Our job is now to transform it's shape such that all the grid lines (or any possible line) remain lines. So you see at best what we can do is *stretch it with equal forces on opposite sides. You are free to choose where to stretch and Of course we can simply rotate it also after or before all the stretching.
It's not difficult to imagine that we can only get an ellipse (or bigger circle if we stretched from all sides with equal force).
Other interesting points:
(1): The direction of stretch is eigen vector and extent to which it is stretched is eigen value;
(2): Each rectangle from the gridlines is now parallelogram.
