Geometry of $2 \times 2$ matrix mappings Suppose $\mathbf{A}$ is a $2 \times 2$ matrix. I'm interested in the geometry of the mapping $T(\mathbf{x}) = \mathbf{A}\mathbf{x}$ from $\mathbb{R}^2$ to $\mathbb{R}^2$. The effect on the vectors $\mathbf{e_1} = (1,0)$ and $\mathbf{e_2} = (0,1)$ is clear -- $T(\mathbf{e_1})$ and $T(\mathbf{e_2})$ are just the columns of $\mathbf{A}$. So, by linearity, a unit square gets mapped to a parallelogram. The green square gets mapped to the pink parallelogram in the pictures below.
Now let's consider the effect on  a unit circle. It gets mapped to an ellipse, and I'm interested in how the geometry of this ellipse is related to the matrix $\mathbf{A}$.
One case seems clear: if $\mathbf{A}$ is symmetric, then the axes of the ellipse are the eignevectors of $\mathbf{A}$, and its semi-axis lengths are the eigenvalues. The "stretching" of the circle to form the ellipse is nicely related to eigenvalues and eigenvectors. Fabulous. This is illustrated in the following picture:

Another case is also clear: if the eigenvalues of $\mathbf{A}$ are not real, then presumably there is no relationship whatsoever to the geometry of the ellipse.
Now the case that's puzzling me: what if $\mathbf{A}$ is not symmetric, but still has real eigenvalues. What sort of geometric relationship exists in this case, if any? This case is illustrated in the following picture:

 A: The relevant keyword to read about is "singular values". For a map $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$, you can always find an orthonormal basis $(v_1,v_2)$ of the domain (with respect to the standard Euclidean metric) and an orthonormal basis $(w_1,w_2)$ of the range such that $T(v_i) = \sigma_i w_i$ for $\sigma_i \geq 0$. The numbers $\sigma_i$ are called the singular values of $T$ and are the eigenvalues of $\sqrt{T^{*}T}$ or $\sqrt{TT^{*}}$ (in your cases, $\sqrt{A^TA}$ or $\sqrt{AA^T}$). 
Since the bases $(v_i),(w_i)$ are orthonormal, this means that assuming $T$ is invertible (and then $\sigma_i > 0$ for all $i$), a unit circle will be mapped to an ellipse whose axes are $w_1,w_2$ of lengths $\sigma_1,\sigma_2$ respectively. It $T$ is not invertible, it will map the unit circle to a "circle" of lower dimension.
If $T$ is positive definite, then $T$ is orthogonally diagonalizable, you can get $v_i = w_i$ and the $\sigma_i$ will be the eigenvalues of $T$. If $T$ is symmetric, the singular values will be the absolute values of the eigenvalues of $T$. If $T$ is not symmetric, then the only relation you can expect between the singular values and the eigenvalues is that $\sigma_1 \cdot \sigma_2 = |\lambda_1 \cdot \lambda_2| = |\det(A)|$. For example, consider the family of matrices
$$ A = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}. $$
The only eigenvalue of $A$ is $1$ and if $a \neq 0$ then $A$ is not diagonalizable. The singular values of $A$ are given by
$$ \sigma_1 = \sqrt{\frac{2 + a^2 + a\sqrt{4 + a^2}}{2}}, \sigma_2 = \sqrt{\frac{2 + a^2 - a\sqrt{4 + a^2}}{2}}. $$
They satisfy $\sigma_1 \sigma_2 = 1$ and $\sigma_1,\sigma_2 > 0$ and when $a$ runs in $(0,\infty)$, the singular value $\sigma_1$ runs between $(1,\infty)$ while $\sigma_2 = \frac{1}{\sigma_1}$ runs between $1$ and $0$ so you get all possible pairs of singular values subject to the constraint $\sigma_1 \sigma_2 = 1$. Geometrically, since $\det(A) = 1$, the matrix $A$ must map the unit disc to an ellipse of area $1$. As $\sigma_1 \to \infty$, the matrix $A$ maps the unit disc to an ellipse in which one axis (corresponding to the singular value $\sigma_1$) becomes longer and longer while the second axis becomes shorter and shorter to keep the area of ellipse constant.
