What is wrong with my deformation gradient calculation? I created two ellipses,$\hspace{150px}$,where the red ellipsis is as the blue one, except translated to the right and rotated by ${30}^{\circ} .$ Using rotation matrix, 
$$
\left[
  \begin{array}{cc}
    x  \\[2px]
    y   \end{array}
\right]
\
\phantom{F} ~=~
\left[
  \begin{array}{cc}
    cos(30^{\circ})  & -sin(30^{\circ})  \\[2px]
    sin(30^{\circ}) & \phantom{}cos(30^{\circ})
  \end{array}
\right]
\
*
\left[
  \begin{array}{cc}
    X  \\[2px]
    Y   \end{array}
\right]
\
$$
Next I tried to calculate deformation gradient by using principal axes(a and b for initial ellipsis,c and d for translated one), and then I decomposed $\vec{c}$ and $\vec{d}$ vectors,$$
\begin{alignat}{7}
\vec{c} &~~=~~ 1.0021 \, \vec{b} &~&+ &~&0.8654 \, \vec{a} \\[5px]
\vec{d} &~~=~~ 0.8654 \, \vec{b} &&- &&0.2505 \, \vec{a}
\end{alignat}
$$and created the deformation gradient as$$
F ~=~
\left[
  \begin{array}{cc}
    \phantom{-} 0.8654  &    1.0021  \\[2px]
             -  0.2505  &    0.8654
  \end{array}
\right]
\,.
$$But it is obvious that, this is not the same as the ${30}^{\circ}$ rotation matrix that I had expected it to be,$$
\phantom{F} {\llap{\textsf{rotation matrix}}} ~=~
\left[
  \begin{array}{cc}
    0.8660  &            -  0.5000  \\[2px]
    0.5000  &   \phantom{-} 0.8660
  \end{array}
\right]
\,.
$$
Question:  Why aren't the deformation gradient, $F ,$ and and the rotation matrix, $\textsf{rotation matrix} ,$ the same?
 A: Your method for determining the coefficients in the deformation gradient matrix is described in these notes, so I assume you are using something similar. However, the formulae are only valid if the starting vectors describe a unit square (or cube in 3D). They must act as if they are unit basis vectors in the original coordinate system (more generally, the deformation is defined relative to the lengths of these vectors, but they do need to be the same length). Your vector $\vec{a}$ is not a unit vector, and this leads to the incorrect result for $F$.
If you repeat the calculation, with $\vec{a}$ (and hence $\vec{c}$ in this simple example) replaced by unit vectors, then the deformation gradient matrix $F$ will be identical with the rotation matrix $R$ (provided you correct the sign error described below), as it should be, up to the numerical precision of your calculation. Of course, having determined the deformation gradient matrix, it can be used to transform any vectors through $\vec{x}=F\vec{X}$, not just unit vectors (I'm ignoring the translation, and sticking to the case of a homogeneous deformation, which is what we are discussing here).
I believe you got the signs wrong in the off-diagonal elements of the rotation matrix because the rotation angle in your example is $-30^\circ$, not $+30^\circ$. You can check this yourself by writing out $\vec{c}=R\vec{a}$ in components. But of course, rotation matrix sign conventions always need double checking to be sure, they are a common source of error.
A: In component form,
$$x=F_{xx}X+F_{xy}Y$$
$$y=F_{yx}X+F_{yy}Y$$
In your example, there is a mapping of 2 vectors:
X = 0, Y= 2 ---> x = 1, y = 1.732
and
X = 1, Y = 0 ---> x = 0.866, y = - 0.5
Solving for the components of the deformation gradient tensor gives:
$F_{xx}=0.866$, $F_{yx}=-0.5$, $F_{xy}=0.5$, and $F_{yy}=0.866$
