Visualizing this Matrix Transformation on the Unit Square The matrix in question is:
$$
p=\begin{pmatrix}
-1 & 1 \\ -1 & 0
\end{pmatrix}
$$
acting on vectors $\begin{pmatrix} x \\ y \end{pmatrix}$ in the unit square.
Is there an  intuitive way to interpret this transfomation geometrically? Or summarize it with a nice pictorial mapping from $[0,1]^2 \to \mathbb{R}^2$?
EDIT: Of course it won't take $[0,1]^2 \to [0,1]^2$
EDIT: For context, this is for work on a project I'm doing. I want to be able to explain that the attached plot reveals symmetries consistent with the transformation. The plot below will cover $\mathbb{R}^2$ due to the periodic nature of the trig functions I'm working with. 

So if I take a subset of this plot (say the triangle defined by $A(\pi,0)$ $B(2\pi,0)$ and $C(\pi,\pi)$; it should map back onto another area of the plot in a way consisent with the rotation + shear. Not sure if that makes sense or not. Let me know if clarification is needed.
 A: Generally, the linear transformation with standard matrix $A = \left[\begin{array}{@{}cc@{}}
    a & c \\
    b & d \\
  \end{array}\right]$ sends $\left[\begin{array}{@{}c@{}}
    x \\
    y \\
  \end{array}\right]$ to
$$
\left[\begin{array}{@{}cc@{}}
    a & c \\
    b & d \\
  \end{array}\right] \left[\begin{array}{@{}c@{}}
    x \\
    y \\
  \end{array}\right]
  = x \left[\begin{array}{@{}cc@{}}
    a \\
    b \\
  \end{array}\right] + y \left[\begin{array}{@{}cc@{}}
    c \\
    d \\
  \end{array}\right].
$$
Particularly,
$$
A(\mathbf{e}_{1}) = \left[\begin{array}{@{}cc@{}}
    a \\
    b \\
  \end{array}\right],\qquad
A(\mathbf{e}_{2}) = \left[\begin{array}{@{}cc@{}}
    c \\
    d \\
  \end{array}\right].
$$
The geometric effect on the plane can be depicted using an F (the first Roman letter with no non-trivial symmetries) inscribed in the unit square

A: The red square is the unit square and the green square is the transformation.

A: sure... where to your prinincipal component vector go?
$\hat i = (1,0) \to (-1,-1)\\
\hat j = (0,1) \to (1,0)$
So, $\hat i$ is rotating $135$ degrees clockwise, and stretching by a factor of $\sqrt 2$
$\hat j$ is rotating $90$ degrees clockwise, with no stretch.
This may be sufficent for you, but you could then try to break this into rotations and shears -- A 90 degree clockwise rotation followed by a horizontal shear.  
Areas are preserved: |det (A)| = 1
Can you see it yet?
A: It's quite a funny transformation. If you change coordinates a bit it is simply a rotation by $-2\pi/3$ and $p^3={\rm id}$. The orbit of 
$e_1=\left( \begin{matrix} 1\\ 0 \end{matrix}\right)$ under $p$ is 
$$
 \left( \begin{matrix} 1\\ 0 \end{matrix}\right) \mapsto 
\left( \begin{matrix} -1\\ -1 \end{matrix}\right) \mapsto 
\left( \begin{matrix} 0\\ 1 \end{matrix}\right) \mapsto 
\left( \begin{matrix} 1\\ 0 \end{matrix}\right). 
 $$
