Visualizing this Matrix Transformation on the Unit Square

Question

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.

Answer 1

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

Answer 2

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?

Answer 3

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 $$ \left( \begin{matrix} 1\\ 0 \end{matrix}\right) \mapsto \left( \begin{matrix} -1\\ -1 \end{matrix}\right) \mapsto \left( \begin{matrix} 0\\ 1 \end{matrix}\right) \mapsto \left( \begin{matrix} 1\\ 0 \end{matrix}\right). $$

Answer 4

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