2
$\begingroup$

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. enter image description here

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.

$\endgroup$
  • $\begingroup$ Maybe $[0,1]^2 \to \mathbb R^2$? *(non-surjective) $\endgroup$ – ahorn Oct 14 '16 at 21:37
  • $\begingroup$ Ah, yes of course. Fixed. $\endgroup$ – David D. Oct 14 '16 at 22:18
3
$\begingroup$

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

The geometric action of matrix multiplication

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, a rotation and a shear! What about subsets of $[0,1]^2$ such suggested in my edit above. Easy to visualize? $\endgroup$ – David D. Oct 14 '16 at 22:27
1
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes! I agree that it is a 90 degree rotation followed by a shear :) I guess my concerns are more specifically related to an application of this transformation (see my edits above) . Thanks for your help! $\endgroup$ – David D. Oct 14 '16 at 22:29
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.