Non-orthonormal "standard" vector basis

55 Views Asked by At

First of all, this is more of a "philosophical" question I've been pondering for a while now, and I don't think I have the tools to solve it myself. I apologize beforehand for the somewhat lengthy description, but I'm afraid I don't have the mathematical vocabulary to be more concise, and I am by no means a mathematician.

Setup

I'm imagining standing on an infinite 2D plane, with no concept of coordinates - I'm about to choose them in two separate ways. Say I'm facing one direction, and I know about the concept of distance (that is, I can tell, whether a which points $A$ and $B$ is closer/farther to me, or whether they are the same distance.

First, I decide to create a set of basis vectors such, that I choose:

  • $e_1$ to be a point to my right side of distance $d_1$
  • $e_2$ to be a point to the front of me of distance $d_1$

Second, I choose:

  • $e'_1$ to be the same as last time
  • $e'_2$ to be a point not to the front of me, but shifted slightly to the left, of some distance $d_2$

I now define a vector basis/coordinate system as follows:

  • The point at which I am standing is the origin
  • $e_1 = (1, 0)$; $e_2 = (0, 1)$
  • $e'_1 = (1, 0)$; $e'_2 = (0, 1)$

In the first case, I've just setup the basic Cartesian coordinate system, the second case is slightly different, as illustrated:

Illustration of the two cases

Question

In the first case, a parallelogram given by the points $\{(1, 1), (1, 2), (2, 1), (2, 2)\}$ is obviously a square, and especially its angles at each point are $90°$.

I am wondering whether in the second case, a parallelogram given by those same points is still a square, or rather a rhombus.

I guess the more general question is: are those two systems equivalent, and is the second case just some projection of the first one (which I suspect), or is it some different structure? Or is there maybe some serious flaw with the question itself that I don't see?

P.S.: This is my first question ever on Math StackExchange, so I will welcome any comments as to how to improve my questions in the future!

1

There are 1 best solutions below

2
On BEST ANSWER

In any case, the given four points always form a parallelogram, namely the translation of the parallelogram $(0,0),\ (1,0),\ (0,1),\ (1,1)$ by the vector $(1,1)$.

It will be a rhombus if and only if $d_1=d_2$, i.e. if the basis vectors $e_1,e_2$ have the same length.

It will be a rectangle if and only if $e_1\perp e_2$.

Yes, the two coordinate systems are equivalent, in the sense that we can transform the coordinates of a given vector (point) between them:
Say, $e_1'=e_1$ and $e_2'=a\cdot e_1+b\cdot e_2$ (meaning that its coordinates in the original basis are $(a,b)$) with $b\ne 0$, then $$x\cdot e_1'+y\cdot e_2'=(x+ay)\cdot e_1+by\cdot e_2\\ x\cdot e_1+y\cdot e_2=(x-\frac{ay}b)\cdot e_1'+\frac yb\cdot e_2'$$