Makayla Boyd

2022-06-28

Transformation matrix for matrix indices to cartesian coordinates

(1,1) (1,2) (1,3)(2,1) (2,2) (2,3)(3,1) (3,2) (3,3)

This is an example 3x3 matrix. In corresponding cartesian coordinate system, the representation would be:

(-1,1) (0,1) (1,1)(-1,0) (0,0) (1,0)(-1,1) (0,-1) (1,-1)

Say, I have any square matrix with dimension-N, where N is odd. I need a generic transformation matrix such that I can get a vector as cartesian coordinates from matrix indices. Does such a function already exist? How should I go ahead in implementing this?

(1,1) (1,2) (1,3)(2,1) (2,2) (2,3)(3,1) (3,2) (3,3)

This is an example 3x3 matrix. In corresponding cartesian coordinate system, the representation would be:

(-1,1) (0,1) (1,1)(-1,0) (0,0) (1,0)(-1,1) (0,-1) (1,-1)

Say, I have any square matrix with dimension-N, where N is odd. I need a generic transformation matrix such that I can get a vector as cartesian coordinates from matrix indices. Does such a function already exist? How should I go ahead in implementing this?

nuvolor8

Beginner2022-06-29Added 32 answers

The transformation of indices is the following:

$(x,y)=f(i,j)=(j-\frac{n+1}{2},-i+\frac{n+1}{2})\text{}.$

Here $i$ is the index for the rows, $j$ the one for the columns and $n$ the order of your square matrix.

$(x,y)=f(i,j)=(j-\frac{n+1}{2},-i+\frac{n+1}{2})\text{}.$

Here $i$ is the index for the rows, $j$ the one for the columns and $n$ the order of your square matrix.

glycleWogry

Beginner2022-06-30Added 8 answers

Interchange indices $i$ and $j$ in initial matrix, then flip it upside down to get the same orientation like a usual coordinate system and then subtract $(2,2)$ or $(\frac{n+1}{2},\frac{n+1}{2})$ in general to shift the center.

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