matrices - How to rotate the positions of a matrix by 90 degrees - Mathematics Stack Exchange

Excerpt

I have a 5x5 matrix of values. I’m looking for a simple formula that I can use to rotate the position of the values (not the values themselves) 90 degrees within the matrix.

For example, here is the

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The idea here is to find out where the values get shifted to. A location is given by two indices (i,j). How to transform the indices?

Let us first try to rotate the indices around the center element:

Matrix Elements   Coordinates
01 02 03 04 05    (1,5) (2,5) (3,5) (4,5) (5,5)
06 07 08 09 10    (1,4) (2,4) (3,4) (4,4) (5,4)
11 12(13)14 15    (1,3) (2,3)((3,3))(4,3) (5,3)
16 17 18 19 20    (1,2) (2,2) (3,2) (4,2) (5,2)
21 22 23 24 25    (1,1) (2,1) (3,1) (4,1) (5,1)

Rotation by θ=−90∘ around the origin is performed by

R=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)=(01−10)

Plus we need coordinate translations before and after the rotation, because our index center is (3,3) and not (0,0), so we need homogeneous coordinates to express translations by (Δx,Δy) in matrix form. First we extend the rotation to homogeneous coordinates

R=(010−100001)

and the translation is

T=(10Δx01Δy001)

It acts on coordinates like this:

(x′y′1)=(10Δx01Δy001)(xy1)=(x+Δxy+Δy1)

Combining these operations we get:

A=T−1RT=(103013001)(010−100001)(10−301−3001)=(103013001)(01−3−103001)=(010−106001)

The final step is to change the y order: y′=6−y:

Matrix Elements   Coordinates
01 02 03 04 05    (1,1) (2,1) (3,1) (4,1) (5,1)
06 07 08 09 10    (1,2) (2,2) (3,2) (4,2) (5,2)
11 12(13)14 15    (1,3) (2,3)((3,3))(4,3) (5,3)
16 17 18 19 20    (1,4) (2,4) (3,4) (4,4) (5,4)
21 22 23 24 25    (1,5) (2,5) (3,5) (4,5) (5,5)

This affine transformation can be written in matrix form as

Y=(1000−16001)=Y−1

This leads to

B=YT−1RTY=(1000−16001)(010−106001)(1000−16001)=(1000−16001)(0−16−106001)=(0−16100001)

A matrix element at indices (i,j) (i-th row, j-th column) has coordinates (j,i). Thus the index transformation (i,j)→(i′,j′) is

(j′i′1)=(0−16100001)(ji1)

or

i′=jj′=6−i

A possible algorithm to calculate the rotated matrix b from a given matrix a is:

(1,5).each do |i| 
  (1,5).each do |j| 
    b[j, 6-i] = a[i, j]
  end
end

Here is some test code online: (link) Click the tab marked “Execute” to run it.

Example:

Let us do the calculation for the 18-th element:

Matrix Elements   Matrix Indices
01 02 03 04 05    (1,1) (1,2) (1,3) (1,4) (1,5)
06 07 08 09 10    (2,1) (2,2) (2,3) (2,4) (2,5)
11 12 13 14 15    (3,1) (3,2) (3,3) (3,4) (3,5)
16 17(18)19 20    (4,1) (4,2)((4,3))(4,4) (4,5)
21 22 23 24 25    (5,1) (5,2) (5,3) (5,4) (5,5)

This entry is the matrix element a43. The indices are (i,j)=(4,3).

The rotated matrix is

21 16 11 06 01
22 17 12 07 02
23(18)13 08 03
24 19 14 09 04
25 20 15 10 05

The value is now at b32, thus (i′,j′)=(3,2). This agrees with the found transformation.

What did the transformation do? First, if the lower left element of the original matrix (the one containing value 21) is at coordinates (x,y)=(1,1), we want to have element 18 at coordinates (x,y)=(3,2). So original (i,j)=(4,3) is interpreted as coordinates (x,y)=(3,4) and flipped by Y to coordinates (3,2). Then T translates it to coordinates (0,−1), which is fine as it sits under the central element (center now at (0,0)). R rotates it to (−1,0), which is left from the central element (center now at (0,0)), fine again. T−1 translates it back to (x′.y′)=(2,3) which is still left to the center. The final Y−1=Y changes nothing, as we are on the middle row. Final translation to matrix indices gives (i′,j′)=(3,2).