Hi Daniel,
Thanks for the suggestion!
a) It seems that when there is only rotation the upper and lower triangle (3x3 matrix) are symmetric aside from the sign flip.
1dApar2mat 0 0 0 -0.081944 -0.103117 -0.169312 1.000000 1 1.000 0 0 0
# mat44 1dApar2mat 0 0 0 -0.081944 -0.103117 -0.169312 1.000000 1 1.000 0 0 0 :
0.999995 -0.001425 0.002955 0.000000
0.001430 0.999997 -0.001800 0.000000
-0.002952 0.001804 0.999994 0.000000
Polar decomposition is the same:
0.999995 -0.00142484 0.00295479 0
0.00143016 0.999997 -0.00179989 0
-0.00295221 0.00180411 0.999994 0

b) If there is shearing only then this is only in the lower triangle:

1dApar2mat 0 0 0 0 0 0 1.000000 1 1.000000 -0.000783 0.02345 0.000108
# mat44 1dApar2mat 0 0 0 0 0 0 1.000000 1 1.000000 -0.000783 0.02345 0.000108 :
1.000000 0.000000 0.000000 0.000000
-0.000783 1.000000 0.000000 0.000000
0.023450 0.000108 1.000000 0.000000

Polar decomposition would be the following:
0.999931 0.000391157 -0.0117242 0
-0.00039179 1 -5.17014e-05 0
0.0117242 5.62912e-05 0.999931 0

c) If there is shearing and rotation you get the upper and lower triangles not matching:
1dApar2mat 0 0 0 -0.081944 -0.103117 -0.169312 1.000000 1 1.000000 -0.000783 0.02345 0.000108 > test1.txt
# mat44 1dApar2mat 0 0 0 -0.081944 -0.103117 -0.169312 1.000000 1 1.000000 -0.000783 0.02345 0.000108 :
0.999995 -0.001425 0.002955 0.000000
0.000647 0.999999 -0.001802 0.000000
0.020498 0.001879 1.000063 0.000000

Polar Decomposition would be the following:
cat_matvec test1.txt -P
0.999961 -0.00105473 -0.00877026 0
0.00103852 0.999998 -0.0018528 0
0.00877219 0.00184362 0.99996 0


So in general it seems that doing the polar decomposition and applying the 3x3 matrix would give the rotational components, whereas if you use the original full matrix then the upper triangle is the rotational component while the lower triangle is a mixture of shearing and rotation.


This definitely helps a lot, thanks for pointing that new tool out.
Thanks,
Ajay



Edited 2 time(s). Last edit at 05/14/2019 04:24PM by AjaySK.