Can we rotate sobel operator and still get the same results for gradient image

I have a 5*5 image and I applied imgradient() on it and got the directions(just focussing on directions). I also tried to calculate the directions for the same image on paper using the following sobel operators.

gx=[-1 0 1; -2 0 2; -1 0 1]     gy=[-1 -2 -1; 0 0 0; 1 2 1]

I found theta using

Gdir = atan2(-gy,gx)*180/pi  %Note: gy is negative as y moves from top to bottom

and got same answers as were obtained using imgradient().

But when I used

gx=[1 0 -1; 2 0 -2; 1 0 -1]     gy=[1 2 1; 0 0 0; -1 -2 -1]

I got different answers. Why?

Can't we rotate sobel operator and still get the same results. Does MATLAB uses a fixed set of sobel operators to find gradient and never uses the rotated version? What is the problem. Please explain.

Answer 1

[Anand]

The imgradient function uses the following kernels for the 'sobel':

hx = -fspecial('sobel')'

hx =

      -1     0     1
      -2     0     2
      -1     0     1

hy = -fspecial('sobel')

hy =

      -1    -2    -1
       0     0     0
       1     2     1

You get different results with the rotated kernels because the Sobel kernel is not symmetric.