# Understanding the resulting image matrix when differentiating image

Let $$A$$ be a image matrix and let $$P(i,j)$$ be the gray level of pixel $$i,j$$. Let $$0$$ be black and $$255$$ be white Assume I want to differentiate this image with respect to the columns $$(x)$$ as in I want $$P(i,j)=P(i,j+1)-P(i,j)$$ in the new image.

I get that I can achieve this with convolution of $$A$$ with the filter $$f = [1 \ -1]$$ since you flip the filter in the $$x-$$direction (so pixel $$i,j$$ would be multiplied by $$-1$$ and $$i,j+1$$ with $$1$$ which gives me what I want.

My understanding is that this process is supposed to highlight differences in the $$x-$$direction.

Two questions:

1. How does the gray-level scale in the resulting image work? If one pixel had gray level $$240$$ and the next pixel had gray level $$5$$, this would result in the new pixel having the gray level $$-235$$, what does that mean in the scale of $$0$$ to $$255$$? Does my scale change from $$0...255$$ to $$-255...255$$?

2. Does it matter if I convolve $$A$$ with the filter $$[-1 \ 1]$$ instead of $$[1 \ -1]$$?. What would be the result?

• The [-1 1] is a discrete approximation of first derivative.The sign of the outcome define the slope of variation in intensity. Also the the 0-255 is a limitation by uint8 type (values outside 0-255 considered as overflow). And for your second question, these two would be exactly negative of each other. – Mohammad M Sep 20 at 17:01

Your interpretation is correct: directional derivation operators highlight variation in a given direction. Here, you use the $$2$$-point discrete derivative in the $$x$$-direction (along image rows). It may emphasize vertical features.
Second, as $$[1 \;-1] = -[-1\; 1]$$, and since convolution is a linear operator, you will get the opposite result as in the previous case. And it won't matter if you take the absolute value.