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I have a question about image gradient magnitude, given an image $I$. The horizontal and vertical derivatives resulting from Sobel filters are defined as: $$K_x= \begin{bmatrix} -1 & 0 & 1 \\ -2 & 0 & 2 \\ -1 & 0 & 1 \\ \end{bmatrix}*I; K_y= \begin{bmatrix} -1 & -2 & -1 \\ 0 & 0 & 0 \\ 1 & 2 & 1 \\ \end{bmatrix}*I $$

where $*$ denotes the convolution operator. The gradient magnitudes can be computed as

$$I_{\delta}=\sqrt{(K_x^2+K_y^2)}$$.

Now, I want to "invert" the result of $I_{\delta}$, so that maximum magnitude gradient values become zero. Then the "inverting" function $g$ is $$g(I_{\delta})=\max (I_{\delta}) - I_{\delta}\,.$$

My question is: is $I_{\delta}$ scalar or vector? What do you think? If it is possible, could you correct help me.

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  • $\begingroup$ I have strongly edited your question. I hope it looks more precise now. I have changed the $|I|$ notation into $I_\delta$ because I found it confusing. Feel free to re-edit if necessary $\endgroup$ – Laurent Duval Nov 15 '15 at 10:32
  • $\begingroup$ @LaurentDuval: What is different betwen $I_{\delta}$ and $|I|$, If I put above function $g$ in an integrate as $\int_\Omega g(|I|) dx$ or $\int_\Omega g(I_{\delta}) dx$, which one is correct? $\endgroup$ – Jame Apr 16 '16 at 14:09
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Pretty much by definition, the magnitude of anything is a scalar value. Though, since you are working on an image, the result is a 2D array of scalar values.

Given Kx and Ky, it's also possible to work out the directions of the edges, as well as the magnitudes, but that's a whole different bit of maths.

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  • $\begingroup$ Thank Simon B. The above magnitude just Sobel magnitude. I just show it in the question to make more clear. At each pixel $x$, we have a own magnitude value, it will be a scalar value. Hence, for whole image, it will be 2D array of scalar values. You are right. So, Which is correct for $g$ function? $g(|I|)=\max(|I|)-|I|$ or $g(|I(x)|)=\max(|I(x)|)-|I(x)|, x \in \Omega$ $\endgroup$ – John Nov 4 '15 at 15:40
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    $\begingroup$ @user8264 I don't really know what the x is meant to represent here. To me the first version is correct, as well as being simpler to read. That said, it would be simpler still to set J = |*I*|, and then write g (J) = max(J) - J. $\endgroup$ – Simon B Nov 4 '15 at 16:04
  • $\begingroup$ $x$ denotes a pixel in image domain. I write the second version because i want to said that maximum of 2D scalar value, not just maximum of a single value. The $g$ function tried to invert the value in the 2D scalar, so that maximum value becomes zeros. Hence, max (.) must take over whole the 2D scalar. Therefore, if i write max(J), the reader may be confused that max take over one value $\endgroup$ – John Nov 4 '15 at 16:09
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Your question is strongly related to the discussion in How to represent a given equation more clear, profressional and short form?, and a matter of speaking formally or loosely. First, an image convolved by a filter remains an image, which can be interpreted as a matrix or a vector.

Yet the resulting image might not be of the same size, depending on filter size and border management. So scalar additions and subtractions could be ill-defined. Luckily here, the two gradient Sobel filters have the same size, hence $K_x$ and $K_y$ as well. Powers and square-roots should be understood as scalar too in that case, hence pixel by pixel. So assuming only scalar operations, $I_\delta$ is an image.

The $\max$ is hardly a matrix, unless "very lossely" speaking. The subtraction, that might work in Matlab, is only understandable if you allow a scalar to be a constant matrix with the same size as the others matrices.

Your notation $x$ does not seem much clearer to me. So apart from notations suggested in How to represent a given equation more clear, profressional and short form?, you can defined an "inverted gradient image" $I_g$ as:

$$\forall (i,j)\in [1,\ldots,M]\times [1,\ldots,N],\; I_g(i,j) = \max_{ (i,j)\in [1,\ldots,M]\times [1,\ldots,N]} I_\delta (i,j) - I_\delta(i,j)\,. $$

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