I want to prove that 2D Gaussian filter is separable and we can separate it into two dimensions, my problem is about the size of filters. we should prove that $G(x,y)*I$(where $G(x,y)=$$\begin{bmatrix}0.01 & 0.1 & 0.01 \\0.1 & 1 & 0.1 \\ 0.01 & 0.1 & 0.01\end{bmatrix}$,I is image and $*$ is convolution operator) is equal to $G(x)*I*G(y)$ where $G(x)=$$\begin{bmatrix}0.1 & 1 & 0.1 \end{bmatrix}$ $,G(y)=$$\begin{bmatrix}0.1\\1\\ 0.1\end{bmatrix}$ and I is image. in other words we should prove that $G(x,y)=G(x)*G(y)$ but I don't know how to convolve these filters with different sizes.


Let's have a different perspective on that.

Let's say our 2D Linear Operator is given by the Matrix $ G \in {\mathbb{R}}^{n \times n} $.
Using the SVD Decomposition the operator can be written as:

$$ G = \sum_{i = 1}^{n} {\sigma}_{i} {u}_{i} {v}_{i}^{T} $$

Separable Linear 2D Operator is defined as operator which can be composed by Outer Product of 2 vectors.
Looking at the SVD Decomposition of $ G $ we can conclude that $ G $ is separable operator if and only if $ \forall i > 1 \; {\sigma}_{i} = 0 $ and it is given by:

$$ G = {\sigma}_{1} {u}_{1} {v}_{1}^{T} $$

Usually LPF 2D Linear Operators, such as the Gaussian Filter, in the Image Processing world are normalized to have sum of 1 (Keep DC) which suggests $ {\sigma}_{1} = 1 $ moreover, they are also symmetric and hence $ {u}_{1} = {v}_{1} $ (If you want, in those cases, it means you can use the Eigen Value Decomposition instead of the SVD).

So basically, to prove that a Linear 2D Operator is Separable you must show that it has only 1 non vanishing singular value.

Note: I was not rigorous in the claims moving form general SVD to the Eigen Decomposition yet the intuition holds for most 2D LPF operators in the Image Processing world.

| improve this answer | |

Given that $G(x)$ is a row vector, while $G(y)$ is a column one, their convolution will be identical to the matrix product $G(x,y)=G(x)*G(y)=G(x)G(y)$. For this reason, as soon as $G(x,y)$ is rank-1, the convolution kernel can be separated (decomposed into two 1-D filters). This is because the other columns of the matrix could be written as a linear combination of the elements of the first.

So to prove that a kernel is separable, just check the rank:

s = svd(G);
sum(s > eps('single'))
| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.