For each eigenvalue, there will be eigenvectors for which the eigenvalue equations are true. We will only handle the case of n distinct roots through which they may be repeated. These roots are called the eigenvalue of A. This equation is called the characteristic equations of A, and is a n th order polynomial in λ with n roots. If vis a non-zero, this equation will only have the solutions if The eigenvalues problem can be written as The vector, v, which corresponds to this equation, is called eigenvectors. It is also called the characteristic value. Any value of the λ for which this equation has a solution known as eigenvalues of the matrix A. In this equation, A is a n-by-n matrix, v is non-zero n-by-1 vector, and λ is the scalar (which might be either real or complex). This is currently a quite exciting field of research applied to multidimensional arrays (tensors).Next → ← prev Eigenvalues and EigenvectorsĪn eigenvalues and eigenvectors of the square matrix A are a scalar λ and a nonzero vector v that satisfy In terms of data compression, both domains offer competitive approaches depending on the case, there is no one better than the other. The more vertical and horizontal variability in the image the more components you'll need in order to reproduce the image. SVD provides a linear decomposition of the vertical and horizontal features of an image, in which each of the components (ranks) can contain high and low frequencies. To sum up, the relationship between SVD and spectral decomposition of images is vague. The result for the previous image would be a low frequency version which looks like a "blurry" version of the initial image: The "equivalent" operation to lower the rank in SVD decomposition would be to apply a low pass filter in the frequency domain. If we now look at the transpose of the first image $A^\intercal$, $U$ and $V$ change places in the equation following the rules of transposition of products and we'll have the opposite case: we only need to keep the first column of $V$ to recover the image ( $U$ doesn't contain any information, as the image has no variability in the vertical axis). Which means that we only need to keep the first row of $U$ to recover the image ( $V$ doesn't contain any information, as the image has no variability in the horizontal axis). For the previous image we have a perfect decomposition for k-rank equals to 1. $U$ operates in the column space of $A$ and $V$ in its row space. I'll use the $A = USV^\intercal$ notation for SVD (notice that $U$ is actually a different matrix to $V^\intercal$ as in your notation). To illustrate the point let's see how both approaches are applied to a very simple image:
SVD performs a decomposition based on the spatial structure of a matrix (image) whereas a spectral filters look at its frequency components. The similarity between both techniques stops there as they operate over different domains. In this sense both SVD and image filtering perform a decomposition on images based on a change of basis. Image filtering (in the frequency domain) is performed by decomposing an image in its frequency components and removing part of the spectrum. Do the smaller eigenvalues contribute to high-frequency components of the signal? So is the compression algorithm acting like a low pass filter and depending on the threshold set is essentially stopping the high frequency signals to pass through and basically acting as a smoothing oiperator or is there no relationship there? The main question that I have and this relates to eigenvalue decomposition as well as SVD is whether there is some relationship to the frequency content of the signal. In this case, the smaller eigenvalues will have a relatively shrinking effect on the rows of $V^$ and will overall contribute less. So, I can perform compression using eigenvalue decomposition by setting the eigenvalues under some threshold to 0. Where $V$ is the matrix where each column corresponds to an eigenvector of $A$ and $D$ is the diagonal matrix where the diagonal entry corresponds to the corresponding eigenvector. So, the eigenvalue decomposition of a square matrix can be written as: So, before we discuss SVD, I want to check if my understanding of eigenvalue decomposition is correct. I have been studying the SVD algorithm recently and I can understand how it might be used for compression but I am trying to figure out if there is a perspective of SVD where it can be seen as a low pass filter.
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