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In the geometrical interpretation of SVD, the data points that we have need to be imagined as points in high dimensional space (say $d$-dimensional space). But we need to find a hyperplane in $k-$dimensional subspace that best fits the given data points

To gain insight into the SVD, treat the rows of an $n \times d$ matrix $A$ as $n$ points in a $d$-dimensional space and consider the problem of finding the best $k$-dimensional subspace with respect to the set of points.

My doubt here is about the uniqueness of $k$. Can we do decomposition for any $k \le d$ or for only certain values of $k$ or only for an unique $k$?


The paragraph is taken from the material on Singular Value Decomposition available here.

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The number of singular vectors we need to find during SVD is not unique. The possible values for k are from 1 to $r$. Here, $r$ is the rank of matrix $A$, on which we are performing decomposition.

The same pdf says that

First, in many applications, the data matrix $A$ is close to a matrix of low rank and it is useful to find a low rank matrix which is a good approximation to the data matrix . We will show that from the singular value decomposition of $A$, we can get the matrix $B$ of rank $k$ which best approximates $A$; in fact we can do this for every $k$. Also, singular value decomposition is defined for all matrices (rectangular or square)unlike the more commonly used spectral decomposition in Linear Algebra.

So, the value of $k$ is up to the designer. If the designer selects the value of $k$ smaller than the rank $r$ of the matrix $A$, then it is called as truncated SVD.

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