# How many singular vectors do we need to calculate for SVD?

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.

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.