# To what extent is the Singular Value Decomposition unique?

In Adam Koranyi's article "Around the finite dimensioal spectral theorem", in Theorem 1 he says that there exist unique orthogonal decompositions.

What is meant here by unique?

We know that the Polar Decomposition and the SVD are equivalent, but the polar decomposition is not unique unless the operator is invertible, therefore the SVD is not unique.

What is the difference between these uniquenesses?

• – glS Jun 12 at 9:05

SVD is unique up to permutations of columns of the $U,V$ matrices. Usually one asks for the singular values to appear in decreasing order on the main diagonal so that uniqueness is up to permutations of singular vectors with the same singular values.