Is Singular Value Decomposition (SVD) a linear function?

Question

I know SVD is a linear algebra topic, but i've seen multiple times in different articles that SVD is a nonlinear function. So i don't know what to think..

Answer 1

In higher mathematics, a function is said to be linear if $f(x+y)=f(x)+f(y)$. The process of SVD decomposition can be seen as a function, which we will call $S$, which take in a matrix and returns three matrices: $S(A)=(U,\Sigma,V)$. The three matrices that are returned have the property that $A=U\Sigma V^T$. However, given two matrices of the same size, say $A$ and $B$, $S(A+B)\neq S(A)+S(B)$. In other words, the $U$ in the SVD decomposition for $A+B$ is not equal to the $U$ in the SVD decomposition for $A$ plus the $U$ in the SVD decomposition for $B$. The same is true for the $\Sigma$ and the $V$.

Linear algebra has the moniker "linear" because all linear functions that act on vectors of finite dimension can be represented by matrices, and all matrices represent linear functions acting on vectors. However, certain aspects of linear algebra are nonlinear. For example, the function that accepts a matrix and returns its square: $f(A)=A^2$ is nonlinear because $(A+B)^2\neq A^2+B^2$ in general. So SVD is a linear algebra topic because it involves breaking up a single linear action into three simpler linear actions, but the function that takes each matrix to its decomposition is nonlinear.