[Math] the Singular Value Decomposition for the Zero Matrix

linear algebrasvd

I am interested in the singular value decomposition of a matrix: $\mathbf{M} = \mathbf{U} \mathbf{S} \mathbf{V}^T$.

Suppose $\mathbf{M} = \mathbf{0}$ (zero matrix) and square. Clearly, $\mathbf{S} = \mathbf{0}$, but what about $\mathbf{U}$ and $\mathbf{V}^T$ — Do they have defined values?

When use MATLAB, $\mathbf{U} = \mathbf{V} = \mathbf{I}$, but is this a definition or pure luck? If so, why is it defined as such, since any matrix for $\mathbf{U}$ or $\mathbf{V}$ will satisfy the decomposition – it does not even need to be orthogonal. Is there some sort of proof that it must indeed fall to a specific value?

>> [U, S, V] = svd(zeros(3))

U =

     1     0     0
     0     1     0
     0     0     1


S =

     0     0     0
     0     0     0
     0     0     0


V =

     1     0     0
     0     1     0
     0     0     1

Best Answer

A matrix has not necessarily only one decomposition in singular value, so $U$, $V$, and $S$ aren't well defined, Matlab give you one of the solutions...

For example, if $(U,S,V)$ is a solution, $(-U, S, -V)$ is one too.

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