Are SVD (Singular Value Decomposition) values always positive? Is there a relation between the maximum SVD value and the original data?

Question

Assuming it's the standard SVD (no variation of it) with $A = USV^T$, would the $A$ matrix always have positive values (0 to $\infty$)? I noticed that the $U$ and $V^T$ matrices had some negative values with the sample data I used, but I want to be sure that the $A$ matrix has only positive values so that I can choose the proper normalization technique.

Also, is there is a mathematical relation between values in the original matrix and values in the $A$ matrix of the SVD? For example, if your original data's mean and range are so and so, then the maximum value in the SVD will be some function of that.

Answer 1

You can legitimately perform SVD on a matrix that has some negative values. Here's an example in R:

> (A=matrix(c(1,-.5,-.5,1),nr=2))
     [,1] [,2]
[1,]  1.0 -0.5
[2,] -0.5  1.0
> svd(A)
$d
[1] 1.5 0.5

$u
           [,1]      [,2]
[1,] -0.7071068 0.7071068
[2,]  0.7071068 0.7071068

$v
           [,1]      [,2]
[1,] -0.7071068 0.7071068
[2,]  0.7071068 0.7071068

That doesn't necessarily mean it doesn't do what you want if you have a matrix that's all positive.

Note that the singular values (the diagonal of $\Sigma$ in $A=U\Sigma V^T$, which is $S$ in your notation) should always be non-negative. The vector d in the R example above contains that diagonal for the example. Since $\Sigma$ is diagonal, all the entries in it will be non-negative.

Perhaps you should say more about what you're trying to do and why. It seems difficult to give much helpful advice with what you have said so far.