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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.

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  • $\begingroup$ Please reformulate the question, as it is rahther unclear what you mean exactly. Btw: It is the diagonal matrix $\mathbf S$ of singular values which has only non-negative entries, the orthogonal matrices $\mathbf U$ and $\mathbf V$ of course might contain also negative values, and the relation you asked for is given by directions of maximal variance. $\endgroup$ – davidhigh Jul 10 '14 at 23:02
  • $\begingroup$ Frobenius-Perron theorem is what you probably seek for? $\endgroup$ – ttnphns Jul 11 '14 at 5:27
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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.

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