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In non-negative matrix factorization (NMF), the problem is to minimize $A - WH$. Dimensions are $A$ (m x n), $W$ (m, k) and $H$ (k, n). The matrix $H$ reveals soft clustering assignments of $n$ items over $k$ clusters, and is called clustering indicator matrix. Values in $H$ are constrained to have nonnegative numbers.

I am wondering how to properly interpret this $H$ matrix. There doesn't seem to be constraint (other than nonnegativity constraint) on the range of values that entries in $H$ can take. I'd like to perhaps have a row-wise sum of 1 for all rows in $H$. So, for a given row (cluster), I could perhaps interpret values probabilitisticallly. Would it be correct to simply divide each row's elements by row-sum? I am worried that it's not correct interpretation but I am not able to figure out why.

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    $\begingroup$ This is a nice question (+1) but can you please provide some references on how you came across this characterisation of $H$? I suspect you referr to the associations of NNMF and spectral clustering but these are usually associated with symmetric NNMF so your second paragraph threw me a bit off about the "no constraints" part. $\endgroup$ – usεr11852 Mar 18 '17 at 21:53
  • $\begingroup$ Welcome to the CV community. $\endgroup$ – usεr11852 Mar 18 '17 at 21:54
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Let $(W, H)$ be a solution to the NMF problem. Define $D$ to be the diagonal matrix with $D_ii = \sum_{j=1}^n H_{ij}$, so that (assuming $H$ has no all-0 rows), $D^{-1} H$ is the version of $H$ scaled to have row sums 1. But then $W H = W (D D^{-1}) H = (W D) (D^{-1} H)$, so $(W D, D^{-1} H)$ is an equally-good NMF solution (up to the regularization terms).

So yes, it's fine to rescale $H$ to have row sums 1 as long as long as you also scale the columns of $W$ accordingly. Whether a probabilistic interpretation is appropriate, though, depends on the problem setting.

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    $\begingroup$ While not wrong, I fail to see how this answers this question by the OP mostly because the OP's question is unclear... You seem to address one of the general comments ("I'd like to perhaps have a row-wise sum of 1 for all rows in $H$) but then again given that the OP has done clustering an integral part of her post this seems to miss the mark. Anyway, +1 for your time and for writing something hopefully relevant! $\endgroup$ – usεr11852 Apr 9 '17 at 22:57

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