How to Decompose a Separable Filter?
Indeed you can do that.
You may look on my answer to How to Prove a 2D Filter Is Separable?   

By the SVD for any filter $ A $:

$$ A = \sum_{i = 1}^{n} {\sigma}_{i} {u}_{i} {v}_{i}^{T} $$

Since we're talking about separable filter then:

$$ A = {\sigma}_{1} {u}_{1} {v}_{1}^{T} $$

So the columns filter is $ \sqrt{\sigma}_{1} {u}_{1} $ and the rows filter is $ \sqrt{\sigma}_{1} {v}_{1}^{T} $.

Since what's important is that $ A = {\sigma}_{1} {u}_{1} {v}_{1}^{T} $ you can actually choose any 2 vectors which their outer product is $ {\sigma}_{1} {u}_{1} {v}_{1}^{T} $. The choice isn't unique. For example you could take for columns $ {\sigma}_{1}^{2} {u}_{1} $ and $ \frac{1}{{\sigma}_{1}} {v}_{1}^{T} $ for the rows.