I am just hoping to confirm my hypothesis, that a single MLP (untrained and randomly initialized) can be used for random projection for dimensionality reduction.

If a random MLP layer with no activation is:

$$Y = X \ W $$

is there really any difference to the random projection method, especially if its weights are sampled from a distribution with a variance following the Johnson–Lindenstrauss lemma ?


1 Answer 1


A single MLP layer with no activation function is a linear function, as you already seem to know, and hence trivially equivalent to a projection. It is equivalent to a random projection if you choose the entries in the matrix randomly from the appropriate distribution.

Normally a single-layer MLP is understood to include an activation. That is not equivalent to a random projection.

The value of this equivalence is unclear. If you want a random projection, it seems clearer to simply use a random projection, without needing to introduce a MLP.

  • $\begingroup$ Thanks for the response. I'm asking because I want to use a random projection in a Tensorflow project, and as far as I can tell, TF doesn't have it yet. As for an MLP without an activation, I've seen some projects use it for linear operations (like here for example). $\endgroup$
    – Liam F-A
    Commented Jul 11, 2023 at 2:32

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