I have another question with dimensionality reduction. I have a matrix $S \in R^{k \times d}$ and S is in {$- \frac{1}{\sqrt k}, \frac{1}{\sqrt k}$} and i have two vector $u,v \in R^d $.

I need to understand why $E[u'^T v' ]= u^Tv$ where $u'=Su$.

I just have a intution, that maybe i just prove the left(or right span vector), but the hint is compute $E[S^TS]$ that for me this expectation is expectation of E[ I ] that if i m wrong will be 0...

I m a bit lost with this stuff.

i m thinking to do this

$E[u^t v] = E[u^T \sum vi ]$

another question is about the interval ... using $1 /\sqrt k $ is the same if i use a -1,1 with scaled factor of $\sqrt k $?


The idea is that $$ \mathbb{E}[u^{\prime T} v'] = \mathbb{E}[u^T S^T S v] = u^T \mathbb{E}[S^T S] v, $$ due to linearity of expectation. You can take it from here.

  • $\begingroup$ So the $E[S^T S] = 1$ because the variables are independent ? $\endgroup$ – theantomc Jan 16 '19 at 8:59
  • $\begingroup$ Note $S^T S$ is a matrix. $\endgroup$ – Yuval Filmus Jan 16 '19 at 9:02
  • $\begingroup$ Yes, for me $S^T S = I $ the identity matrix, and S have entries $S_{ij}$ $\endgroup$ – theantomc Jan 16 '19 at 9:20

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