Expectation of $u'^t v$ = $u^t v$ 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.