# 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.
• So the $E[S^T S] = 1$ because the variables are independent ? – theantomc Jan 16 at 8:59
• Note $S^T S$ is a matrix. – Yuval Filmus Jan 16 at 9:02
• Yes, for me $S^T S = I$ the identity matrix, and S have entries $S_{ij}$ – theantomc Jan 16 at 9:20