normalized gaussian random matrix preserves geometry in expectation

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Important

Let GN(0,1)d×m and YRm×n. Now, define

G^:=1dG,Law(G^)=N(0,1d)d×m

Then

\begin{align} \mathbb{E}[\,\lvert \lvert \hat{G}y \rvert \rvert { #2} \,] &= \lvert \lvert y \rvert \rvert { #2} \\ \mathbb{E}[\,\langle \hat{G}y_{1}, \hat{G}y_{2} \rangle \,] &= \langle y_{1}, y_{2} \rangle \\ \mathbb{E}\left[ \,(\hat{G}Y)^{\intercal}(\hat{G}Y)\, \right] &= Y^{\intercal}Y \end{align}

ie, in expectation, G^ preserves the lengths, angles, and Gram matrices. That is, it preserves all the geometry of Rm

Line of Reasoning

Note that for any column of Y, say y=yi, we have

Expected Magnitude

\begin{align} \mathbb{E}[\lvert \lvert Gy \rvert \rvert { #2} ] &= \mathrm{Tr}(\text{Cov}(Gy) ) \\ &= d \lvert \lvert y \rvert \rvert { #2}

\end{align}$$

and for any y1,y2 columns of Y

Expected Direction

E[Gy1,Gy2]=Tr(Cov(Gy1,Gy2))=dy1,y2

And, looking at the Gram matrix,

Expected Gram Matrix

E[(GY)(GY)]=dYY

Caveats

Note

Note that this holds for any d, including dm and d=1. This is because we are taking expectations.

If y is selected first before we realize G^, then we hope that ||G^y|| will concentrate about ||y||. Our calculations for the joint show that

\text{Law}(\hat{G}y) = {\cal N}\left( 0, \frac{1}{d}\lvert \lvert y \rvert \rvert { #2} I_{d} \right)

So (with rescaling, obviously), our concentration inequality for magnitude of standard gaussian random vector holds and we should see the desired behavior.

If y does not depend on G^

Assuming we ensure G^y, the probability of actually preserving geometry depends on d.

Example

If d=1, then G^=g for some gN(0,Id). We still have

\mathbb{E}[\lvert \lvert \hat{G}y_{i} \rvert \rvert { #2} ] = \mathbb{E}[\langle g,y_{i} \rangle { #2} ] =\lvert \lvert y_{i} \rvert \rvert { #2}

But clearly g,yi2||yi||2 cannot hold simultaneously for many yi regardless of if we choose them before realizing G^.

Exercise

Construct some examples of this case

If y depends on G^...

If dm, then once G^ is realized, we can pick yNull(G^) so that 0=||G^y||||y||.

Review

#flashcards/math/rmt

Normalized gaussian random matrices preserve geometry (in expectation), BUT

Assuming we ensure G^y, the probability of {1||preserving geometry} depends on {2||the "output" dimension d}.

For G^N(0,1d)d×m and yRm, what is the (co)variance matrix of G^y?
-?-

\frac{1}{d} \lvert \lvert y \rvert \rvert { #2} I_{d}$$ <!--SR:!2025-09-18,2,230--> If $\hat{G} \sim {\cal N}\left( 0, \frac{1}{d} \right)^{\otimes d\times m}$, why do we need $y\in \mathbb{R}^m$ to be independent of $\hat{G}$ to preserve geometry if $d<m$? -?- Otherwise, we can choose $y \in \text{Null}(\hat{G})$ and then the norm is not preserved. <!--SR:!2025-09-18,3,250--> # References >[!references] ## See Also - ### Mentions > [!mentions]+ > | File | Last Modified | > | ------------------------------------------------------------------------------------------------------------------------------------------------------------ | ------------- | > | normalized gaussian random matrix preserves geometry in expectation | 2025-09-13 | > | Random Matrix Lecture 02 | 2025-09-13 | > { .block-language-dataview} <span class="dv-modified"><span>Created 2025-09-13 ֍ Last Modified 2025-09-18</span></span>