# Expected norm of linear maps

I want to compute the expected norm of a vector-matrix multiplication. I have a vector $$x \in \mathbb{R}^n$$ with norm one and a matrix $$M \in \mathbb{R}^{n \times n}$$, whose entries are iid taken from the Gaussian distribution with mean zero and variance $$2/n$$: $$M_{i,j} \sim \mathcal{N}(0,2/n)$$.

I need to compute the expected value of the norm of the product, i.e. $$\mathbb{E}[||xM||].$$

I have computed $$\mathbb{E}[||xM||^2] = 2$$, but i have no idea on how to get rid of that square sign.

I could use the chi or Gamma functions, but I'm somehow stuck:

I'd have $$\sqrt{\sum_i\sum_j^n x_i^2 M_{i,j}^2 + \sum_i \sum_j \sum_k x_i x_j M_{i,j} M_{i,k}}$$

I know I can use the gamma function for the first sum, and that the expected value of the second part goes to zero. The problem is that I can't compute the expected value of just the second part since it's under the square root. Any suggestions?

Thank you!

The key is the simple observation that, for any orthogonal matrix $$Q$$, the matrix $$QM$$ equals $$M$$ in distribution. Let now $$Q$$ be any orthogonal matrix such that $$xQ=e_1:=[1,0,\dots,0]$$. Then $$xM$$ equals $$$$xQM=e_1M=[M_{1,1},\dots,M_{1,n}]$$$$ in distribution. So, $$\|xM\|$$ equals $$\sqrt{\frac2n\,X}$$ in distribution, where $$X\sim\chi^2_n=\text{Gamma}(\frac n2,2)$$. It follows that $$$$E\|xM\|=\frac{2\Gamma(\frac{n+1}2)}{\sqrt n\,\Gamma(\frac n2)},$$$$ which is close to $$\sqrt 2=\sqrt{E\|xM\|^2}$$ for large $$n$$, as should be expected, in view of the measure concentration phenomenon.
• Sorry, I'm trying to plug numbers inside the equation you provided, but even for large $n$, the final output is not close to $\sqrt2$. How is this possible? Thanks! – Alfred May 28 '19 at 10:55
• @Alfred : I guess you made a mistake somewhere. I have rechecked the limit manually using Striling's formula, and it is $\sqrt2$ indeed. Look also at the section "Series expansion at $n=\infty$" at wolframalpha.com/input/… – Iosif Pinelis May 28 '19 at 17:17