TITLE: What is the expected distortion of a linear transformation?
QUESTION [5 upvotes]: Let $A: \mathbb{R}^n \to \mathbb{R}^n$. I am interested in the "average distortion" caused by the action of $A$ on vectors. (i.e stretching or contraction of the norm).
Consider for instance the uniform distribution on $\mathbb{S}^{n-1}$, and the random variable $X:\mathbb{S}^{n-1} \to \mathbb{R}$ defined by $X(x)=(\|A(x)\|_2)^2$. 
What is the expectation of $X$?
Using SVD, it is easy to check that the problem reduces to $A$ being a diagonal matrix with non-negative entries. So, the question amounts to calculating $$\int_{\mathbb{S}^{n-1}} \sum_{i=1}^n (\sigma_ix_i)^2 $$ (and dividing by the volume of $\mathbb{S}^{n-1}$).

Is there a closed formula for this integral?
Also, one could take the expected value of the norm, and not its square (I thought this should be easier if there are no sqaure roots involved...)

REPLY [4 votes]: Using the comment given by Hagen von Eitzen, we get: 
$$ \int_{\mathbb{S}^{n-1}} \sum_{i=1}^n (\sigma_ix_i)^2 = \sum_{i=1}^n \int_{\mathbb{S}^{n-1}} (\sigma_ix_i)^2= \sum_{i=1}^n \sigma_i^2\int_{\mathbb{S}^{n-1}} x_i^2$$
By symmetry, we see that $\int_{\mathbb{S}^{n-1}} x_i^2$ is independent of the index $i$, hence:
$$\int_{\mathbb{S}^{n-1}} x_i^2=\frac{1}{n} \sum_{j=1}^n \int_{\mathbb{S}^{n-1}} x_j^2 = \frac{1}{n} \int_{\mathbb{S}^{n-1}} \sum_{j=1}^n x_j^2=\frac{1}{n} \int_{\mathbb{S}^{n-1}} 1=\frac{1}{n} \operatorname{Vol}(\mathbb{S}^{n-1})$$
Hence the average distortion is:
$$E(X)=\frac{1}{n}\sum_{i=1}^n \sigma_i^2 $$