2
$\begingroup$

Let $A: \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation. I am interested in the "average distortion" caused by the action of $A$ on vectors. (i.e stretching or contraction of the norm).

Consider the uniform distribution on $\mathbb{S}^{n-1}$, and the random variable $X:\mathbb{S}^{n-1} \to \mathbb{R}$ defined by $X(x)=\|A(x)\|_2^2$.

It is easy to see that $E(X)=\frac{1}{n}\sum_{i=1}^n \sigma_i^2$, where the $\sigma_i$ are the singular values of $A$.

Question: What is the variance of $X$?


Remark:

Using SVD, the problem reduces to $A$ being a diagonal matrix with non-negative entries, and the question amounts to calculating $$E(X^2)=\int_{\mathbb{S}^{n-1}} \big(\sum_{i=1}^n \sigma_i^2x_i^2 \big)^2$$ (and dividing by the volume of $\mathbb{S}^{n-1}$).

Is there a closed formula for this integral?


Motivation:

I am interested in measuring the

"deviation of a linear transformation from being an isometry".

I think something like $\text{Dev}(X):=(E(X)-1)^2 + \operatorname{Var}(X)$ might serve as a nice measure. Therfore, I want to be able to calculate the variance.

If not, estimates might also be useful. In particular, I am interested in comparing $\text{Dev}(X)$ and $d(A,O_n)=\sqrt{\sum_{i=1}^n (\sigma_i-1)^2}$, where

$O_n$ is orthogonal group, $d(A,O_n)$ is the distance of $A$ from $O_n$, measured w.r.t the Euclidean distance $d(A,B)=\|A-B\|_2$. This is the standard measure in the theory of Elasticity for the deviation of a transformation from being an isometry.

The "right power" of $d(A,O_n) \,$, suitable for comparison with $\text{Dev}(X)$ is $4$, since the highest powers of the $\sigma_i$ in both expression is the same. (i.e we should compare $\text{Dev}(X)$ VS $d^4(A,O_n)$).

$\endgroup$
1
$\begingroup$

First, I provide a way to generate a uniform random vector on $S^{n-1}$ and it will be used in our proof later.

Fact: Let $X_1$, $X_2$, $\cdots$, $X_n \sim \mathcal{N}(0, 1)$ and be independent. The the vector $$ X = (\frac{X_1}{Z}, \frac{X_2}{Z}, \cdots, \frac{X_n}{Z}) $$ is a uniform random vector on $S^{n-1}$ where $Z = \sqrt{X_1^2 + \cdots + X_n^2}$ is a normalization factor.


Assume $n > 1$.

  • How to get $\mathbb{E}(X)$ in your question?

Let $A = U\Sigma V^T$ be the SVD of $A$. Then $$ X = \|Ax\|_2^2 = x^TV\Sigma U^TU\Sigma V^T x = (V^Tx)^T\Sigma^2(V^Tx) = y^T\Sigma^2y $$ where $y = V^Tx$ is still a uniform random vector on $S^{n-1}$ since $V^Tx$ just rotates $x$.

Therefore, $$ X = \sum_i y_i^2\sigma_i^2 \quad \text{and}\quad \mathbb{E}(X) = \sum_i \mathbb{E}(y_i^2)\sigma_i^2 $$ Note that $y_i^2$ follows the same distribution with $\frac{z_i^2}{z_1^2 + \cdots + z_n^2}$ for independent standard normal random variables $z_1$, $\cdots$, $z_n$. We have $\mathbb{E}(y_i^2) = \frac{1}{n}$ by symmetry (Note that $\mathbb{E}(\sum_i y_i^2) = 1$). Therefore, $$ \mathbb{E}(X) = \frac{1}{n}\sum_i\sigma_i^2 $$


  • How to obtain $\mathbb{E}(X^2)$?

Similarly, $$ X^2 = \sum_{i,j}y_i^2y_j^2\sigma_i^2\sigma_j^2 $$ We have $$ \mathbb{E}(y_i^4) = \mathbb{E}(\frac{z_i^4}{(z_1^2 + \cdots + z_n^2)^2}) = \frac{3}{n(n+2)} $$ For a proof of the equality above, see here. Moreover, by the equality $$ \sum_{i\neq j}\mathbb{E}(\frac{z_i^2z_j^2}{(z_1^2 + \cdots + z_n^2)^2}) + \sum_i \mathbb{E}(\frac{z_i^4}{(z_1^2 + \cdots + z_n^2)^2}) = 1 $$ and by symmetry, we have $$ \mathbb{E}(y_i^2y_j^2) = \mathbb{E}(\frac{z_i^2z_j^2}{(z_1^2 + \cdots + z_n^2)^2}) = \frac{1}{n(n+2)} \quad \text{for } i \neq j $$


Therefore, $$ \mathbb{E}(X^2) = \sum_{i \neq j} \frac{1}{n(n+2)}\sigma_i^2\sigma_j^2 + \sum_i \frac{3}{n(n+2)} \sigma_i^4 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.