# The variance of the expected distortion of a linear transformation

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 $d^4(A,O_n)$).

## 1 Answer

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