Let's assume I have a $m \times m$ matrix $M$ with Frobenius norm $1$ and a unit vector $x \in S^{m-1}$. I also have a second $m \times m$ matrix $M^\*$ which is obtained from the first one plus some injected noise $\eta$, where every entry $\eta\_{i,j}$ is i.i.d. and comes from a normal distribution with mean $0$ and variance such that $||\eta||\_F = 1/10$. In other words, as Dirk pointed out in the answer,the entries of η (viewed vectorized) come from the uniform distribution on the scaled sphere in $n^2$ dimensions.


What can we say about the following expected value $$\mathbb{E}[| Mx - M^\*x|]?$$


I know that $\mathbb{E}|\eta(x)| <1/10$ since, because of the frobenius norm of the noise $|\eta(x)| < 1/10$, so the equality only holds when vector $x$
 is aligned with the highest eigenvector of the noise matrix $\eta$, but this happens with low probability, and this probability should (intuitively) change with respect to $m$: in a higher dimensional space the probability of two random vectors having a similar direction is lower than in a 2-dimensional one. 


Is there a way to improve the bound on this expected value?



Are Linear Maps resistant to Noise?
I'm assuming the vectorized $\eta$ is uniformly distributed on the sphere of radius $1/10$ in $\mathbb R^{n^2}$.


For fixed $x$, 
$$ \mathbb E \| \eta x \|\_2^2 = \mathbb E
\sum\_i \sum\_j \sum\_k \eta\_{ij} \eta\_{ik} x\_j x\_k = \sum\_j \sum\_k \mathbb E (\eta^T \eta)\_{jk} x\_j x\_k $$


Now for $j = k$, 
$$ \mathbb E(\eta^T \eta)\_{kk} = \sum\_i \mathbb E \eta\_{ik}^2 = \frac{1}{n} \sum\_{i}\sum\_j \eta\_{ik}^2 = \frac{1}{100 n}$$
while for $j \ne k$, since the conditional distribution of $\eta\_{ij}$ given $\eta\_{ik}$ is symmetric about $0$,
$$\mathbb E(\eta^T \eta)\_{jk} = \sum\_i \mathbb E(\eta\_{ij} \eta\_{ik}) = 0$$
Thus 
$$ \mathbb E \|\eta x\|\_2^2 = \sum\_k \frac{1}{100 n}x\_k^2 = \frac{\|x\|\_2^2}{100n}$$


I suspect more can be said about the distribution of $\|\eta x \|$ using concentration of measures, but I'll leave that to someone else.


