Markov chains - Why is this a distribution? I read about markov chains here : https://www.cs.princeton.edu/courses/archive/fall05/cos521/markov.pdf .
There was something I didn't understand at the beginning of page 6, 
the last inequality, with the $A_iE_i$ and $L_2$ norm.
From what I understand, they mean that $x - v_1 (\frac{1}{n})$ (where $v_1 $is the $(1,1,...,1)$ vector), is a distribution.  But how come? $x$ is a distribution and we deduct a vector whose coordinates sum to $1$. 
Why is $x - v_1(\frac{1}{n})$ (or sum of  $A_iE_i$ from $i=2$ to $n$) , a distribution? 
 A: Since $x$ is a distribution, $Mx$ is too, since $M$ is a transition matrix.
Note that its stationary distribution is $s=(1/n)\vec{1}$. But $s$ is the first eigenvector of $M$ (by definition of stationary distribution given), with eigenvalue $1$ (so $Ms=s$). In other words $e_1=s$ and since the eigenvectors form a complete system, you can write:
$$
x = \sum_{i=1}^n \alpha_i e_i = s + \sum_{i=2}^n \alpha_i e_i
$$
using the observation (in the notes) that $\alpha_1=1$.
Now:
$$
Mx = M\left(s+\sum_{i=2}^n \alpha_i e_i\right)=
Ms+\sum_{i=2}^n \alpha_i M e_i
= s+\sum_{i=2}^n \alpha_i \lambda_i e_i
$$
is a distribution (since $Mx$ is one). So $M^tx$ is also one.
Repeatedly applying our rule above gives:
$$
M^tx = s + \sum_{i=2}^n \alpha_i \lambda_i^t e_i
$$
The author is interested in how fast $ M^tx $ converges to $s$. So he looks at and bounds
$$
||M^tx - s||_2 = ||s + \sum_{i=2}^n \alpha_i \lambda_i^t e_i - s||_2
= ||\sum_{i=2}^n \alpha_i \lambda_i^t e_i||_2
$$
which is the $\ell_2$ distance of the distribution $M^tx$ from $s$, for some $t$.
The quantity in the norm is not a distribution; rather, it is the distance between two discrete distributions.

Edit: it seems the asker's actual question is why 
$||\sum_{i=2}^n \alpha_i e_i||_2 \leq 1$. I'll show that now:
\begin{align}
|| \sum_{i=2}^n \alpha_i e_i  ||_2^2
&=
||x - s||_2^2 \\
&= 
\sum_i \left( x_i - \frac{1}{n} \right)^2 \\
&=
\sum_i x_i^2 - \frac{2x_i}{n} +\frac{1}{n^2}\\
&=
||x_i||_2^2 - \frac{2}{n}\sum_i x_i +\frac{1}{n}\\
&=
||x_i||_2^2 - \frac{1}{n}\\
&\leq
||x_i||_1^2 - \frac{1}{n}\\
&\leq
1 - \frac{1}{n}\\
&\leq 1
\end{align}
