Proof that the largest eigenvalue of a stochastic matrix is $1$ The largest eigenvalue of a stochastic matrix (i.e. a matrix whose entries are positive and whose rows add up to $1$) is $1$.
Wikipedia marks this as a special case of the Perron-Frobenius theorem, but I wonder if there is a simpler (more direct) way to demonstrate this result.
 A: If we can show that $A$ doesn't increase the 1-norm, i.e.,
$$\|Ax\|_1\leq\|x\|_1$$
Then $$\|Ax\|_1=\|\lambda x\|_1=|\lambda|\|x\|_1\leq\|x\|_1$$
which is $|\lambda|\leq 1$, we are done, but how to show above inequality? For convenience, let's set stochastic matrix
$$A=\begin{pmatrix}a_{11}& a_{12}\\a_{21}& a_{22}\end{pmatrix}$$
Then
\begin{eqnarray*}\|Ax\|_1&=&|a_{11}x_1+a_{12}x_2|+|a_{21}x_1+a_{22}x_2|\\&\leq& a_{11}|x_1|+a_{12}|x_2|+a_{21}|x_1|+a_{22}|x_2|\\&=&|x_1|+|x_2|\\&=&\|x\|_1\end{eqnarray*}
For n-dimensional matrix, it can be shown in same manner.
A: Here's a really elementary proof (which is a slight modification of Fanfan's answer to a question of mine). As Calle shows, it is easy to see that the eigenvalue $1$ is obtained.  Now, suppose $Ax = \lambda x$ for some $\lambda > 1$.  Since the rows of $A$ are nonnegative and sum to $1$, each element of vector $Ax$ is a convex combination of the components of $x$, which can be no greater than $x_{max}$, the largest component of $x$.  On the other hand, at least one element of $\lambda x$ is greater than $x_{max}$, which proves that $\lambda > 1$ is impossible.
A: It is an elementary exercise that a matrix $A \in \textsf{M}_n (\mathbb{C})$ has row sums equal to one if and only if $Ae = e$ (here, $e$ denotes the all-ones column vector of size $n$). Thus, if $A$ is stochastic, then $1 \in \sigma(A)$.
Now, let $(\lambda,v)$ be an eigenpair of $A$, in which $A$ is stochastic. Without loss of generality, we may assume that 
$$ \vert \vert v \vert \vert_\infty := \max_{1 \le k \le n} \{ \vert v_k \vert \} = 1.$$
Thus, there is a positive integer $i$, $1 \le i \le n$, such that $\vert v_i \vert = 1$. Since $Av = \lambda v$, by the mechanics of matrix multiplication and the above, we have 
$$\vert \lambda \vert = 
\vert \lambda \vert\cdot 1 =  
\vert \lambda \vert \cdot \vert v_i \vert = 
\vert \lambda v_i \vert = \left\vert \sum_{j=1}^n a_{ij} v_j \right \vert 
\le \sum_{j=1}^n a_{ij} \vert v_j \vert
\le \sum_{j=1}^n a_{ij} = 1,
$$
i.e., $\vert \lambda \vert \le 1$. 
A: Call A a $n \times n$ stochastic matrix and denote with $(\lambda,\textbf{x})$ one of its eigenpair.
Obviously $1$ is an eigenvalue for $A$, indeed follow directly from the definition of row-stochastic [column-stochastic] matrix that $\textbf e$ $=(1\dots1)^{T}$ is a right [left] eigenvector associated with $1$.
Using induced matrix norm $\parallel\parallel_{1}$ or $\parallel\parallel_{\infty}$ ,  it's easy to prove that the spectral radius $\rho(A)\leq 1$ :
$$
|\lambda|= \frac{||A x||}{||x||} \leq max_{||x||=1} ||Ax||= ||A||
$$
Now, since $||A||_{1}$  [respectively $||A||_{\infty}$ ] is the maximum absolute column [row] sum of the matrix, we have 
$||A||_{1}=1$ if $A$ is a column-stochastic matrix and
$||A||_{\infty}=1$ if $A$ is a row-stochastic matrix,
and then in any case $|\lambda|\leq 1$.
A: Say $A$ is a $n \times n$ row stochastic matrix. Now:
$$A \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} = 
\begin{pmatrix}
\sum_{i=1}^n a_{1i} \\ \sum_{i=1}^n a_{2i} \\ \vdots \\ \sum_{i=1}^n a_{ni}
\end{pmatrix}
=
\begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}
$$
Thus the eigenvalue $1$ is attained.
To show that the this is the largest eigenvalue you can use the Gershgorin circle theorem. Take row $k$ in $A$. The diagonal element will be $a_{kk}$ and the radius will be $\sum_{i\neq k} |a_{ki}| = \sum_{i \neq k} a_{ki}$ since all $a_{ki} \geq 0$. This will be a circle with its center in $a_{kk} \in [0,1]$, and a radius of $\sum_{i \neq k} a_{ki} = 1-a_{kk}$. So this circle will have $1$ on its perimeter. This is true for all Gershgorin circles for this matrix (since $k$ was taken arbitrarily). Thus, since all eigenvalues lie in the union of the Gershgorin circles, all eigenvalues $\lambda_i$ satisfy $|\lambda_i| \leq 1$.
A: If $A \in \textsf{M}_n (\mathbb{C})$, then $\begin{Vmatrix} A \end{Vmatrix}_\infty := \max\limits_{1\le i \le n} \sum_{j=1}^n \vert a_{ij} \vert$ is a matrix norm. Furthermore, $\rho(A) \le \begin{Vmatrix} A \end{Vmatrix}$ for any matrix norm $\vert \vert \cdot \vert\vert$ (e.g., Theorem 5.6.9 in the first-edition of Horn and Johnson's Matrix Analysis). If $A$ is stochastic, then $\begin{Vmatrix} A \end{Vmatrix}_\infty = 1$ so that $\vert \lambda \vert \le \rho(A) \le 1$.   
