Hint:
try to define an appropriate norm on the matrix so that it is 1 for every doubly stochastic matrix and use the usual equation Av=pv.Now apply norm on this eqn and derive the fact that |p|<1 or you can use the spectral radius formula to directly get the result.1 is the largest eigenvalue so by perron-frobenius theorem,every other eigenvalue has absolute value strictly less than 1
I don't think a convention is well-established: in some contexts, I see "different eigenvalues" refer to a set of distinct values with associated algebraic multiplicities, while in other contexts, I see "different eigenvalues" refer to the set of $n$ eigenvalues, possibly with repetitions due to multiplicity. Typically one can either discern which convention is being used, or the author should take care to clarify what is meant.
In your case, I think you just have to read the definition of "dominant eigenvalue" carefully. Based on the problem writing "dominant eigenvalue $\lambda_1$," I suspect the definition is written as
if $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$, then $\lambda_1$ is considered dominant if $|\lambda_1| > |\lambda_i|$ for all $i \ne 1$
or something like that, which is unambiguous compared to
$|\lambda| > |\gamma|$ for all other eigenvalues $\gamma$
which is very ambiguous for the reasons you raise.
Now that we know that the context of your question is the power method, then my above guess on what "dominant eigenvalue" means is incorrect.
Let $\lambda_1, \ldots, \lambda_m$ be the distinct eigenvalues of $A$ with multiplicities $n_1, \ldots, n_m$. If $|\lambda_1| > |\lambda_i|$ for all $i \ne 1$, then $\lambda_1$ is said to be the dominant eigenvalue. The power method will converge to something in the eigenspace corresponding to $\lambda_1$. To ensure that it does not converge to zero, the initial vector must not be orthogonal to the eigenspace.
Best Answer
EDIT. This old proof proves nothing. Sorry for the OP.
Let $A$ be a stochastic matrix. There exists a permutation matrix $P$ s.t. $PAP^{-1} = \left( \begin{smallmatrix} B_1 & * & * & \cdots & * \\ 0 & B_2 & * & \cdots & * \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & * \\ 0 & 0 & 0 & \cdots & B_h \end{smallmatrix} \right)$ where the $B_i$'s are either irreducible or zero. $spectrum(A)=\{sp(B_1),\cdots,{s}p(B_h)\}$. If $1\in{s}p(B_i)$ then $B_i$ is irreducible and therefore $1$ is a SIMPLE eigenvalue of $B_i$.
Unfortunately, we can only conclude that the number of indices $i$ s.t. $1\in sp(B_i)$ is equal to the algebraic multiplicity of $1$.
A correct proof is as follows.
$\textbf{Proposition.}$ If $A$ is stochastic, then all the eigenvalues of $A$ with modulus $1$ are semi simple.
$\textbf{Proof.}$ For every $k$, $A^k$ is stochastic; then $||A^k||_{\infty}=1$ and the sequence $(A^k)_k$ is bounded. Then each Jordan block of $A$ associated to an eigenvalue of modulus $1$ is $1\times 1$. $\square$