Algebraic multiplicity of the eigenvalue $1$ in a positive Markov matrix is equal to $1$ I want to show that the following linear system has only one non-trivial positive solution,
\begin{equation}
\begin{bmatrix}
\mu_{1,1}^* & \mu_{2,1}^* &\dots &\mu_{N,1}^* \\
\mu_{1,2}^* & \mu_{2,2}^* &\dots &\mu_{N,2}^*\\
\dots &\dots &\dots &\dots\\
\mu_{1,N}^* & \mu_{2,N}^* &\dots &\mu_{N,N}^*\\
1 & 1 &\dots &1
\end{bmatrix}
\cdot
\begin{bmatrix}
L_1  \\
L_2\\
\dots\\
L_N
\end{bmatrix}
=
\begin{bmatrix}
L_1  \\
L_2\\
\dots\\
L_N\\
\bar{L}
\end{bmatrix}.
\end{equation}
The only assumptions I have are that the entries are positive and the sums of each column in the first matrix are all equal to 2. In other words, the matrix consisting of $\mu^*$s is a positive Markov matrix. Can anyone provide some theorems related to this question? Thanks.
 A: Your conditions are not sufficient. Here are some properties you need. Firstly, forget about the positiveness of the solution and just assume that $\overline L\ne0$. Break down the condition into two:
Let $M$ be the $N\times N$ matrix consisting of $\mu^*_{i,j}$s. We know that sum of every column on $M$ is equal to $1$. First of all, you want $1$ to be an eigenvalue of your matrix, which already is true by the fact every column adds up to $1$, (think of the vector $(1,1,\ldots,1)$ as the eigenvector for $M^t$, and use the fact that $M^t$ and $M$ has the same eigenvalues.)
Let $V$ be the set of vectors satisfying $Mv=v$. Then, for a given non-zero $\overline L$, you want a unique $v$ in $V$ satisfying $L_1+L_2+\ldots+L_n=\overline L$. Observe that if $\dim V=1$ and the basis $\{v_0\}$ for $V$ is not satisfying $[1,1,\ldots,1]v_0=0$, you have what you want, as any vector in $V$ is of the form $cv_0$, for scalar $c$ and you can choose $c$ as appropriately. 
However, if $\dim V>1$ you can show that the uniqueness is not longer valid. For example, let $v_1,v_2$ be two linearly independent vectors in $V$ such that $[1,1,\ldots,1]v_1\ne0$, (if such $v_1$ does not exists, there is no solution to your equation). Then, one can choose a scalar $c_!$ such that $c_1v_1$ is a solution. If $$[1,1,\ldots,1]v_2\ne0$$ then one cn choose a $c_2$ similarly and we would have two solutions $c_2v_1,c_2v_2$. So, suppose $$[1,1,\ldots,1]v_2=0$$ 
then, $c_1v_1+v_2\ne c_1v_1$ is also a solution. Contradiction.
About positiveness, I believe some extra conditions, maybe positiveness etc, on $\mu^*_{i,j}$s are necessary to achieve it.
