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.

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.
