Column stochastic matrix/Positive row/One-dimensional eigenspace/Fact/Proof

Proof

1. The transposed matrix is row stochastic; therefore, it has an eigenvector ${}{\begin{pmatrix}1\\1\\\vdots \\1\end{pmatrix}}$ to the eigenvalue ${}1$. Due to fact, the characteristic polynomial of the transposed matrix has a zero at ${}1$. Because of exercise, this also holds for the characteristic polynomial of the matrix we have started with. Hence, ${}M$ has an eigenvector to the eigenvalue ${}1$.
2. We now assume also that all entries of the ${}k$-th row are positive, and let ${}v\in V$ denote a vector with (at least) a positive and a negative entry. Then

   ${}{\begin{aligned}\mid \mid \!Mv\!\mid \mid _{\operatorname {sum} }&=\sum _{i=1}^{n}\vert {(Mv)_{i}}\vert \\&=\sum _{i=1}^{n}\vert {\sum _{j=1}^{n}a_{ij}v_{j}}\vert \\&=\sum _{i\neq k}\vert {\sum _{j=1}^{n}a_{ij}v_{j}}\vert +\vert {\sum _{j=1}^{n}a_{kj}v_{j}}\vert \\&<\sum _{i\neq k}\sum _{j=1}^{n}a_{ij}\vert {v_{j}}\vert +\sum _{j=1}^{n}\vert {a_{kj}v_{j}}\vert \\&=\sum _{i=1}^{n}\sum _{j=1}^{n}a_{ij}\vert {v_{j}}\vert \\&=\sum _{j=1}^{n}\vert {v_{j}}\vert {\left(\sum _{i=1}^{n}a_{ij}\right)}\\&=\sum _{j=1}^{n}\vert {v_{j}}\vert \\&=\mid \mid \!v\!\mid \mid _{\operatorname {sum} }\end{aligned}}$

   holds.
3. As in the proof of (2), let all entries of the ${}k$-th row be positive. For any eigenvector ${}v$ to the eigenvalue ${}1$, according to (2), either all entries are non-negative, or non-positive. Hence, for such a vector, because of ${}Mv=v$, its ${}k$-th entry is not ${}0$. Let ${}v,w$ be such eigenvectors. Then ${}{\frac {w_{k}}{v_{k}}}v-w$ belongs to the fixed space. However, the ${}k$-th component of this vector equals ${}0$; therefore, it is the zero vector. This means that ${}v$ and ${}w$ are linearly dependent. Therefore, this eigenspace is one-dimensional. Because of (2), there exists an eigenvector to the eigenvalue ${}1$ with non-negative entries. By normalizing, we get a stationary distribution.

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