TITLE: Why isn't every eigenvalue of a stochastic matrix equal to 1?
QUESTION [8 upvotes]: In this question, we see a proof that the largest eigenvalue of a stochastic matrix is equal to 1:
Proof that the largest eigenvalue of a stochastic matrix is 1
However, I think I've found a proof that every eigenvalue of a stochastic matrix is equal to 1.  Can you tell me where my proof is wrong?
Proof: Suppose ${\bf r}$ is an eigenvector of the column stochastic matrix $M$ (i.e. $M{\bf r} = \lambda {\bf r}$ for some $\lambda$), and assume without loss of generality that the entries of ${\bf r}$ sum to $1$.  Then $$M{\bf r} =
\begin{bmatrix}
M_{11}\\
M_{21}\\
\vdots\\
M_{n1}
\end{bmatrix} r_1 +
\begin{bmatrix}
M_{12}\\
M_{22}\\
\vdots\\
M_{n2}
\end{bmatrix} r_2
+ \dots +
\begin{bmatrix}
M_{1n}\\
M_{2n}\\
\vdots\\
M_{nn}
\end{bmatrix} r_n$$
Since $M$ is column stochastic, each column must sum to $1$, so the sum of the entries in $M {\bf r}$ is just $1 \cdot r_1 + 1 \cdot r_2 + \dots + 1 \cdot r_n = 1$.  Therefore, $\lambda$ must be $1$, and $M$ can only have one eigenvalue.
Thanks!

REPLY [3 votes]: The problem is when you assume WLOG that the entries of $\bf{r}$ sum to one. Consider what happens when entries of $\bf r$ sum to zero. 
What your proof shows is that if $(\lambda,\bf{r})$ is an eigenpair, and $\lambda\neq 1$, then the entries of $\bf r$ sum to zero. For example, the matrix 
$$
\begin{bmatrix}
0&1\\1&0
\end{bmatrix}
$$
is stochastic, with eigenvalues $\pm1$. The eigenspace for $+1$ is spanned by $(1/2,1/2)$, and for $-1$ is spanned by $(1/2,-1/2)$, where the latter's entries sum to zero.