How can a markov transition matrix have eigenvalues other than 1?

Question

A Markov transition matrix has all nonnegative entries and so by the Perron-Frobenius theorem has real, positive eigenvalues. In particular the largest eigenvalue is 1 by property 11 here. Furthermore in these notes (sec 10.3) it says that the eigenvalues of $P$ are $1 = \lambda_1 > \lambda_2 \geq \dots \geq \lambda_N \geq -1$.

But how can a transition matrix $P$ have eigenvalues less than 1? Since the matrix is acting on probability distributions $v$, which have to have $\sum_i v_i = 1$, we cannot have $Pv = cv$ with $c\neq 1$ since $\sum_i cv_i = c$.

Answer 1

Maybe a good starting point would be to look at a simple example. The eigenvalues of the transition matrix $\begin{bmatrix} \frac{1}{2} & 1 \\ \frac{1}{2} & 0\end{bmatrix}$ are:

• $1$, with eigenvector $\begin{pmatrix} \frac{2}{3} \\ \frac{1}{3} \end{pmatrix}$ (the principal eigenvalue giving the stationary Markov distribution)
• $-\frac{1}{2}$, with eigenvector $\begin{pmatrix} -1 \\ 1\end{pmatrix}$.

Of course, the second one cannot be a probability distribution no matter how we renormalize it, because it has negative entries!

Answer 2

The eigenvectors corresponding to the non-one eigenvalues simply do not correspond to probability distributions; they have both negative and positive entries. For example, $$\begin{bmatrix}0.1&0.9\\0.9&0.1\end{bmatrix}$$ has an eigenvalue of $-0.8$ corresppnding to the eigenvector $(1,-1)$.

Answer 3

The $n$-vectors representing a probability distribution live in a $n$-space.
Since their components sum to 1, they represent a point on the diagonal plane $x_1+x_2+ \cdots + x_n=1$ .
So, if one eigen-vector lies on the plane, and within the positive "octant", the others (being independent, i.e. normal to each other) cannot all lie in that octant, apart from the case in which they correspond to the base vectors $(0,0,\cdots, 1,0, \cdots)$.