Transition Matrix eigenvalues constraints

Question

I have a Transition Matrix, i.e. a matrix whose items are bounded between 0 and 1 and either rows or columns sum to one. I would like to know if it is possible that in any such matrices the eigenvalues or eigenvectors could contain an imaginary part. Thanks in advance.

Answer 1

The matrix $$ A=\left(\begin{array}{ccc} 0&1&0\\ 0&0&1 \\ 1&0&0 \end{array}\right) $$ has row and column sum equal to one, but has complex eigenvalues. The characteristic polynomial is $$ det(\lambda I-A)= \lambda^3-1. $$

Answer 2

I think it is worth pointing out that, having daw's fine upon which to build, we can easily construct a family of irreducible transition matrices which also have a pair of complex conjugate eigenvalues, where by "irreducible" I mean there is a non-zero probability of any state transiting to any other state in one step, that is, all matrix entries lie strictly 'twixt $0$ and $1$. (Not sure if this terminology is standard, but I know the underlying concept is important and useful.) For if we let

$A = \begin{bmatrix} p_1 & 1 - p_1 - q_1 & q_1 \\ p_2 & q_2 & 1 - p_2 - q_2 \\ 1- p_3 - q_3 & p_3 & q_3 \end{bmatrix} \tag{1}$

with the $p_i, q_i$ sufficiently small albeit nonvanishing positive real numbers, the the resulting $A$ will still have a pair of complex conjugate eigenvalues, by virtue of the well-known fact that the zeroes of a polynomial depend continuously on its coefficients. The polynomial in question here is of course the characteristic polynomial $p_A(\lambda)$ of $A$. Since the complex eigenvalues in the case $p_i = q_i = 0$ are distinct, they remain so for small $p_i, q_i$. Of course, $1$ remains a real eigenvalue of $A$, since

$A\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix} \tag{2}$

for any $A$ the row sums of which are $1$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!