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.


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 $$ det(\lambda I-A)= \lambda^3-1. $$


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!!!

  • 1
    $\begingroup$ Though it is confusing terminology, transition matrices with all $p_{ij}>0$ are typically called "ergodic", because they satisfy the "ergodic theorem" (i.e., there exists an invariant measure and any initial measure will converge to it). "Irreducible" typically means that any state is accessible from any other state with positive probability (but possibly in more than one time step). $\endgroup$ – Ian Jun 24 '14 at 3:06
  • $\begingroup$ OK, thanks . . . it's been awhile since I worked on Markov chains, so my recollection of the terminology is perhaps a little rusty. I guess I can save my posterior in this case by pointing out that I pointed out what I meant by "irreducible" in the post . . . must have been a little uncertain! Thanks! $\endgroup$ – Robert Lewis Jun 24 '14 at 3:37

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