Suppose that I have two positive matrices, $A$, and $B$, and I know their Frobenius-Perron eigenvalues ($\lambda_A$, $\lambda_B$) and eigenvectors ($v_A$, $v_B$). I'm interested in what I can say about the Frobenius-Perron eigenvalue and eigenvector of $C=A+B$. I'm sure that this question must have been considered before, but I haven't found a good source.

Here are two particular questions:

(1) One clear bound on $\lambda_C$ is that $\max(\lambda_A,\lambda_B) \le \lambda_C$. Is there a nice upper bound? Also, if $v_A=v_B$, then $\lambda_C = \lambda_A + \lambda_B$. But can we do better than this if we know $v_A$ and $v_B$?

(2) What can we say about $v_C$? I expect it to be "somewhere between" $v_A$ and $v_B$, but that's just vague intuition. (Assume all the eigenvectors are normalized.) Is there a simple bound on the elements of $v_C$ given knowledge of $\lambda_A$, $\lambda_B$, $v_A$, and $v_B$?

I'd prefer not to assume that $A$ and $B$ are symmetric, but if the question has only been solved for the symmetric case, that's fine.

Edit: I realized that I should clarify that all of the $v$'s are right eigenvectors. If it helps to know the left Frobenius-Perron eigenvectors $u_A$ and $u_B$, then feel free to use this too.

Edit 2: In light of David's answer, then I've corrected my "intuitive" bound on $\lambda_C$ which was completely wrong.


1 Answer 1


First, the definition of $positive$ should be clarified. It could mean all entries strictly positive, or merely nonnegative, or that the matrix is primitive (all entries nonnegative and some power strictly positive).

I will use $\rho(M)$ to denote the spectral radius of a matrix $M$, which in case $M$ is nonnegative, is the Perron eigenvalue.

Note a weird example: if $M$ is strictly upper triangular with all entries above the diagonal strictly positive, then the spectral radius is $0$, and the same for $M^T$. However, $M+M^T$ is primitive, so its spectral radius exceeds zero. We can modify this to obtain strictly positive $M$ with $\rho(M)$ small, but $\rho(M+M^T)$ big, simply by adding a tiny amount to all the zero entries. In this case, $\rho(C)$ can be made much larger than $\rho(A) + \rho(B) $.

A well-known result ($well-known$ means I can't give a reference off the top of my head) is that if $C \geq A$ entrywise, $C \neq A$, and $A$ is strictly positive (or merely primitive), then $\rho(C) > \rho (A)$. (But the argument is easy: use an H-transform to convert $A$ to one with all column sums equal; adding anything will increase the spectral radius.) So if $A$ and $B$ are primitive, then $\rho(A+B) > \max \{\rho(A), \rho(B) \}$.

I doubt there's much of anything beside the obvious that can be said about the Perron eigenvectors.

Perhaps you have matrices in special forms in mind?

  • $\begingroup$ Thanks. I corrected the bound above. I don't have special restrictions in mind right now except that $A_{ij}$ and $B_{ij}$ are in $(0,1]$. But if the problem can't be solved in general, then maybe it can be solved for random matrices with entries in $(0,1]$, and the result can be used as a heuristic (although not a rigorous result). A good heuristic is really all I need. $\endgroup$
    – sasquires
    Commented Nov 13, 2013 at 13:49

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