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Let $M$ be an irreducible non-negative square matrix with spectral radius $\lambda$. Due to the Perron-Frobenius theorem, we know that $\lambda > 0$ and $\lambda$ is an eigenvalue of $M$. Also, all other eigenvalues of $M$ are strictly smaller than $\lambda$.

Now consider such a matrix with an unknown entry, e.g. $$M = \begin{pmatrix} 0 & 0 & 40 & 60 & 80\\ p & 0 & 0 & 0 & 0\\ 0 & 0.35 & 0 & 0 & 0\\ 0 & 0 & 0.16 & 0 & 0\\ 0 & 0 & 0 & 0.08 & 0 \end{pmatrix}.$$ Question: How can I determine $p$ such that $\lambda = 1$ (in general)? I'm not sure how simple finding an exact solution is, I would already be happy with a numerical solution. I thought about guessing some values for $p$ such that $\lambda \approx 1$ and using interpolation, but am not sure which interpolation would be appropriate, since I don't know how $p$ and $\lambda$ should relate to each other.

Motivation: Leslie matrices are often used to study changes of a population. The entries must be estimated and the case $\lambda = 1$ is the equilibrium state of the population.

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It's easy to determine all values of $p$ for which $M$ has any fixed eigenvalue $\lambda$: simply compute the determinant of $M-\lambda I$, which is a linear function of $p$, and find the value of $p$ that makes this determinant equal to $0$. In particular, the unique value of $p$ for which $1$ is an eigenvalue of $M$ is $p=625/11074 \approx 0.0564385$, and one can then check by computer that $1$ is indeed the eigenvalue of largest modulus.

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  • $\begingroup$ Thanks, this is indeed very easy. I'm working with a matrix algebra system which has a lot of trouble with symbolic algebra (no native support). Since I want to do these kind of things in a high school class, I want to avoid any possible problems. Is there a usable numerical way (interpolation)? I could use a different software to compute the determinant, but only as last resort. $\endgroup$
    – Huy
    Feb 21 '19 at 9:12

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