When is the dominant eigenvalue of this matrix greater than one?

Question

So I am trying to figure out when this matrix

$\left[\begin{matrix} a_1 & 0 & b \\ a_2 & a_3 & 0 \\ a_4 & a_5 & a_6 \end{matrix}\right]$

$b, a_i\geq0$ for all $i$, and $a_1,a_3,a_6\leq1$,

has a dominant eigenvalue less than one and when it has a dominant eigenvalue greater than one. I've tried calculating the eigenvalue, explicitly, which is a mess, but I was hoping there is some way to take advantage of the restrictions, to answer this specific question. Consider the case where $b=0$ (for intuition's sake), this would lead to a dominant eigenvalue less than one if none of the diagonal entries are exactly one. Intuitively $b$ must be large enough to make up for this.

The Perron Frobenius theorem gives us some conditions, namely if all of the column or row sums are greater than one the dominant eigenvalue will be greater than one and if they are all less than one the dominant eigenvalue will be less than one. But I'm looking for something a bit stronger.

I think if $a_2a_5b>1$ then the eigenvalue would be greater than one (thinking about number of particles moving from state 1->2->3 and then back to 1 again picking up mass along the way, on average.

Answer 1

You can use the Gerschgorin circle theorem to get a simple estimate: to have the dominant eigenvalue less than $1$ it is enough to have $a_1 + b < 1$, $a_3 + a_2 < 1$, $a_4 + a_5 + a_6 < 1$. This is of course unsatisfactory: if $a_1,a_3,a_6<1$, even if $a_2,a_4,a_5>1$, the dominant eigenvalue will still be less than $1$ for small enough $b$.

Another idea is to estimate the eigenvalues using the implicit function theorem applied to the characteristic polynomial, where the "base point" is the eigenvalues for $b=0$, which you know.

Edit: clarifying the implicit function theorem description. Fix the $a_i$, but let $b$ vary. Then you have three implicit functions $\lambda_i$ defined by $p(\lambda_i(b),b)=0$, where for each $b$, $x \mapsto p(x,b)$ is the characteristic polynomial of your matrix and $\lambda_1(0)=a_1$, $\lambda_2(0)=a_3$, $\lambda_3(0)=a_6$. You can find the derivatives with the implicit function theorem, the result being:

$$\frac{d \lambda_i}{db} = -\frac{\frac{\partial p}{\partial b}}{\frac{\partial p}{\partial \lambda_i}}$$

Note that this fails if any of the $a_1,a_3,a_6$ are the same (because the derivative in the denominator becomes $0$).

Edit 2: Doing the details of this implicit function theorem argument. We have

\begin{eqnarray*} p(x,b) & = & (a_1-x)(x-a_3)(x-a_6) + b (a_2 a_5 - a_3 a_4 + a_4 x) \\ \frac{\partial p}{\partial b} & = & a_2 a_5 - a_3 a_4 + a_4 x \\ \frac{\partial p}{\partial x} & = & -(x-a_3)(x-a_6) + (a_1-x)(x-a_6) + (a_1-x)(x-a_3) + b a_4 \end{eqnarray*}

(The first line is obtained by cofactor expansion across the first row, though I used Mathematica.) This looks worse than it is: for $b=0$ and $x=a_1,a_3,a_6$, only one term survives in the denominator. The easiest case:

$$\frac{d \lambda_2}{d b} = \frac{a_2 a_5}{(a_3-a_1)(a_3-a_6)}$$

So:

$$\lambda_2(b) = a_3 + \frac{a_2 a_5 b}{(a_3 - a_1)(a_3 - a_6)} + O(b^2)$$

as $b \to 0$.

Testing: if $a_1=a_2=a_4=a_5=1$, $a_3=2$, $a_6=3$, $b=0.1$, we get $\lambda_2 \approx 1.908$. The implicit function theorem predicts $\lambda_2 \approx 1.9$.

Answer 2

Maybe Perron-Frobenius Theorem can perhaps help already a little bit $$A = \left[\begin{matrix} a_1 & 0 & b \\ a_2 & a_3 & 0 \\ a_4 & a_5 & a_6 \end{matrix}\right],$$ is irreducible, e.g. if $0<a_i,b,$ then $A^2 > 0$ and thus $A$ is irreducible. Then we know that there exists $v_1,v_2,v_3> 0$ such that $v=(v_1,v_2,v_3)$ has norm $1$ and $Av = \lambda v$ where $\lambda$ is the dominant eigenvalue. Then in particular we have $(Av)_i/v_i = \lambda$ (this is a Collatz-Wielandt ratio type). Let us plug that all in a nice little system $$\left\{\begin{array}{rclc} a_1+b\frac{v_3}{v_1} &=& \lambda & \quad (1) \\ a_2\frac{v_1}{v_2} + a_3 &=& \lambda & \quad (2) \\ \frac{a_4v_1 + a_5v_2}{v_3} +a_6&=& \lambda & \quad (3)\end{array}\right.$$ From $(2)$ we may observe that $$v_1 = \left(\frac{\lambda-a_3}{a_2}\right)v_2,$$ this together with $(1)$ shows $$v_3 = \left(\frac{\lambda-a_1}{b}\right)v_1= \left(\frac{\lambda-a_1}{b}\right) \left(\frac{\lambda-a_3}{a_2}\right)v_2$$ now plug that in $(3)$ to find $$\dfrac{a_4\left(\dfrac{\lambda-a_3}{a_2}\right)v_2 + a_5v_2}{\left(\dfrac{\lambda-a_1}{b}\right)\left(\dfrac{\lambda-a_3}{a_2}\right)v_2}+ a_6 = \lambda $$ now simplify $v_2$ and multiply by $\left(\frac{\lambda-a_1}{b}\right)\left(\frac{\lambda-a_3}{a_2}\right)$ to get $$ a_5 + \left(a_4+(a_6-\lambda)\left(\frac{\lambda-a_1}{b}\right)\right)\left(\frac{\lambda-a_3}{a_2}\right) =0$$ This is a third degree polynomial which roots should be findable.