# Dominant root of a family of polynomials

Let $$f(x)=x^5-x^4-x^3-x^2-x-c$$, where $$c>2$$ is a real number. It is easy to prove that there exists a positive real root $$\alpha>2$$ of $$f(x)$$ and all the other roots are non real.

Also, by using Kakeya-Enestrom theorem, it is possible to prove that if $$\beta$$ is a non real root of $$f(x)$$, then $$|\beta|\leq \alpha$$.

However, I would like to prove that this inequality is strict, but I am not being able to prove this. Someone has some suggestion?

The same problem happens by defining $$f_c(x)=x^k-x^{k-1}-\cdots - x- c$$, where $$c>2$$. So, $$f_c(x)$$ has only one positive root, say $$\alpha$$, which must be dominant, i.e., if $$\beta$$ is another root, then $$|\beta|<\alpha$$.

• The question received two correct (and nice) answers -- any reason not to accept one of them? – Peter Mueller Jan 30 at 16:55

Write the equation in the form $$1=x^{-1}+x^{-2}+x^{-3}+x^{-4}+cx^{-5}:=f(x)$$. If $$|\beta|\geqslant \alpha$$, the RHS has absolute value at most $$f(\alpha) =1$$ with equality if and only if $$|\beta|=\alpha$$ and all five summands $$\beta^{-1}$$ etc are positive reals. That is, $$\beta=\alpha$$.
$$f(x)$$ is the characteristic polynomial of its companion matrix $$A = \pmatrix{0 & 0 & 0 & 0 & c\cr 1 & 0 & 0 & 0 & 1\cr 0 & 1 & 0 & 0 & 1\cr 0 & 0 & 1 & 0 & 1\cr 0 & 0 & 0 & 1 & 1\cr}$$ and $$A^5$$ has all entries $$> 0$$. Therefore by the Perron-Frobenius theorem there is a positive eigenvalue of multiplicity $$1$$ strictly greater in absolute value than all other eigenvalues.
This works for all $$c > 0$$.
• Why all the entries of $A^5$ are positive? I didn't see this. – Jeremy Jan 27 at 22:30
• It's easy to prove. Look at the directed hypergraph of $5$ vertices where arcs correspond to positive entries of $A$ (including a loop $5 \to 5$). Entry $(i,j)$ of $A^k$ is positive if you can get from $i$ to $j$ in $k$ steps. You can always do that here: the first step goes to $5$, then stay at $5$ for $j-1$ steps, then decrease by one $5-j$ times. – Robert Israel Jan 27 at 23:27
• Where can I find this fact: 'Entry $(i,j)$ of $A^k$ is positive if you can get from $i$ to $j$ in $k$ steps?" By steps means a path with possibilty of cycles? Thanks! – Jeremy Jan 28 at 20:24
• @Jeremy: That follows directly from the definition of matrix multiplication and the assumption that all entries of $A$ are non-negative. – Peter Mueller Jan 29 at 8:45