Prove that $\det(\lambda I-A)\le\lambda^n-1$ Let $A$ be an $n\times n$ nonnegative matrix with spectral radius $1$. Suppose $\lambda>1$, then prove $\det(\lambda I-A)\le\lambda^n-1$.
 A: There should be a simpler proof, but at least the inequality can be proved by applying Jacobi's formula twice.
Suppose $A\ge0$ (i.e. it is entrywise nonnegative) and $\rho(A)=1$. In the following list, each statement is either equivalent to or weaker than the subsequent one. Therefore, it suffices to prove the last statement in the list.

*

*$\det(xI-A) \le x^n-1$ for all $x>1$.

*$\frac{d}{dx}\det(xI-A) \le \frac{d}{dx}(x^n-1)$ for all $x>1$. This implies statement 1 because $\det(xI-A)=0=x^n-1$ when $x=1$.

*$\operatorname{tr}\big(\operatorname{adj}(xI-A)\big) \le nx^{n-1}$ for all $x>1$. This is equivalent to statement 2, by Jacobi's formula.

*$\det(xI-B) \le x^{n-1}$ for all $x>1$ and all principal $(n-1)\times(n-1)$ submatrices $B$ of $A$.

*$\det(I-tB)\le1$ for all $0<t<1$ and all nonnegative matrices $B$ with $\rho(B)\le1$.

*$\det(I-tB)$ is decreasing on $(0,1)$ when $B\ge0$ and $\rho(B)\le1$. This implies the previous statement because $\det(I-tB)=1$ when $t=0$.

*$\operatorname{tr}\big(\operatorname{adj}(I-tB)\,(-B)\big) \le 0$ for all $0<t<1$ when $B\ge0$ and $\rho(B)\le1$. This is equivalent to statement 6, by Jacobi's formula.

*$\operatorname{tr}\big(\operatorname{adj}(I-tB)\,(I-tB)\big) \le \operatorname{tr}\big(\operatorname{adj}(I-tB)\big)$ for all $0<t<1$ when $B\ge0$ and $\rho(B)\le1$.

*$\det(I-tB) \le \frac1m\operatorname{tr}\big(\operatorname{adj}(I-tB)\big)$ for all $0<t<1$ when $M_m(\mathbb R)\ni B\ge0$ and $\rho(B)\le1$.

*$\det(I-tB) \le \sqrt[m]{\prod_{i=1}^m\big(\operatorname{adj}(I-tB)\big)_{ii}}$ for all $0<t<1$ when $M_m(\mathbb R)\ni B\ge0$ and $\rho(B)\le1$. Note that the conditions on $B$ imply that each principal minor of $I-tB$ is positive. Therefore the inequality here implies statement 9, by the AM-GM inequality.

*$1\le\prod_i\big(\frac{\operatorname{adj}(I-tB)}{\det(I-tB)}\big)_{ii}$ for all $0<t<1$ when $B\ge0$ and $\rho(B)\le1$.

*$1\le\prod_i \big((I-tB)^{-1}\big)_{ii} $ for all $0<t<1$ when $B\ge0$ and $\rho(B)\le1$.

Now statement 12 is true because all diagonal elements of $(I-tB)^{-1}=I+tB+(tB)^2+(tB)^3+\cdots$ are bounded below by $1$.
