Non-Archimedean Brauer Oval (of Cassini) Theorem and Applications

1. Introduction

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$. For $1\le j\le n$, define

$$\begin{array}{c}\hfill r_j\left(A\right)\sum _{k=1,k\ne j}^n\left|a_{j,k}\right|.\end{array}$$

For $1\le k\le n$, define

$$\begin{array}{c}\hfill c_k\left(A\right)\sum _{j=1,j\ne k}^n\left|a_{j,k}\right|.\end{array}$$

Let $\sigma \left(A\right)$ be the set of all eigenvalues of A. In 1931, Gershgorin proved the following breakthrough result known as the Gershgorin circle/disk theorem which uses single row/column for determining the radius of the disk.

Theorem 1.  

[1,2,3,4] (Gershgorin Eigenvalue Inclusion TheoremorGershgorin Disk Theorem) For every $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$,

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j=1}^n\{z\in \mathbb{C}:|z-a_{j,j}|\le r_j\left(A\right)\}\end{array}$$

and

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{k=1}^n\{z\in \mathbb{C}:|z-a_{k,k}|\le c_k\left(A\right)\}.\end{array}$$

A remarkable application of Theorem 1 is the following result of Frobenius (which advances result of Browne).

Theorem 2.  

[5,6,7] Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$. For every $\lambda \in \sigma \left(A\right)$,

$$\begin{array}{c}\hfill \left|\lambda \right|\le min\left\{\underset{1\le j\le n}{max}\sum _{k=1}^n|a_{j,k}|,\underset{1\le k\le n}{max}\sum _{j=1}^n\left|a_{j,k}\right|\right\}\le {\displaystyle \frac{1}{2}}\left(\underset{1\le j\le n}{max}\sum _{k=1}^n|a_{j,k}|+\underset{1\le k\le n}{max}\sum _{j=1}^n\left|a_{j,k}\right|\right).\end{array}$$

In particular,

$$\begin{array}{cc}\hfill \left|\det \right(A\left)\right|& \le min\left\{{\left(\underset{1\le j\le n}{max}\sum _{k=1}^n\left|a_{j,k}\right|\right)}^n,{\left(\underset{1\le k\le n}{max}\sum _{j=1}^n\left|a_{j,k}\right|\right)}^n\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({\left(\underset{1\le j\le n}{max}\sum _{k=1}^n\left|a_{j,k}\right|\right)}^n+{\left(\underset{1\le k\le n}{max}\sum _{j=1}^n\left|a_{j,k}\right|\right)}^n\right).\hfill \end{array}$$

Another remarkable application of Theorem 1 is on the bounds for the zeros of polynomials. Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right]$. A direct observation reveals that the zeros of p are the eigenvalues of the Frobenius companion matrix

$$\begin{array}{c}\hfill C_p\left(\begin{array}{ccccccc}0& 1& 0& \dots & 0& 0& 0\\ 0& 0& 1& \dots & 0& 0& 0\\ 0& 0& 0& \dots & 0& 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮& ⋮\\ 0& 0& 0& \dots & 0& 1& 0\\ 0& 0& 0& \dots & 0& 0& 1\\ -c_0& -c_1& -c_2& \dots & -c_{n-3}& -c_{n-2}& -c_{n-1}\end{array}\right)\in {\mathbb{M}}_n\left(\mathbb{C}\right)\end{array}$$

and the eigenvalues of $C_p$ are the same as the zeros of p[8,9]. In 1965, Bell derived following bounds for the zeros of p using Theorem 1.

Theorem 3.  

[10] Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right]$. If λ is a zero of p, then

$$\begin{array}{c}\hfill \left|\lambda \right|\le 1\end{array}$$

or

$$\begin{array}{c}\hfill |\lambda +c_{n-1}|\le |c_0|+\dots +|c_{n-2}|.\end{array}$$

In particular,

$$\begin{array}{c}\hfill (\mathrm{Lagrange}\mathrm{bound})\left|\lambda \right|\le max\{1,|c_0|+\dots +|c_{n-1}\left|\right\}.\end{array}$$

(1)

Note that Inequality (1) is a generalization of famous Montel bound [2] which says that

$$\begin{array}{c}\hfill \left|\lambda \right|\le 1+|c_0|+\dots +|c_{n-1}|.\end{array}$$

It is interesting to note that one can give a proof of Inequality (1) without using companion matrix [11].

Theorem 4.  

[10] Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right]$. If λ is a zero of p, then

$$\begin{array}{c}\hfill \left|\lambda \right|\le |c_0|\end{array}$$

or

$$\begin{array}{c}\hfill \left|\lambda \right|\le 1+|c_j|,\mathrm{for}\mathrm{some}1\le j\le n-2\end{array}$$

or

$$\begin{array}{c}\hfill |\lambda +c_{n-1}|\le 1.\end{array}$$

In particular,

$$\begin{array}{c}\hfill (\mathrm{Bell}\mathrm{bound})\left|\lambda \right|\le max\left\{\right|c_0|,1+|c_1|,\dots ,1+|c_{n-1}\left|\right\}.\end{array}$$

(2)

Note that Inequality (2) is a generalization of famous Cauchy bound [12] which says that

$$\begin{array}{c}\hfill \left|\lambda \right|\le 1+max\left\{\right|c_0|,\dots ,|c_{n-1}\left|\right\}.\end{array}$$

Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right]$ with $c_0\ne 0$. Define

$$\begin{array}{c}\hfill q\left(z\right){\displaystyle \frac{1}{c_0}}{z}^{n}p\left({\displaystyle \frac{1}{z}}\right)={\displaystyle \frac{1}{c_0}}+{\displaystyle \frac{c_{n-1}}{c_0}}z+\dots +{\displaystyle \frac{c_1}{c_0}}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right].\end{array}$$

We note that $\lambda \in \mathbb{C}\setminus \left\{0\right\}$ satisfies $p\left(\lambda \right)=0$ if and only if $q(1/\lambda )=0$. Frobenius companion matrix of q is

$$\begin{array}{c}\hfill C_q\left(\begin{array}{ccccccc}0& 1& 0& \dots & 0& 0& 0\\ 0& 0& 1& \dots & 0& 0& 0\\ 0& 0& 0& \dots & 0& 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮& ⋮\\ 0& 0& 0& \dots & 0& 1& 0\\ 0& 0& 0& \dots & 0& 0& 1\\ {\displaystyle \frac{-1}{c_0}}& {\displaystyle \frac{-c_{n-1}}{c_0}}& {\displaystyle \frac{-c_{n-2}}{c_0}}& \dots & {\displaystyle \frac{-c_3}{c_0}}& {\displaystyle \frac{-c_2}{c_0}}& {\displaystyle \frac{-c_1}{c_0}}\end{array}\right)\in {\mathbb{M}}_n\left(\mathbb{C}\right).\end{array}$$

Now by applying earlier two results to the polynomial q and rearranging, we get following results.

Theorem 5.  

[2,10] Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right]$ with $c_0\ne 0$. If λ is a zero of p, then

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le 1\end{array}$$

or

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{c_1}{c_0}}\right|\le {\displaystyle \frac{1}{|c_0|}}+{\displaystyle \frac{|c_2|}{|c_0|}}+\dots +{\displaystyle \frac{|c_{n-1}|}{|c_0|}}.\end{array}$$

In particular,

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le max\left\{1,{\displaystyle \frac{1}{|c_0|}}+{\displaystyle \frac{|c_1|}{|c_0|}}+{\displaystyle \frac{|c_2|}{|c_0|}}+\dots +{\displaystyle \frac{|c_{n-1}|}{|c_0|}}\right\}\le 1+{\displaystyle \frac{1}{|c_0|}}+{\displaystyle \frac{|c_1|}{|c_0|}}+\dots +{\displaystyle \frac{|c_{n-1}|}{|c_0|}}\end{array}$$

and

$$\begin{array}{cc}\hfill (\mathrm{Lagrange}\mathrm{lower}\mathrm{bound})\left|\lambda \right|& \ge {\displaystyle \frac{|c_0|}{max\left\{\right|c_0|,1+|c_1|+\dots +|c_{n-1}\left|\right\}}}\hfill \\ \hfill (\mathrm{Montel}\mathrm{lower}\mathrm{bound})& \ge {\displaystyle \frac{|c_0|}{1+|c_0|+|c_1|+\dots +|c_{n-1}|}}.\hfill \end{array}$$

Theorem 6.  

[2,10] Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right]$ with $c_0\ne 0$. If λ is a zero of p, then

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le {\displaystyle \frac{1}{|c_0|}}\end{array}$$

or

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le 1+{\displaystyle \frac{|c_j|}{|c_0|}},\mathrm{for}\mathrm{some}2\le j\le n-1\end{array}$$

or

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{c_1}{c_0}}\right|\le 1.\end{array}$$

In particular,

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le max\left\{{\displaystyle \frac{1}{|c_0|}},1+{\displaystyle \frac{|c_1|}{|c_0|}},\dots ,1+{\displaystyle \frac{|c_{n-1}|}{|c_0|}}\right\}\le 1+max\left\{{\displaystyle \frac{1}{|c_0|}},{\displaystyle \frac{|c_1|}{|c_0|}},\dots ,{\displaystyle \frac{|c_{n-1}|}{|c_0|}}\right\}\end{array}$$

and

$$\begin{array}{cc}\hfill (\mathrm{Bell}\mathrm{lower}\mathrm{bound})\left|\lambda \right|& \ge {\displaystyle \frac{|c_0|}{max\{1,|c_0|+|c_1|,\dots ,|c_0|+|c_{n-1}\left|\right\}}}\hfill \\ \hfill (\mathrm{Cauchy}\mathrm{lower}\mathrm{bound})& \ge {\displaystyle \frac{|c_0|}{|c_0|+max\{1,|a_1|,\dots ,|a_{n-1}\left|\right\}}}.\hfill \end{array}$$

A result much older than Gershgorin is the following.

Theorem 7.  

[2,3,13,14] (Strict Diagonal Dominance TheoremorLevy-Desplanques Theorem) If $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$ satisfies

$$\begin{array}{c}\hfill |a_{j,j}|>r_j\left(A\right),\forall 1\le j\le n\end{array}$$

or

$$\begin{array}{c}\hfill |a_{k,k}|>c_k\left(A\right),\forall 1\le k\le n,\end{array}$$

then A is invertible.

Following results say that Theorems 1 and 7 are equivalent.

Theorem 8.  

[3] Let $n\in \mathbb{N}$. Following two statements are equivalent.

(i)

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$. Then

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j=1}^n\{z\in \mathbb{C}:|z-a_{j,j}|\le r_j\left(A\right)\}.\end{array}$$

(ii)

If $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{j,j}|>r_j\left(B\right),\forall 1\le j\le n,\end{array}$$

then B is invertible.

Theorem 9.  

[3] Let $n\in \mathbb{N}$. Following two statements are equivalent.

(i)

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$. Then

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{k=1}^n\{z\in \mathbb{C}:|z-a_{k,k}|\le c_k\left(A\right)\}.\end{array}$$

(ii)

If $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{k,k}|>c_k\left(B\right),\forall 1\le k\le n,\end{array}$$

then B is invertible.

In 1947, Brauer derived following generalization of Theorem 1, known as the Brauer oval (of Cassini) theorem which uses two rows/columns for determining the radius of the oval.

Theorem 10.  

[3,6,15] (Brauer Eigenvalue Inclusion TheoremorBrauer Oval (of Cassini) Theorem) For every $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$,

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{C}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le r_j\left(A\right)r_k\left(A\right)\}\end{array}$$

and

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{C}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le c_j\left(A\right)c_k\left(A\right)\}.\end{array}$$

By applying Brauer’s theorem, we get following results.

Theorem 11.  

Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right]$. If λ is a zero of p, then

$$\begin{array}{c}\hfill \left|\lambda \right|\le 1\end{array}$$

or

$$\begin{array}{c}\hfill \left|\lambda \right||\lambda +c_{n-1}|\le |c_0|+\dots +|c_{n-2}|.\end{array}$$

Theorem 12.  

Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{C}\left[z\right]$. If λ is a zero of p, then

$$\begin{array}{c}\hfill {\left|\lambda \right|}^2\le |c_0\left|\right(1+|c_j\left|\right),\mathrm{for}\mathrm{some}1\le j\le n-2\end{array}$$

or

$$\begin{array}{c}\hfill \left|\lambda \right||\lambda +c_{n-1}|\le |c_0|\end{array}$$

or

$$\begin{array}{c}\hfill {\left|\lambda \right|}^2\le (1+|c_j\left|\right)(1+|c_k\left|\right),\mathrm{for}\mathrm{some}1\le j,k\le n-2,j\ne k\end{array}$$

or

$$\begin{array}{c}\hfill \left|\lambda \right||\lambda +c_{n-1}|\le 1+|c_j|,\mathrm{for}\mathrm{some}1\le j\le n-2.\end{array}$$

It is known that Brauer theorem cannot be extended by considering three rows/columns [3]. In 1937, Ostrowski derived following generalization of Theorem 7.

Theorem 13.  

[3,16] (Ostrowski Nonsingularity Theorem) If $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$ satisfies

$$\begin{array}{c}\hfill |a_{j,j}\left|\right|a_{k,k}|>r_j\left(A\right)r_k\left(A\right),\forall 1\le j,k\le n,j\ne k\end{array}$$

or

$$\begin{array}{c}\hfill |a_{j,j}\left|\right|a_{k,k}|>c_j\left(A\right)c_k\left(A\right),\forall 1\le j,k\le n,j\ne k,\end{array}$$

then A is invertible.

It is known that Theorems 10 and 13 are again equivalent.

Theorem 14.  

[3] Let $n\in \mathbb{N}$. Following two statements are equivalent.

(i)

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$. Then

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{C}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le r_j\left(A\right)r_k\left(A\right)\}.\end{array}$$

(ii)

If $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{j,j}\left|\right|b_{k,k}|>r_j\left(B\right)r_k\left(B\right),\forall 1\le j,k\le n,j\ne k,\end{array}$$

then B is invertible.

Theorem 15.  

[3] Let $n\in \mathbb{N}$. Following two statements are equivalent.

(i)

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$. Then

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{C}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le c_j\left(A\right)c_k\left(A\right)\}.\end{array}$$

(ii)

If $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{C}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{j,j}\left|\right|b_{k,k}|>c_j\left(B\right)c_k\left(B\right),\forall 1\le j,k\le n,j\ne k,\end{array}$$

then B is invertible.

It is natural to ask what are versions of Theorems 1, 2, 7, 10 and 13 for matrices over non-Archimedean fields? In the paper, $\mathbb{K}$ denotes a non-Archimedean valued field with valuation $|·|$. Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$. For $1\le j\le n$, define

$$\begin{array}{c}\hfill h_j\left(A\right)\underset{1\le k\le n,k\ne j}{max}\left|a_{j,k}\right|.\end{array}$$

For $1\le k\le n$, define

$$\begin{array}{c}\hfill v_k\left(A\right)\underset{1\le j\le n,j\ne k}{max}\left|a_{j,k}\right|.\end{array}$$

In 2023, Nica and Sprague derived the following non-Archimedean analogue of Theorems 1 and 7.

Theorem 16.  

[17] (Non-Archimedean Gershgorin Eigenvalue Inclusion TheoremorNica-Sprague Disk Theorem) For every $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$,

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j=1}^n\{z\in \mathbb{K}:|z-a_{j,j}|\le h_j\left(A\right)\}\end{array}$$

and

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{k=1}^n\{z\in \mathbb{K}:|z-a_{k,k}|\le v_k\left(A\right)\}.\end{array}$$

Theorem 17.  

[17] (Non-Archimedean Strict Diagonal Dominance TheoremorNica-Sprague Nonsingularity Theorem) If $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$ satisfies

$$\begin{array}{c}\hfill |a_{j,j}|>h_j\left(A\right),\forall 1\le j\le n\end{array}$$

or

$$\begin{array}{c}\hfill |a_{k,k}|>v_k\left(A\right),\forall 1\le k\le n,\end{array}$$

then A is invertible.

Nica and Sprague showed that Theorems 16 and 17 are equivalent.

Theorem 18.  

[17] Let $n\in \mathbb{N}$. Following two statements are equivalent.

(i)

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$. Then

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j=1}^n\{z\in \mathbb{K}:|z-a_{j,j}|\le h_j\left(A\right)\}.\end{array}$$

(ii)

If $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{j,j}|>h_j\left(B\right),\forall 1\le j\le n,\end{array}$$

then B is invertible.

Theorem 19.  

[17] Let $n\in \mathbb{N}$. Following two statements are equivalent.

(i)

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$. Then

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{k=1}^n\{z\in \mathbb{K}:|z-a_{k,k}|\le v_k\left(A\right)\}.\end{array}$$

(ii)

If $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{k,k}|>v_k\left(B\right),\forall 1\le k\le n,\end{array}$$

then B is invertible.

Non-Archimedean version of Theorem 2 reads as follows.

Theorem 20.  

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$. For every $\lambda \in \sigma \left(A\right)$,

$$\begin{array}{c}\hfill \left|\lambda \right|\le min\left\{\underset{1\le j\le n}{max}\underset{1\le k\le n}{max}|a_{j,k}|,\underset{1\le k\le n}{max}\underset{1\le j\le n}{max}\left|a_{j,k}\right|\right\}\le {\displaystyle \frac{1}{2}}\left(\underset{1\le j\le n}{max}\underset{1\le k\le n}{max}|a_{j,k}|+\underset{1\le k\le n}{max}\underset{1\le j\le n}{max}\left|a_{j,k}\right|\right).\end{array}$$

In particular,

$$\begin{array}{cc}\hfill \left|\det \right(A\left)\right|& \le min\left\{{\left(\underset{1\le j\le n}{max}\underset{1\le k\le n}{max}\left|a_{j,k}\right|\right)}^n,{\left(\underset{1\le k\le n}{max}\underset{1\le j\le n}{max}\left|a_{j,k}\right|\right)}^n\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({\left(\underset{1\le j\le n}{max}\underset{1\le k\le n}{max}\left|a_{j,k}\right|\right)}^n+{\left(\underset{1\le k\le n}{max}\underset{1\le j\le n}{max}\left|a_{j,k}\right|\right)}^n\right).\hfill \end{array}$$

Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$. Let

$$\begin{array}{c}\hfill C_p\left(\begin{array}{ccccccc}0& 1& 0& \dots & 0& 0& 0\\ 0& 0& 1& \dots & 0& 0& 0\\ 0& 0& 0& \dots & 0& 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮& ⋮\\ 0& 0& 0& \dots & 0& 1& 0\\ 0& 0& 0& \dots & 0& 0& 1\\ -c_0& -c_1& -c_2& \dots & -c_{n-3}& -c_{n-2}& -c_{n-1}\end{array}\right)\in {\mathbb{M}}_n\left(\mathbb{K}\right)\end{array}$$

be the companion matrix of p. By applying Theorem 16 Nica and Sprague obtained following results.

Theorem 21.  

[17] Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$. If λ is a zero of p, then

$$\begin{array}{c}\hfill \left|\lambda \right|\le 1\end{array}$$

or

$$\begin{array}{c}\hfill |\lambda +c_{n-1}|\le max\{|c_0|,|c_1|,\dots ,|c_{n-2}\left|\right\}.\end{array}$$

In particular,

$$\begin{array}{c}\hfill (\mathrm{Nica}-\mathrm{Sprague}\mathrm{bound})\left|\lambda \right|\le max\{1,|c_0|,|c_1|,\dots ,|c_{n-1}\left|\right\}.\end{array}$$

Theorem 22.  

[17] Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$. If λ is a zero of p, then

$$\begin{array}{c}\hfill \left|\lambda \right|\le |c_o|\end{array}$$

or

$$\begin{array}{c}\hfill \left|\lambda \right|\le max\{1,|c_j\left|\right\},\mathrm{for}\mathrm{some}1\le j\le n-2\end{array}$$

or

$$\begin{array}{c}\hfill |\lambda +c_{n-1}|\le 1.\end{array}$$

In particular,

$$\begin{array}{c}\hfill (\mathrm{Nica}-\mathrm{Sprague}\mathrm{bound})\left|\lambda \right|\le max\{1,|c_0|,|c_1|,\dots ,|c_{n-1}\left|\right\}.\end{array}$$

(3)

Theorem 23.  

[17] Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$ with $c_0\ne 0$. If λ is a zero of p, then

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le 1\end{array}$$

or

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{c_1}{c_0}}\right|\le max\left\{{\displaystyle \frac{1}{|c_0|}},{\displaystyle \frac{|c_2|}{|c_0|}},\dots ,{\displaystyle \frac{|c_{n-1}|}{|c_0|}}\right\}.\end{array}$$

In particular,

$$\begin{array}{c}\hfill (\mathrm{Nica}-\mathrm{Sprague}\mathrm{lower}\mathrm{bound})\left|\lambda \right|\ge {\displaystyle \frac{|c_0|}{max\{1,|c_0|,|c_1|,\dots ,|c_{n-1}\left|\right\}}}.\end{array}$$

Theorem 24.  

[17] Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$ with $c_0\ne 0$. If λ is a zero of p, then

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le {\displaystyle \frac{1}{|c_0|}}\end{array}$$

or

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le max\left\{1,{\displaystyle \frac{|c_j|}{|c_0|}}\right\},\mathrm{for}\mathrm{some}2\le j\le n-1\end{array}$$

or

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{c_1}{c_0}}\right|\le 1.\end{array}$$

In particular,

$$\begin{array}{c}\hfill (\mathrm{Nica}-\mathrm{Sprague}\mathrm{lower}\mathrm{bound})\left|\lambda \right|\ge {\displaystyle \frac{|c_0|}{max\{1,|c_0|,|c_1|,\dots ,|c_{n-1}\left|\right\}}}.\end{array}$$

Before passing, we give a different and direct proof of Inequality (3). Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$ and $\lambda$ be a zero of p. If $\left|\lambda \right|\le 1$, then clearly we have Inequality (3). So we assume that $\left|\lambda \right|>1$. Since $p\left(\lambda \right)=0$, we have

$$\begin{array}{c}\hfill \left({\displaystyle \frac{c_0}{{\lambda }^n}}+{\displaystyle \frac{c_1}{{\lambda }^{n-1}}}+\dots +{\displaystyle \frac{c_{n-1}}{\lambda }}+1\right){\lambda }^n=0.\end{array}$$

Since $\lambda \ne 0$,

$$\begin{array}{c}\hfill {\displaystyle \frac{c_0}{{\lambda }^n}}+{\displaystyle \frac{c_1}{{\lambda }^{n-1}}}+\dots +{\displaystyle \frac{c_{n-1}}{\lambda }}+1=0.\end{array}$$

Rearranging,

$$\begin{array}{c}\hfill -1={\displaystyle \frac{c_0}{{\lambda }^n}}+{\displaystyle \frac{c_1}{{\lambda }^{n-1}}}+\dots +{\displaystyle \frac{c_{n-1}}{\lambda }}.\end{array}$$

By taking absolute value and noticing $\left|\lambda \right|>1$, we get

$$\begin{array}{cc}\hfill 1& \le \left|{\displaystyle \frac{c_0}{{\lambda }^n}}+{\displaystyle \frac{c_1}{{\lambda }^{n-1}}}+\dots +{\displaystyle \frac{c_{n-1}}{\lambda }}\right|\le max\left\{{\displaystyle \frac{|c_0|}{{\left|\lambda \right|}^n}},{\displaystyle \frac{|c_1|}{{\left|\lambda \right|}^{n-1}}},\dots ,{\displaystyle \frac{|c_{n-1}|}{\left|\lambda \right|}}\right\}\hfill \\ \hfill & \le max\left\{{\displaystyle \frac{|c_0|}{\left|\lambda \right|}},{\displaystyle \frac{|c_1|}{\left|\lambda \right|}},\dots ,{\displaystyle \frac{|c_{n-1}|}{\left|\lambda \right|}}\right\}={\displaystyle \frac{1}{\left|\lambda \right|}}max\left\{\right|c_0|,|c_1|,\dots ,|c_{n-1}\left|\right\}.\hfill \end{array}$$

Rearranging above inequality completes the argument.

In 2025, Li and Li derived the following non-Archimedean analogue of Theorem 13.

Theorem 25.  

[18] (Non-Archimedean Ostrowski Nonsingularity TheoremorLi-Li Nonsingularity Theorem) If $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$ satisfies

$$\begin{array}{c}\hfill |a_{j,j}\left|\right|a_{k,k}|>h_j\left(A\right)h_k\left(A\right),\forall 1\le j,k\le n,j\ne k\end{array}$$

or

$$\begin{array}{c}\hfill |a_{j,j}\left|\right|a_{k,k}|>v_j\left(A\right)v_k\left(A\right),\forall 1\le j,k\le n,j\ne k,\end{array}$$

then A is invertible.

In this article, we derive non-Archimedean version of Theorem 10. We also show that our result is equivalent to Theorem 25. We give applications for bounding the zeros of polynomials over non-Archimedean fields.

2. Non-Archimedean Brauer Oval (of Cassini) Theorem

We start with non-Archimedean Brauer eigenvalue inclusion theorem. Our proof is motivated from the proof of Brauer [15].

Theorem 26.(Non-Archimedean Brauer Oval (of Cassini) Theorem)

For every $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$,

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{K}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le h_j\left(A\right)h_k\left(A\right)\}\end{array}$$

and

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{K}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le v_j\left(A\right)v_k\left(A\right)\}.\end{array}$$

Proof. 

Let $\lambda \in \sigma \left(A\right)$. Then there exists a $0\ne \mathbf{x}={\left(x_j\right)}_{j=1}^n\in {\mathbb{K}}^n$ such that

$$\begin{array}{c}\hfill \lambda \mathbf{x}=A\mathbf{x}.\end{array}$$

(4)

Choose $1\le j\le n$ such that

$$\begin{array}{c}\hfill |x_j|=\underset{1\le l\le n}{max}\left|x_l\right|.\end{array}$$

Now choose $1\le k\le n$ with $k\ne j$ such that

$$\begin{array}{c}\hfill |x_k|=\underset{1\le l\le n,l\ne j}{max}\left|x_l\right|.\end{array}$$

Then we have $1\le j,k\le n$ with $j\ne k$ and

$$\begin{array}{c}\hfill |x_j|\ge |x_k|\ge \underset{1\le l\le n,l\ne j,l\ne k}{max}\left|x_l\right|.\end{array}$$

(5)

We have two cases. Case (i): $|x_k|=0$. Considering the j-th coordinate in Equation (4) gives

$$\begin{array}{c}\hfill \lambda x_j=\sum _{p=1}^n{a}_{j,p}{x}_p.\end{array}$$

Rewriting previous equation gives

$$\begin{array}{c}\hfill (\lambda -a_{j,j})x_j=\sum _{p=1,p\ne j}^n{a}_{j,p}{x}_p.\end{array}$$

Therefore using (5) we get

$$\begin{array}{cc}\hfill |(\lambda -a_{j,j})x_j|& =\left|\sum _{p=1,p\ne j}^n{a}_{j,p}{x}_p\right|\le \underset{1\le p\le n,p\ne j}{max}\left|a_{j,p}{x}_p\right|\hfill \\ \hfill & \le \left(\underset{1\le p\le n,p\ne j}{max}\left|a_{j,p}\right|\right)\left(\underset{1\le p\le n,p\ne j}{max}\left|x_p\right|\right)=\left(\underset{1\le p\le n,p\ne j}{max}\left|a_{j,p}\right|\right)\left|x_k\right|\hfill \\ \hfill & =0.\hfill \end{array}$$

Since $|x_j|\ge |x_k|$ for all $1\le k\le n$, $|x_k|=0$ and $\mathbf{x}\ne 0$, we must have $|x_j|\ne 0$. Previous inequality then gives $|\lambda -a_{j,j}|=0$. So

$$\begin{array}{c}\hfill \lambda =a_{j,j}\in \bigcup _{p,q=1,p\ne q}^n\{z\in \mathbb{K}:|z-a_{p,p}\left|\right|z-a_{q,q}|\le h_p\left(A\right)h_q\left(A\right)\}.\end{array}$$

Case (ii): $|x_k|>0$. Considering j-th and k-th coordinates in Equation (4) give

$$\begin{array}{c}\hfill (\lambda -a_{j,j})x_j=\sum _{p=1,p\ne j}^n{a}_{j,p}{x}_p\end{array}$$

(6)

and

$$\begin{array}{c}\hfill (\lambda -a_{k,k})x_k=\sum _{q=1,q\ne k}^n{a}_{k,q}{x}_q.\end{array}$$

(7)

Multiplying Equations (6) and (7) and taking non-Archimedean valuation gives

$$\begin{array}{cc}\hfill |(\lambda -a_{j,j})x_j(\lambda -a_{k,k})x_k|& =\left|\sum _{p=1,p\ne j}^n{a}_{j,p}{x}_p\right|\left|\sum _{q=1,q\ne k}^n{a}_{k,q}{x}_q\right|\hfill \\ \hfill & \le \left(\underset{1\le p\le n,p\ne j}{max}\left|a_{j,p}{x}_p\right|\right)\left(\underset{1\le q\le n,q\ne k}{max}\left|a_{k,p}{x}_q\right|\right)\hfill \\ \hfill & \le \left(\underset{1\le p\le n,p\ne j}{max}\left|a_{j,p}\right|\right)\left(\underset{1\le p\le n,p\ne j}{max}\left|x_p\right|\right)\left(\underset{1\le q\le n,q\ne k}{max}\left|a_{k,p}\right|\right)\left(\underset{1\le q\le n,q\ne k}{max}\left|x_q\right|\right)\hfill \\ \hfill & =\left(\underset{1\le p\le n,p\ne j}{max}\left|a_{j,p}\right|\right)|x_k|\left(\underset{1\le q\le n,q\ne k}{max}\left|a_{k,p}\right|\right)\left|x_j\right|\hfill \\ \hfill & =h_j\left(A\right)|x_k|h_k\left(A\right)\left|x_j\right|.\hfill \end{array}$$

Therefore

$$\begin{array}{c}\hfill |(\lambda -a_{j,j})(\lambda -a_{k,k})\left|\right|x_j{x}_k|\le h_j\left(A\right)h_k\left(A\right)|x_j\left|\right|x_k|.\end{array}$$

Since $|x_j{x}_k|\ne 0$, we have

$$\begin{array}{c}\hfill |(\lambda -a_{j,j})(\lambda -a_{k,k})|\le h_j\left(A\right)h_k\left(A\right).\end{array}$$

Previous inequality says that

$$\begin{array}{c}\hfill \lambda \in \bigcup _{p,q=1,p\ne q}^n\{z\in \mathbb{K}:|z-a_{p,p}\left|\right|z-a_{q,q}|\le h_p\left(A\right)h_q\left(A\right)\}.\end{array}$$

Second inclusion in the statement follows by considering the transpose of A and noting that the spectrum of a matrix and its transpose are equal. □

By applying Theorem 26, we get following results.

Theorem 27.  

Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$. If λ is a zero of p, then

$$\begin{array}{c}\hfill \left|\lambda \right|\le 1\end{array}$$

or

$$\begin{array}{c}\hfill \left|\lambda \right||\lambda +c_{n-1}|\le max\{|c_0|,\dots ,|c_{n-2}\left|\right\}.\end{array}$$

Theorem 28.  

Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$. If λ is a zero of p, then

$$\begin{array}{c}\hfill {\left|\lambda \right|}^2\le |c_0|max\{1,|c_j\left|\right\},\mathrm{for}\mathrm{some}1\le j\le n-2\end{array}$$

or

$$\begin{array}{c}\hfill \left|\lambda \right||\lambda +c_{n-1}|\le |c_0|\end{array}$$

or

$$\begin{array}{c}\hfill {\left|\lambda \right|}^2\le \left(max\{1,|c_j\left|\right\}\right)\left(max\{1,|c_k\left|\right\}\right),\mathrm{for}\mathrm{some}1\le j,k\le n-2,j\ne k\end{array}$$

or

$$\begin{array}{c}\hfill \left|\lambda \right||\lambda +c_{n-1}|\le max\{1,|c_j\left|\right\},\mathrm{for}\mathrm{some}1\le j\le n-2.\end{array}$$

Theorem 29.  

Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$ with $c_0\ne 0$. If λ is a zero of p, then

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\le 1\end{array}$$

or

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\left|{\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{c_1}{c_0}}\right|\le max\left\{{\displaystyle \frac{1}{|c_0|}},{\displaystyle \frac{|c_2|}{|c_0|}},\dots ,{\displaystyle \frac{|c_{n-1}|}{|c_0|}}\right\}.\end{array}$$

Theorem 30.  

Let $p\left(z\right)c_0+c_{1}z+\dots +c_{n-1}{z}^{n-1}+z^n\in \mathbb{K}\left[z\right]$ with $c_0\ne 0$. If λ is a zero of p, then

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{\left|\lambda \right|}^2}}\le {\displaystyle \frac{1}{|c_0|}}max\left\{1,{\displaystyle \frac{|c_j|}{|c_0|}}\right\},\mathrm{for}\mathrm{some}2\le j\le n-1\end{array}$$

or

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\left|{\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{c_1}{c_0}}\right|\le {\displaystyle \frac{1}{|c_0|}}\end{array}$$

or

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{\left|\lambda \right|}^2}}\le \left(max\left\{1,{\displaystyle \frac{|c_j|}{|c_0|}}\right\}\right)\left(max\left\{1,{\displaystyle \frac{|c_k|}{|c_0|}}\right\}\right),\mathrm{for}\mathrm{some}2\le j,k\le n-1,j\ne k\end{array}$$

or

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\left|\lambda \right|}}\left|{\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{c_1}{c_0}}\right|\le max\left\{1,{\displaystyle \frac{|c_j|}{|c_0|}}\right\},\mathrm{for}\mathrm{some}2\le j\le n-1.\end{array}$$

Like the complex case, Theorem 26 cannot be extended by considering three rows/columns. An example given for the scalar case in [3] (also see [19,20]) extends to non-Archimedean case. Consider the following matrix over any non-Archimedean field $\mathbb{K}$.

$$\begin{array}{c}\hfill A\left(\begin{array}{cccc}1& 1& 0& 0\\ 1& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).\end{array}$$

Then $\sigma \left(A\right)=\{0,1,1,2\}$ and $h_1\left(A\right)=1$, $h_2\left(A\right)=1$, $h_3\left(A\right)=0$, $h_4\left(A\right)=0$. Hence

$$\begin{array}{c}\hfill \sigma \left(A\right)⊈\bigcup _{j,k,l=1,j\ne k,j\ne l,k\ne l}^4\{z\in \mathbb{K}:|z-a_{j,j}\left|\right|z-a_{k,k}\left|\right|z-a_{l,l}|\le h_j\left(A\right)h_k\left(A\right)h_l\left(A\right)\}=\left\{0\right\}.\end{array}$$

Next we show that Theorem 26 improves Theorem 16. Our proof is motivated from the complex case, given in [3].

Theorem 31.  

For every $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$,

$$\begin{array}{c}\hfill \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{K}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le h_j\left(A\right)h_k\left(A\right)\}\subseteq \bigcup _{j=1}^n\{z\in \mathbb{K}:|z-a_{j,j}|\le h_j\left(A\right)\}\end{array}$$

and

$$\begin{array}{c}\hfill \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{K}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le v_j\left(A\right)v_k\left(A\right)\}\subseteq \bigcup _{j=1}^n\{z\in \mathbb{K}:|z-a_{j,j}|\le v_j\left(A\right)\}.\end{array}$$

Proof. 

We prove the first inclusion, proof of second inclusion is similar. Set

$$\begin{array}{c}\hfill Z\bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{K}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le h_j\left(A\right)h_k\left(A\right)\}\end{array}$$

and let $z\in Z$. Then there exist $1\le j,k\le n$ with $j\ne k$ such that

$$\begin{array}{c}\hfill |z-a_{j,j}\left|\right|z-a_{k,k}|\le h_j\left(A\right)h_k\left(A\right).\end{array}$$

We have to consider two cases. Case (i): $h_j\left(A\right)h_k\left(A\right)=0$. Then $z=a_{j,j}$ or $z=a_{k,k}$. Now clearly we have

$$\begin{array}{c}\hfill z\in \{x\in \mathbb{K}:|x-a_{j,j}|\le h_j\left(A\right)\}\cup \{y\in \mathbb{K}:|y-a_{k,k}|\le h_k\left(A\right)\}\subseteq \bigcup _{l=1}^n\{z\in \mathbb{K}:|z-a_{l,l}|\le v_l\left(A\right)\}.\end{array}$$

Case(ii): $h_j\left(A\right)h_k\left(A\right)>0$. Then

$$\begin{array}{c}\hfill |z-a_{j,j}|\le h_j\left(A\right)\end{array}$$

or

$$\begin{array}{c}\hfill |z-a_{k,k}|\le h_k\left(A\right).\end{array}$$

Now clearly we have

$$\begin{array}{c}\hfill z\in \{x\in \mathbb{K}:|x-a_{j,j}|\le h_j\left(A\right)\}\cup \{y\in \mathbb{K}:|y-a_{k,k}|\le h_k\left(A\right)\}\subseteq \bigcup _{l=1}^n\{z\in \mathbb{K}:|z-a_{l,l}|\le v_l\left(A\right)\}.\end{array}$$

□

Now we show that Theorems 25 and 26 are equivalent.

Theorem 32.  

Let $n\in \mathbb{N}$. Following two statements are equivalent.

(i)

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$. Then

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{K}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le h_j\left(A\right)h_k\left(A\right)\}.\end{array}$$

(ii)

If $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{j,j}\left|\right|b_{k,k}|>h_j\left(B\right)h_k\left(B\right),\forall 1\le j,k\le n,j\ne k,\end{array}$$

then B is invertible.

Proof. 

(i) 

⇒ (ii) Let $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{j,j}\left|\right|b_{k,k}|>h_j\left(B\right)h_k\left(B\right),\forall 1\le j,k\le n,j\ne k.\end{array}$$

(8)

We need to show that B is invertible. Let us assume that B is not invertible. Then $0\in \sigma \left(B\right)$. By assumption (i), there exist $1\le j,k\le n$ with $j\ne k$ such that

$$\begin{array}{c}\hfill |0-b_{j,j}\left|\right|0-b_{k,k}|\le h_j\left(B\right)h_k\left(B\right).\end{array}$$

(9)

Inequalities (8) and (9) contradict each other. Hence B is invertible.

(ii) 

⇒ (i) Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$ and $\lambda \in \sigma \left(A\right)$. We claim that

$$\begin{array}{c}\hfill |\lambda -a_{j,j}\left|\right|\lambda -a_{k,k}|\le h_j\left(A\right)h_k\left(A\right),\mathrm{for}\mathrm{some}1\le j,k\le n,j\ne k.\end{array}$$

Let us suppose that claim fails. Then

$$\begin{array}{c}\hfill |\lambda -a_{j,j}\left|\right|\lambda -a_{k,k}|>h_j\left(A\right)h_k\left(A\right),\forall 1\le j,k\le n,j\ne k.\end{array}$$

(10)

Let $I_n$ be the identity matrix in ${\mathbb{M}}_n\left(\mathbb{K}\right)$. Define $B\lambda I_n-A=:{\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}$. Then $0\in \sigma \left(B\right)$, hence B is not invertible. Note that $h_j\left(A\right)=h_j\left(B\right)$ for all $1\le j\le n$. But we also have from (10)

$$\begin{array}{c}\hfill |b_{j,j}\left|\right|b_{k,k}|>h_j\left(B\right)h_k\left(B\right),\forall 1\le j,k\le n,j\ne k.\end{array}$$

Assumption (ii) says that B is invertible which is not possible. Hence claim holds.

□

By considering the transpose of a matrix, we easily get following result.

Theorem 33.  

Let $n\in \mathbb{N}$. Following two statements are equivalent.

(i)

Let $A={\left[a_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$. Then

$$\begin{array}{c}\hfill \sigma \left(A\right)\subseteq \bigcup _{j,k=1,j\ne k}^n\{z\in \mathbb{K}:|z-a_{j,j}\left|\right|z-a_{k,k}|\le v_j\left(A\right)v_k\left(A\right)\}.\end{array}$$

(ii)

If $B={\left[b_{j,k}\right]}_{1\le j\le n,1\le k\le n}\in {\mathbb{M}}_n\left(\mathbb{K}\right)$ satisfies

$$\begin{array}{c}\hfill |b_{j,j}\left|\right|b_{k,k}|>v_j\left(B\right)v_k\left(B\right),\forall 1\le j,k\le n,j\ne k,\end{array}$$

then B is invertible.