Ostrowski theorem

Let $A$ be a complex $n\times n$ matrix, $R_i={\sum }_{j\ne i}\left|a_{ij}\right|,C_j={\sum }_{i\ne j}\left|a_{ij}\right|\mathit{\hspace{1em}}1\le i\le n,1\le j\le n$. Let’s consider, for any $\alpha \in (0,1)$, the circles of this kind: $O_i=\{z\in 𝐂:\left|z-a_{ii}\right|\le R_i^{\alpha }{C}_i^{1-\alpha }\}\mathit{\hspace{1em}}1\le i\le n$.

Theorem (A. Ostrowski): For any $\alpha \in (0,1)$, all the eigenvalues of $A$ lie in the union of these $n$ circles:$\sigma (A)\subseteq {\bigcup }_i{O}_i$.

Proof.

If $R_i=0$, the theorem says $a_{ii}$ is an eigenvalue, whichis obviously true. Let’s then concentrate on the $R_i\ne 0$. By eigenvalue definition, we have:

$$(\lambda -a_{ii})x_i=\sum _{j\ne i}{a}_{ij}{x}_j$$

so that, recalling Hölder’s inequality with $p=1/\alpha$ and $q=1/(1-\alpha )$ (to have $p,q>1$, we must have $\alpha \in (0,1)$)

$\left|\lambda -a_{ii}\right|\left|x_i\right|$$\le$${\displaystyle \sum _{j\ne i}}\left|a_{ij}\right|\left|x_j\right|$

$=$${\displaystyle \sum _{j\ne i}}{\left|a_{ij}\right|}^{\alpha }{\left|a_{ij}\right|}^{1-\alpha }\left|x_j\right|$

$\le$${\left({\displaystyle \sum _{j\ne i}}{({\left|a_{ij}\right|}^{\alpha })}^{1/\alpha }\right)}^{\alpha }{\left({\displaystyle \sum _{j\ne i}}{({\left|a_{ij}\right|}^{1-\alpha }\left|x_j\right|)}^{1/(1-\alpha )}\right)}^{1-\alpha }$

$=$${\left({\displaystyle \sum _{j\ne i}}\left|a_{ij}\right|\right)}^{\alpha }{\left({\displaystyle \sum _{j\ne i}}\left|a_{ij}\right|{\left|x_j\right|}^{1/(1-\alpha )}\right)}^{1-\alpha }$

$=$$R_i^{\alpha }{\left({\displaystyle \sum _{j\ne i}}\left|a_{ij}\right|{\left|x_j\right|}^{1/(1-\alpha )}\right)}^{1-\alpha }$

which means

$$\frac{{\left|\lambda -a_{ii}\right|}^{1/(1-\alpha )}}{R_i^{\alpha /(1-\alpha )}}{\left|x_i\right|}^{1/(1-\alpha )}\le \sum _{j\ne i}\left|a_{ij}\right|{\left|x_j\right|}^{1/(1-\alpha )}$$

Summing over all $i$, one obtains

$$\sum _{i=1}^n\frac{{\left|\lambda -a_{ii}\right|}^{1/(1-\alpha )}}{R_i^{\alpha /(1-\alpha )}}{\left|x_i\right|}^{1/(1-\alpha )}\le \sum _{i=1}^n\sum _{j\ne i}\left|a_{ij}\right|{\left|x_j\right|}^{1/(1-\alpha )}=\sum _{j=1}^n{C}_j{\left|x_j\right|}^{1/(1-\alpha )}$$

If, for each $i$, the coefficient of ${\left|x_i\right|}^{1/(1-\alpha )}$ in the firstsum would be greater than the coefficient of the same term in the right-handside, inequality couldn’t hold. So we can conclude that at least one index $p$ existssuch as

$$\frac{{\left|\lambda -a_{pp}\right|}^{1/(1-\alpha )}}{R_p^{\alpha /(1-\alpha )}}\le C_p$$

that is

$$\left|\lambda -a_{pp}\right|\le R_p^{\alpha }{C}_p^{1-\alpha }$$

which is the thesis.∎

Remarks:

The Gershgorin theorem is obtained as a limit for $\alpha \to 0$ orfor $\alpha \to 1$; in other words, Ostrowski’s theorem represents a kind of ”continuous deformation” between the two Gershgorin rows and columns sets.