# Ostrowski theorem

Let $A$ be a complex $n\times n$ matrix, $R_{i}=\sum_{j\neq i}\left|a_{ij}\right|,C_{j}=\sum_{i\neq j}\left|a_{ij}\right% |\quad 1\leq i\leq n,1\leq j\leq n$. Let’s consider, for any $\alpha\in(0,1)$, the circles of this kind: $O_{i}=\left\{z\in\mathbf{C}:\left|z-a_{ii}\right|\leq R_{i}^{\alpha}C_{i}^{1-% \alpha}\right\}\quad 1\leq i\leq n$.

###### Proof.

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

 $(\lambda-a_{ii})x_{i}=\sum_{j\neq i}a_{ij}x_{j}$

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

 $\displaystyle\left|\lambda-a_{ii}\right|\left|x_{i}\right|$ $\displaystyle\leq$ $\displaystyle\sum_{j\neq i}\left|a_{ij}\right|\left|x_{j}\right|$ $\displaystyle=$ $\displaystyle\sum_{j\neq i}\left|a_{ij}\right|^{\alpha}\left|a_{ij}\right|^{1-% \alpha}\left|x_{j}\right|$ $\displaystyle\leq$ $\displaystyle\left(\sum_{j\neq i}(\left|a_{ij}\right|^{\alpha})^{1/\alpha}% \right)^{\alpha}\left(\sum_{j\neq i}(\left|a_{ij}\right|^{1-\alpha}\left|x_{j}% \right|)^{1/(1-\alpha)}\right)^{1-\alpha}$ $\displaystyle=$ $\displaystyle\left(\sum_{j\neq i}\left|a_{ij}\right|\right)^{\alpha}\left(\sum% _{j\neq i}\left|a_{ij}\right|\left|x_{j}\right|^{1/(1-\alpha)}\right)^{1-\alpha}$ $\displaystyle=$ $\displaystyle R_{i}^{\alpha}\left(\sum_{j\neq i}\left|a_{ij}\right|\left|x_{j}% \right|^{1/(1-\alpha)}\right)^{1-\alpha}$

which means

 $\frac{\left|\lambda-a_{ii}\right|^{1/(1-\alpha)}}{R_{i}^{\alpha/(1-\alpha)}}% \left|x_{i}\right|^{1/(1-\alpha)}\leq\sum_{j\neq i}\left|a_{ij}\right|\left|x_% {j}\right|^{1/(1-\alpha)}$

Summing over all $i$, one obtains

 $\sum_{i=1}^{n}\frac{\left|\lambda-a_{ii}\right|^{1/(1-\alpha)}}{R_{i}^{\alpha/% (1-\alpha)}}\left|x_{i}\right|^{1/(1-\alpha)}\leq\sum_{i=1}^{n}\sum_{j\neq i}% \left|a_{ij}\right|\left|x_{j}\right|^{1/(1-\alpha)}=\sum_{j=1}^{n}C_{j}\left|% x_{j}\right|^{1/(1-\alpha)}$

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

 $\frac{\left|\lambda-a_{pp}\right|^{1/(1-\alpha)}}{R_{p}^{\alpha/(1-\alpha)}}% \leq C_{p}$

that is

 $\left|\lambda-a_{pp}\right|\leq R_{p}^{\alpha}C_{p}^{1-\alpha}$

which is the thesis. ∎

Remarks:

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

## References

• 1 R. A. Horn, C. R. Johnson, Matrix Analysis, Cambridge University Press, 1985
Title Ostrowski theorem  OstrowskiTheorem 2013-03-22 15:36:29 2013-03-22 15:36:29 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 22 Andrea Ambrosio (7332) Theorem msc 15A42