Gershgorin circle theorem: A bound on eigenvalues

 

Let ${\displaystyle A}$ be a complex ${\displaystyle n\times n}$ matrix, with entries ${\displaystyle a_{ij}}$. For ${\displaystyle i\in \{1,\dots ,n\}}$ let ${\displaystyle R_i}$ be the sum of the absolute values of the non-diagonal entries in the ${\displaystyle i}$-th row:

${\displaystyle R_i=\sum _{j\ne i}\left|a_{ij}\right|.}$

Let ${\displaystyle D(a_{ii},R_i)\subseteq \mathbb{C}}$ be a closed disc centered at ${\displaystyle a_{ii}}$ with radius ${\displaystyle R_i}$. Such a disc is called a Gershgorin disc.

Theorem. Every eigenvalue of ${\displaystyle A}$ lies within at least one of the Gershgorin discs ${\displaystyle D(a_{ii},R_i).}$ 

Corollary. The eigenvalues of A must also lie within the Gershgorin discs C_j corresponding to the columns of A.

 

Note: The theorem does not claim that there is one disc for each eigenvalue. 

 

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This theorem can be used to solve the following problem, asked in the PhD entrance exam of Pondicherry university in the year 2011.

 

All eigenvalues of the matrix $\begin{pmatrix} 1 & 2 & 0 \\ 2 & 1  & 0 \\ 0 & 0 & -1 \end{pmatrix}$ lie in the disc

a) $|\lambda +1| \le 1$

a) $|\lambda -1| \le 1$

a) $|\lambda +1| \le 0$

a) $|\lambda -1| \le 2$