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It says in Wikipedia:

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

Before that, a Gershgorin disc is defined as.

$D(a_{ii}, R_i)$ be the closed disc centered at aii with radius $R_i$. Such a disc is called a Gershgorin disc.

A Gershgorin disc is defined for a row, so what does "the Gershgorin discs $C_j$ corresponding to the columns of $A$" mean?

Thanks in advance!

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1 Answer 1

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As the proof of the corollary suggests, consider the transpose matrix $A^T$. The rows of $A^T$ "are" the columns of $A$.

In other words, if $$R_i=\sum_{j\ne i}|a_{ij}| \qquad R'_j=\sum_{i\ne j}|a_{ij}|$$ and $\lambda$ is an eigenvalue, then $$\lambda \in \bigcup_i D(a_{ii},R_i)\cap \bigcup_j D(a_{jj},R'_j)$$

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