TITLE: Gerschgorin's Theorem (Round 1)
QUESTION [6 upvotes]: In looking for a proof to the Gerschgorin's Theorem, I stumbled across this paper: http://buzzard.ups.edu/courses/2007spring/projects/brakkenthal-paper.pdf
I don't quite buy the proof for Theorem 2.1 in that paper (yet).

Theorem 2.1: every eigenvalue $\lambda$ of a square matrix $A\in \mathbb{C}^{n\times n}$ satisfies 
  $$
\vert \lambda - A_{ii}\vert \leq \sum_{j\neq i}\vert A_{i,j}\vert , i\in \{1,2,\ldots,n\}
$$

And the proof offered:

If Theorem 2.1 is not satisfied, then $\lambda I-A$ is strictly diagonally dominant (SDD).
$\Rightarrow$ $\lambda I - A$ is non-singular
$\Rightarrow$ $\lambda$ is not an eigenvalue of $A$

My problem lies with 

Theorem not satisfied $\Rightarrow$ $\lambda I - A$ is SDD

part.
I mean, the author also states that 

the matrix $\lambda I - A$ is SDD if $\vert \lambda - A_{ii}\vert > \sum_{j\neq i}\vert A_{i,j}\vert$ for every i

But the way I see it, the theorem is already not satisfied if
$$\vert \lambda - A_{ii}\vert > \sum_{j\neq i}\vert A_{i,j}\vert$$
were to hold for some $i$ while $$\vert \lambda - A_{kk}\vert \leq \sum_{j\neq k}\vert A_{k,j}\vert$$ for some $k\neq i$, in which case $\lambda I - A$ would not be SDD.
What am I missing?

REPLY [4 votes]: The issue is the quantifiers, which this proof has suppressed. The statement of the theorem is: for any eigenvalue $\lambda$ there exists $i$ such that $|\lambda - A_{ii}| \leq \sum_{j \neq i} |A_{ij}|$. The negation of that would be that there exists an eigenvalue $\lambda$ such that for all $i$ we have $|\lambda-A_{ii}| > \sum_{j \neq i} |A_{ij}|$. But this means $\lambda I - A$ is SDD, which means it is nonsingular, which contradicts the assumption that $\lambda$ is an eigenvalue.