I have been struggling to comprehend the proof of Gershgorin Circle Theorem for a long time now, but I think I have come upon a counterexample. I'm probably wrong, but please tell me where I'm wrong...
Theorem: Every eigenvalue of A lies within at least one of the Gershgorin discs $D(a_{ii},R_i)$.
Proof: $$Av=\lambda v \implies \sum_{j=1}^n a_{ij}x_j = \lambda x_i \ \ \forall \ \ i \in \{1,2,...,n\}$$ where $|x_i|=\max_{j} {x_j}$ for $x_j$ elements of the eigenvector.
And so on.
But what if $$A=\left(\begin{matrix} 1 & 0 \\ 1 & -1 \end{matrix}\right), \ \ \lambda_1 = -1, \lambda_2=1 \ \ , v_1=(0,1)^T?$$ Then $x_1=0$ always, $x_2$ is free. Then $x_i = 1$, and $a_{11}+a_{12}=0 \neq (-1)(1) = 1$.
What am I missing?
It turns out that I was getting confused by the last line in this paragraph, in this proof on Wikipedia:
http://en.wikipedia.org/wiki/Gershgorin_circle_theorem
Proof: Let $λ$ be an eigenvalue of A and let $x = (x_j)$ be a corresponding eigenvector. Let $i∈{1,…,n}$ be chosen so that $|x_i| = \max_j |x_j|$. (That is to say, choose $i$ so that $x_i$ is the largest (in absolute value) number in the vector $x$). Then $|x_i| > 0$, otherwise $x = 0$. Since $x$ is an eigenvector, $Ax = λx$, and thus: $\sum \limits_j a_{ij} x_j = \lambda x_i \quad \forall i \in \{1, \ldots, n\}$.
Since $i$ is a bounded variable, it is ok to write it the way it as has been written and it can rewritten as $\sum \limits_j a_{kj} x_j = \lambda x_k \quad \forall k \in \{1, \ldots, n\}$. On the following line what they do is particularize that universal statement to $k=i$ (remember $i$ was already fixed). Therefore, everything falls in place.