[proofplan]
We take an eigenvalue $\lambda$ with eigenvector $x \neq 0$ and select the component $x_i$ of maximum absolute value. The $i$-th row of the eigenvalue equation $Ax = \lambda x$ isolates $(\lambda - a_{ii})x_i = \sum_{j \neq i} a_{ij} x_j$. Taking absolute values, applying the triangle inequality to the right-hand side, and using the maximality of $|x_i|$ to bound each $|x_j|$ produces $|\lambda - a_{ii}| \le r_i$, which places $\lambda$ in the Gershgorin disc $\Gamma_i$.
[/proofplan]
[step:Extract the $i$-th row of the eigenvalue equation, choosing $i$ to maximise $|x_i|$]
Let $\lambda$ be an eigenvalue of $A$ with associated eigenvector $x \in \mathbb{C}^n \setminus \{0\}$, so $Ax = \lambda x$. Choose an index $i \in \{1, \ldots, n\}$ such that
\begin{align*}
|x_i| = \max_{1 \le j \le n} |x_j|.
\end{align*}
Since $x \neq 0$, at least one component is nonzero, so $|x_i| > 0$.
The $i$-th component of the vector equation $Ax = \lambdax$ reads
\begin{align*}
\sum_{j=1}^{n} a_{ij} x_j = \lambda x_i.
\end{align*}
Isolating the diagonal term $a_{ii} x_i$ on the left-hand side:
\begin{align*}
(\lambda - a_{ii}) x_i = \sum_{j \neq i} a_{ij} x_j.
\end{align*}
[/step]
[step:Apply the triangle inequality and the maximality of $|x_i|$ to localise $\lambda$]
Taking absolute values on both sides of the equation $(\lambda - a_{ii}) x_i = \sum_{j \neq i} a_{ij} x_j$ and applying the triangle inequality to the right-hand side:
\begin{align*}
|\lambda - a_{ii}| \cdot |x_i| = \left|\sum_{j \neq i} a_{ij} x_j\right| \le \sum_{j \neq i} |a_{ij}| \cdot |x_j|.
\end{align*}
By the choice of $i$, we have $|x_j| \le |x_i|$ for every $j \in \{1, \ldots, n\}$. Substituting this bound into the right-hand side:
\begin{align*}
|\lambda - a_{ii}| \cdot |x_i| \le \sum_{j \neq i} |a_{ij}| \cdot |x_i| = |x_i| \cdot r_i,
\end{align*}
where $r_i := \sum_{j \neq i} |a_{ij}|$ is the radius of the $i$-th Gershgorin disc. Since $|x_i| > 0$, we divide both sides by $|x_i|$:
\begin{align*}
|\lambda - a_{ii}| \le r_i.
\end{align*}
This means $\lambda \in \Gamma_i = \{z \in \mathbb{C} : |z - a_{ii}| \le r_i\}$. In particular, $\lambda \in \bigcup_{i=1}^{n} \Gamma_i$.
Since $\lambda$ was an arbitrary eigenvalue of $A$, every eigenvalue lies in $\bigcup_{i=1}^n \Gamma_i$, i.e., $\sigma(A) \subset \bigcup_{i=1}^n \Gamma_i$.
[guided]
The proof strategy hinges on a single idea: extract one scalar equation from the vector eigenvalue system by choosing the "largest" component. This is a maximum modulus argument — by looking at the component of $x$ with the greatest absolute value, we gain a usable bound on every other component.
Starting from $Ax = \lambdax$, the $i$-th row gives
\begin{align*}
\sum_{j=1}^n a_{ij} x_j = \lambda x_i.
\end{align*}
Moving the diagonal term to the right-hand side:
\begin{align*}
(\lambda - a_{ii}) x_i = \sum_{j \neq i} a_{ij} x_j.
\end{align*}
Taking absolute values and applying the triangle inequality:
\begin{align*}
|\lambda - a_{ii}| \cdot |x_i| \le \sum_{j \neq i} |a_{ij}| \cdot |x_j|.
\end{align*}
Now comes the key step. Because $|x_j| \le |x_i|$ for all $j$ (by the choice of $i$), we can replace each $|x_j|$ by the uniform upper bound $|x_i|$:
\begin{align*}
|\lambda - a_{ii}| \cdot |x_i| \le |x_i| \sum_{j \neq i} |a_{ij}| = |x_i| \cdot r_i.
\end{align*}
Division by $|x_i| > 0$ yields $|\lambda - a_{ii}| \le r_i$, placing $\lambda$ in the disc $\Gamma_i$.
Why choose the maximum-modulus component? If we had chosen an arbitrary component $x_k$, the bound $|x_j| \le |x_k|$ would fail for components larger than $x_k$, and we could not factor out $|x_k|$ from the right-hand side. The maximum-modulus choice is the only one that gives a bound on $|x_j|$ relative to $|x_i|$ for *every* $j$.
Note that the proof identifies a specific disc $\Gamma_i$ containing $\lambda$, not just the union. The index $i$ depends on the eigenvector $x$, which in turn depends on $\lambda$. Different eigenvalues may lie in different Gershgorin discs.
[/guided]
[/step]
