Gershgorin eigenvalue counting ($\#\{\lambda \text{ in }C\}=k\text{with algebraic multiplicity}$)

Let $C$ be a connected component of the union of the row Gershgorin discs of an $n\times n$ matrix. If $C$ comprises $k$ of the indexed Gershgorin discs, then it contains exactly $k$ eigenvalues, counted with algebraic multiplicity. Coincident disks still count separately, once for each row. In particular, a Gershgorin disc disjoint from all the others contains one simple eigenvalue. More generally, any union of $k$ indexed Gershgorin discs disjoint from the remaining $n-k$ contains exactly $k$ eigenvalues, whether or not that union is connected.^[ref]

The separation is essential: overlapping Gershgorin discs do not individually promise one eigenvalue each. Discs that touch at a boundary point belong to the same connected component. For example, the matrix $\left(\begin{array}{ccc}2& 1& 0\\ 0& 2& 0\\ 0& 0& 8\end{array}\right)$ has Gershgorin discs of radii $1,0,0$ with centres $2,2,8$. The component around $2$ comprises two indexed disks and contains the eigenvalue $2$ with algebraic multiplicity two; the isolated point at $8$ contains the remaining eigenvalue.

For intuition, continuously turn on the off-diagonal entries using the diagonal matrix $E=\mathrm{diag}(a_{11},\dots ,a_{nn})$ and $A(t)=E+t(A-E)$, $0\le t\le 1$. At $t=0$ the eigenvalues are the disk centres; at every later time the Gershgorin discs stay inside their final union. The continuity of the eigenvalues as a collection means that no eigenvalue can cross a gap between final components. The count gives a useful certificate that an eigenvalue cluster is complete.^[ref]