= Gershgorin eigenvalue counting
{c}
{title2=$\#\{\lambda\text{ in }C\}=k\quad\text{with algebraic multiplicity}$}
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.https://cklixx.people.wm.edu/teaching/math408/408-note.pdf[Chi-Kwong Li, Math 408 lecture notes, §6.1, Theorem 6.1.1, p. 48]{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> $\left(\begin{smallmatrix}2&1&0\\0&2&0\\0&0&8\end{smallmatrix}\right)$ 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> $E=\operatorname{diag}(a_{11},\ldots,a_{nn})$ and $A(t)=E+t(A-E)$, $0\leq t\leq1$. 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.https://www.stat.uchicago.edu/~lekheng/courses/302/notes2.pdf[Gene H. Golub, CME 302, Lecture 2, §6: turning on the off-diagonal entries continuously]{ref}
Back to article page
