Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / P / 4

The Gershgorin circle theorem states that every complex eigenvalue of a square matrix $A$ lies in at least one closed disc $D_i=\{z:|z-a_{ii}|\le {\sum }_{j\ne i}|a_{ij}|\}$. To prove it, take a nonzero eigenvector $v$ for $\lambda$ and choose $i$ with $|v_i|$ maximal. The $i$th row of $Av=\lambda v$ givesDivision by $|v_i|>0$ proves the claim. The proof applies to real and complex matrices alike.

The second part of the Gershgorin circle theorem states that if a union of $k$ of the discs is disjoint from every remaining disc, that union contains exactly $k$ eigenvalues, counted with algebraic multiplicity. In particular, a single isolated disc contains precisely one eigenvalue. Connected components of the full union therefore carry the corresponding disc counts.

Write $q=1/4+ϵ$ and take $0<ϵ<1/4$, as intended by the small-parameter assumption. The row discs have centers $1,\dots ,n$ and radii $ϵ,q,\dots ,q,1/4$. They are pairwise disjoint because the sum of neighboring radii is below one. The second Gershgorin circle theorem places one eigenvalue in each disc. Since $A$ is real, nonreal eigenvalues occur in conjugate pairs; a disc centered on the real axis would have to contain both members, contradicting its count of one. Thus all eigenvalues are real andAlternatively, a positive diagonal change of basis symmetrizes the tridiagonal matrix, with transformed off-diagonal entries $\sqrt{ϵ}/2$, also proving reality.

The Gershgorin discs of $A^{\mathsf{T}}$ have radii $1/4,q,\dots ,q,ϵ$. They too are disjoint and have one eigenvalue each. The disc centered at $n$ therefore improves the final bound toIt selects the same eigenvalue as before because all other discs lie a fixed positive distance from the disc centered at $n$.

Assume $n\ge 3$, so the corner entry is separate from the superdiagonal. For $A(\delta )$, the row radii areThus every eigenvalue lies in the union ${\bigcup }_{i}D(i,r_i)$. The discs with centers $2,\dots ,n$ remain mutually disjoint. The expanding disc centered at one meets the disc centered at $j$ exactly when $j-1\le \delta +ϵ+r_j$.

Let $I_{\delta }=\{1\}\cup \{j\ge 2:j-1\le \delta +ϵ+r_j\}$. Then the union of discs indexed by $I_{\delta }$ contains exactly $|I_{\delta }|$ eigenvalues; every other disc contains exactly one. Every isolated eigenvalue is real by conjugation and obeys $|\lambda -j|\le r_j$. Once discs merge, the second Gershgorin circle theorem gives a bound and count for their whole component, rather than one eigenvalue per original disc. Eigenvalue branches coming from that component stay in its union, but may no longer be distinguished by individual discs. These statements describe the bounds through all merger thresholds as $\delta$ increases.

The claimed endpoint bounds with radius $ϵ$ are false for positive corner perturbations. For example, take $n=3$, $ϵ=1/1000$ and $\delta =9/10<1-ϵ$. The characteristic polynomial isThroughout $|\lambda -1|\le 1/1000$, the absolute value of the sum of its first two terms is below $0.003$, whereas $9/160=0.05625$. Hence this interval contains no root, contradicting the first requested bound. The actual eigenvalues are approximately $1.029144,1.943599,3.027257$, which also violate the last requested bound.

There is a valid real-eigenvalue conclusion with wider endpoint bounds. For $0<ϵ<1/4$ and $\delta <(n-1)/2-ϵ$, the Gershgorin discs from rows have one enlarged component on the left; the Gershgorin discs from columns have one enlarged component on the right. Each has an enlarged radius $r=\delta +ϵ<(n-1)/2$. A small disc that meets the enlarged left disc has index $j$ satisfying $j-1\le r+q<n/2$. A small disc meeting the enlarged right disc satisfies $n-j<n/2$. Thus, when $n=2m$, every disc with index at most $m$ is isolated in the column family and every disc with index at least $m+1$ is isolated in the row family. Their counts give all $n$ eigenvalues, one in each disjoint small disc, so all are real.

When $n=2m+1$, the same argument isolates the first $m$ discs in the column family and the last $m$ in the row family. It accounts for $2m$ real eigenvalues. The one remaining eigenvalue is real by conjugation, and belongs to the central small disc: the enlarged left and right discs are disjoint, and their two components can intersect only within the disc centered at $m+1$. Therefore the valid uniform bounds areThe sharper endpoint radii $ϵ$ hold in part (d) when $\delta =0$, and cannot simply be carried over to the perturbed matrix.