Past exam of the options I course of the University of Oxford / 2005 / P

For a complex matrix $C=(c_{ij})$, Gershgorin's theorem puts every eigenvalue in the union of the row discs $D_i=\{z:|z-c_{ii}|\le R_i\}$, where $R_i={\sum }_{j\ne i}|c_{ij}|$. To prove it, choose an eigenvector $v$ for an eigenvalue $\lambda$ and an index $i$ maximizing $|v_i|>0$. Its $i$th equation givesDivision proves $\lambda \in D_i$.

The usual counting refinement says that a connected component made of $k$ discs contains exactly $k$ eigenvalues, with algebraic multiplicity. For completeness, deform the matrix from its diagonal part to the original by multiplying all off-diagonal entries by $t\in [0,1]$. Its discs have radii $t{R}_i$ and stay inside the original discs. Continuous variation of the roots of the characteristic polynomial keeps the number inside any contour separating one original component from the others constant. At $t=0$, the diagonal entries give exactly $k$ roots there, proving the refinement.

For the application put $s=-\mu >0$. The matrix isIts three row discs have positive real centres and leftmost real coordinatesAll are strictly positive, so zero lies in none of the discs. It cannot be an eigenvalue. Thus $A-\mu B$ is nonsingular for every real $\mu <0$.

Suppose an approximation has an error expansion $A(h)=I+c{h}^p+O(h^q)$ with $q>p$. With refinement factor $r>1$, combining the two approximations eliminates the leading term:If further expansion terms are known, repeat the elimination at their successive powers. For an even-power trapezium expansion and repeated halving, this gives the Romberg recurrence $R_{k,j}=(4^j{R}_{k,j-1}-R_{k-1,j-1})/(4^j-1)$, starting from $R_{k,0}=T(2^{-k})$.

For the stated cubic the one-panel trapezium value is $T(1)=[0^3+1^3]/2=1/2$. The two-panel value isThe leading power is two, so one extrapolation step givesThis equals the exact integral $[x^4/4{]}_0^1=1/4$; for a cubic the trapezium error has only the leading $h^2$ term, which this combination removes exactly.