Lanczos handles repeated top eigenvalues, absent top components, and exact breakdown

In case (i), take $b=1$. Then $P=I$, $Pb=b\mathrm{\ne }0$, and $K_1=\mathbb{R}$ is invariant under $A=5I_1$. The first Arnoldi residual is zero, so Lanczos breaks down at $d=1$; every nonzero vector has Rayleigh quotient $5$, giving ${\theta }_m=5$ for all $m\ge 1$ by [F1]. In case (ii), $P$ projects onto $\mathrm{span}\{e_1,e_2\}$, so $Pb=e_1\mathrm{/}\sqrt{2}\mathrm{\ne }0$. Here $Ab=(5,0,1{)}^{\mathsf{T}}\mathrm{/}\sqrt{2}$ and $Ab-b=(4,0,0{)}^{\mathsf{T}}\mathrm{/}\sqrt{2}$, so $K_2(A,b)=\mathrm{span}\{e_1,e_3\}$, an invariant subspace. The first Arnoldi residual is nonzero, while $A{q}_2\in K_2=\mathrm{span}\{q_1,q_2\}$ makes the second residual zero; hence Lanczos breaks down at $d=2$. Since $e_1\in K_2$ has Rayleigh quotient $5$, the maximum is ${\theta }_2=5$, as [F1] predicts for the collapsed interval branch.

In case (iii), $P$ projects onto $e_1$, so $Pb=0$; $A{e}_2=2e_2$ gives $K_m(A,e_2)=\mathrm{span}\{e_2\}$ for every $m$, and therefore ${\theta }_m=2={\lambda }_{r+1}$, in agreement with [F1]. At the first Arnoldi column, $A{e}_2$ already lies in $K_1=\mathrm{span}\{e_2\}$, so subtracting its projection leaves zero residual. Thus [F3]–[F4] give exact Lanczos breakdown at $d=1$ without a top component.

In case (iv), $b$ and $Ab=(5e_1+2e_2)\mathrm{/}\sqrt{2}$ are independent, so $K_2=\mathrm{span}\{e_1,e_2\}$; this space is invariant under $A$. At the second Arnoldi column, $A{q}_2\in K_2=\mathrm{span}\{q_1,q_2\}$, so subtracting its two orthogonal projections leaves zero residual. Thus [F3]–[F4] give exact Lanczos breakdown at $d=2$. Since $Pb=e_1\mathrm{/}\sqrt{2}\mathrm{\ne }0$, [F1] gives ${\theta }_2=5$; directly, $Ab-2b=3e_1\mathrm{/}\sqrt{2}\in K_2$, so $e_1\in K_2$ and its Rayleigh quotient is $5$.