A poorly aligned start delays the largest Ritz value and requires the start-angle factor

Since $1\mathrm{/}10000+2(9999\mathrm{/}20000)=1$, the vector $b$ is unit. Also $Ab=(1\mathrm{/}10,2\alpha ,\alpha {)}^{\mathsf{T}}$. Set $v:=(-{\alpha }^2,9\alpha \mathrm{/}100,-8\alpha \mathrm{/}100{)}^{\mathsf{T}}\mathrm{.}$ Direct dot products give $v^{\mathsf{T}}b=0$ and $v^{\mathsf{T}}Ab=-{\alpha }^2\mathrm{/}10+18{\alpha }^2\mathrm{/}100-8{\alpha }^2\mathrm{/}100=0$. The vectors $b,Ab$ are independent because their first coordinates would force a proportionality factor $10$, while the second coordinates would force factor $2$. Hence $K_2(A,b)$ is two-dimensional and equals $v^{\perp }$.

We have $\frac{v_2^2+v_3^2}{v_1^2}=\frac{145(1\mathrm{/}10000)}{{\alpha }^2}=\frac{290}{9999}\mathrm{.}$ For every unit $x\in K_2(A,b)$, $v^{\mathsf{T}}x=0$. By [F2], $\mathrm{\mid }{x}_1{\mathrm{\mid }}^2\le \frac{v_2^2+v_3^2}{v_1^2}(x_2^2+x_3^2)=\frac{290}{9999}(1-\mathrm{\mid }{x}_1{\mathrm{\mid }}^2),$ so $\mathrm{\mid }{x}_1{\mathrm{\mid }}^2\le 290\mathrm{/}10289$. Its Rayleigh quotient satisfies $x^{\mathsf{T}}Ax=10x_1^2+2x_2^2+x_3^2\le 2+8x_1^2\le 2+\frac{2320}{10289}<\frac{9}{4},$ since $4(2320)=9280<10289$. Taking the maximum gives ${\theta }_2<9\mathrm{/}4$, so ${\lambda }_1-{\theta }_2>31\mathrm{/}4$.

The top eigenspace is $\mathrm{span}\{e_1\}$, hence ${\tau }^2=(9999\mathrm{/}10000)\mathrm{/}(1\mathrm{/}10000)=9999$. The unwanted interval is $[1,2]$, and its affine exterior point is $1+2(10-2)\mathrm{/}(2-1)=17$, so $T_1(17)=17$. The theorem's $m=2$ bound is ${\lambda }_1-{\theta }_2\le \frac{9\cdot 9999}{{17}^2}=\frac{89991}{289},$ which is valid but large. If its factor ${\tau }^2$ were deleted, the proposed right side would be $9\mathrm{/}289<31\mathrm{/}4$, contradicting step 1.2. Thus this start cannot support that angle-free bound.