= Solution
Two consecutive steps multiply by the <polynomial> of the symmetric <matrix>
$$
P(A)=(A-s_0I)(A-s_1I),\qquad P(\lambda)=\lambda^2-\lambda+\frac18
=\frac18T_2(2\lambda-1).
$$
The <Chebyshev polynomial> bound on $[0,1]$ gives $|P(\lambda)|\leq1/8$, and the actual endpoints $0,1$ attain it. At the dominant <eigenvalue>, $P(1+\epsilon)=1/8+\epsilon+\epsilon^2$. Thus
$$
\boxed{\rho^2=\frac1{1+8\epsilon+8\epsilon^2},\qquad
\rho=1-4\epsilon+O(\epsilon^2).}
$$
For a small prescribed error $\delta$ relative to a fixed initial-component factor $C$, the number of <matrix>-<vector> iterations is approximately $\log(C/\delta)/(-\log\rho)$. The three leading denominators are $\epsilon,2\epsilon,4\epsilon$. Therefore \b[the iteration counts are in the ratio $1:1/2:1/4$]: the single shift halves and the alternating double shift quarters the leading count. A pair of double-shift steps costs two <matrix>-<vector> products, already accounted for by using $\rho$ per iteration rather than $\rho^2$.
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