Classical column-pivoted QR can fail the strong coefficient bound

Column $j$ of $A$ has entries $-c{s}^{i-1}$ for $i<j$, $s^{j-1}$ for $i=j$, and zero below. Its squared norm is $c^2{\sum }_{r=0}^{j-2}{s}^{2r}+s^{2(j-1)}=1$, using $c^2=1-s^2$. After the first $r$ columns have been selected, they span the first $r$ coordinate vectors because their leading triangular block has nonzero diagonal. The residual of each remaining column $j>r$ has squared norm $s^{2r}(c^2{\sum }_{t=0}^{j-r-2}{s}^{2t}+s^{2(j-r-1)})=s^{2r}$ (the sum is empty when $j=r+1$). Thus all remaining residuals tie at every step, and the declared tie rule makes CPQR select columns in their original order.

By [F1] and step 1.1, the resulting permutation is the identity. The matrix $A$ is upper triangular with positive diagonal, so the QR factors can be taken as $Q=I$ and $R=A$. For $k=3$, solving $R_{11}x=R_{12}$ gives $x_3=-c$, $x_2=-c(1+c)$, and $x_1=-c(1+c{)}^2=-9\mathrm{/}8$. Hence ${\max }_i\mathrm{\mid }{x}_i\mathrm{\mid }\ge 9\mathrm{/}8>1$, violating the coefficient condition in [F2].

This CPQR output therefore fails the strong coefficient test with $f=1$. It does not contradict the strong RRQR procedure: that procedure performs additional swaps when its coefficient or combined residual score exceeds the threshold.