Computed singular values and rank revealing factorizations — Examples

The determinant of $A_{ϵ}$ is $ϵ\mathrm{\ne }0$, so its exact rank is two. By [F2], write $A_{ϵ}=U\mathrm{\Sigma }{V}^{\ast }$; taking determinants and using $\mathrm{\mid }\det U\mathrm{\mid }=\mathrm{\mid }\det V\mathrm{\mid }=1$ (from $U^{\ast }U=V^{\ast }V=I$) gives ${\sigma }_1{\sigma }_2=ϵ$. Its first column has norm one, hence [F4] gives ${\sigma }_1=\mathrm{\parallel }{A}_{ϵ}{\mathrm{\parallel }}_2\ge 1$; by [F3], ${\sigma }_1\le \mathrm{\parallel }{A}_{ϵ}{\mathrm{\parallel }}_F=\sqrt{2+{ϵ}^2}<2$. Therefore $ϵ\mathrm{/}2<{\sigma }_2\le ϵ$.

At $\tau ={10}^{-7}$, ${\sigma }_2\le {10}^{-8}<\tau <1\le {\sigma }_1$, so exactly one singular value exceeds $\tau$. At $\tau ={10}^{-9}$, ${\sigma }_2>5\cdot {10}^{-9}>\tau$ and ${\sigma }_1\ge 1>\tau$, so both exceed $\tau$.

Applying [F1] to the two counts in step 2.1 gives numerical ranks one and two, respectively, although the exact rank remains two.

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.

The singular values of $B$ are $1$ and $ϵ$. Set $E_0=\mathrm{diag}(0,-ϵ)$. Then $B+E_0=\mathrm{diag}(1,0)$, $\mathrm{\parallel }{E}_0{\mathrm{\parallel }}_2=ϵ$, and the small singular value changes by absolute error $ϵ$ and relative error $ϵ\mathrm{/}ϵ=1$. This attains the absolute bound in [F1].

Set $E_1=\eta B$. Each nonzero diagonal entry is multiplied by $1+\eta$ and the zero superdiagonal remains zero, so [F2] applies. Directly, $B+E_1=(1+\eta )B$, whose small singular value is $(1+\eta )ϵ$ and whose relative error is exactly $\eta$.

Since $\eta$ can be fixed while $ϵ$ tends to zero, step 1.2 gives a relative guarantee independent of the small singular value; step 1.1 shows that the absolute error scale alone permits complete loss of that value.