A fully converged LSQR solve can amplify a small singular-direction error

For $A_N$ and $y^{\delta }$, $r_0=(0,N^{-1}{)}^T$, so ${\beta }_1=N^{-1}$, $u_1=(0,1{)}^T$, ${\alpha }_1=N^{-2}$, and $v_1=(0,1{)}^T$. The first bidiagonalization residual is $A_N{v}_1-{\alpha }_1{u}_1=0$, so ${\beta }_2=0$ and the one-dimensional projected least-squares problem is exact.

Its projected matrix is $B_1=(N^{-2},0{)}^T\in {\mathbb{R}}^{2\times 1}$ and right-hand side is ${\beta }_1{e}_1=(N^{-1},0{)}^T\in {\mathbb{R}}^2$, so the unique minimizer is $z_1=N$. Consequently $x_1^{\delta }=x_0+N{v}_1=(0,N{)}^T$, $A_N{x}_1^{\delta }=y^{\delta }$, and $\mathrm{\parallel }{x}_1^{\delta }-x^{†}{\mathrm{\parallel }}_2=N$. The noise is $\mathrm{\parallel }{y}^{\delta }-y{\mathrm{\parallel }}_2=N^{-1}$, so the error-to-noise ratio is $N^2\to \mathrm{\infty }$. This refutes the proposed uniform bound and shows that the projected solve can fit a small noise component in a weak singular direction. The value ${\beta }_2=0$ marks bidiagonalization breakdown; the projected solve still forms the exact one-step iterate before the algorithm tests its final status. For the assigned discrepancy-stopped algorithm, the initial test already returns at index zero because $\mathrm{\parallel }{r}_0{\mathrm{\parallel }}_2={\delta }_N\le \tau {\delta }_N$; this computation concerns continuation to the full projected solve and claims no breakdown return.