I was looking at Proposition 3.2. in Applied Numerical Linear Algebra by JW Demmel, that states that when solving $$\min \|Ax - b\|_2 $$ if in the singular value decomposition $\sigma_{min}>0$ then $x^*= \text{argmin}\ \|Ax-b\|_2$ has $\|x\|_2 \geq |u_n^Tb|\ /\ \sigma_{min}$ where $u_n$ is the last column of $U$ in $A=U\Sigma V^T$.
In the proof he writes that $x^*=A^+b = V \Sigma ^{-1}U^T b$ hence $$\|x\|_2 = \|\Sigma ^{-1}U^T b\| \geq |(\Sigma ^{-1}U^T b)_n| = |u_n^Tb|\ /\ \sigma_{min}.$$ Should'nt there be $\sigma_{max}$ in place of $\sigma_{min}$ or am I missing something? [1]: https://i.stack.imgur.com/fXPUY.jpg
