Does smallest singular value has anything to do with matrix norm?

Question

I don't understand why smallest singular value has anything to do with matrix norm?

a book called "probability in high dimensions"

i.e. tall matrices are well-conditioned.

Answer 1

It might goes this way: $(1-\delta)K\le\frac{\|Ax\|}{\|x\|}\le(1+\delta)K\implies\min_x\frac{\|Ax\|}{\|x\|}\ge(1-\delta)K$ and $\min_x\frac{\|Ax\|}{\|x\|}=\min\sigma_A$. This is because $\min_x\frac{\|Ax\|}{\|x\|}=\min_x\frac{\|V\Sigma U^*x\|}{\|x\|}=\min_x\frac{\|\Sigma x\|}{\|x\|}=\min\sigma_A$.

The last equality: $\frac{\|\Sigma x\|}{\|x\|}=\frac{\sqrt{\sum_i|\sigma_ix_i|^2}}{\sqrt{\sum_ix_i^2}}\ge\frac{\sqrt{\sum_i\min|\sigma_A|^2|x_i|^2}}{\sqrt{\sum_ix_i^2}}=\sigma_A$ and $\frac{\|\Sigma e_1\|}{\|e_1\|}=\sigma_A$.