For ill-conditioned matrices, must it be that the smallest singular value is arbitrarily close to $0$? I know that $K_2(A) = \frac{\sigma_{max}}{\sigma_{min}}$ where $\sigma$ is a singular value of A. Can their be a case of $\sigma_{max}$ being extremely large and $\sigma_{min}$ being arbitrarily close to 1.
Ill-conditioned matrices and their singular values
linear algebranumerical linear algebranumerical methods
Best Answer
Of course. Think of the matrix $A$ defined as $$ A = \mathrm{diag}(1, 2, \ldots, N), $$ where $\mathrm{diag}$ is the diagonal matrix. For any value $N$, the condition number of the matrix will be $K_2(A) = N$ and the smallest singular value $\sigma_{\min} = 1$. The matrix can also be small, for example $$ A = \begin{pmatrix} 1 & 0 \\ 0 & N \end{pmatrix}, $$ has the same spectral properties as the matrix above.