Condition number

Remarks

By convention, $\kappa(A) = \infty$ when $A$ is not invertible. If $\norm{\cdot}$ is an operator norm, then $\kappa(A) \ge \norm{A A^{-1}} = \norm{I} = 1$. For the operator norm of the Euclidean norm, $\kappa_2(A) = \sigma_{\max} / \sigma_{\min}$, the ratio of the largest and smallest singular values. If $Ax = b$ with $b \ne 0$ and $A(x + \delta x) = b + \delta b$, then $\norm{\delta x} / \norm{x} \le \kappa(A) \, \norm{\delta b} / \norm{b}$ for the operator norm of a vector norm. Example: $A = \begin{pmatrix} 1 & 0 \\ 0 & 10^{-3} \end{pmatrix}$ has $\norm{A}_2 = 1$ and $\norm{A^{-1}}_2 = 10^{3}$, so $\kappa_2(A) = 10^3$. Non-example: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ is not invertible, so the formula does not apply.