How to prove for any matrix norm, the condition number $\kappa(B)=O(\epsilon^{-1})$

Question

The inverse of the matrix $$ B=\begin{bmatrix} 1 & -1 \\ 1 & -1+\epsilon \\ \end{bmatrix} $$ is $$ B^{-1}=\begin{bmatrix} 1-\frac{1}{\epsilon} & \frac{1}{\epsilon} \\ -\frac{1}{\epsilon} & \frac{1}{\epsilon} \\ \end{bmatrix} $$

how to prove that for any matrix norm, the condition number $\kappa(B)=O(\epsilon^{-1})$ ?

I know that \begin{align} \kappa(B)&=|||B^{-1}|||\cdot|||B|||\\ & = \frac{1}{\epsilon}\cdot|||\begin{matrix} \epsilon-1 & 1 \\ -1 & 1 \\ \end{matrix}|||\cdot|||\begin{matrix} 1 & -1 \\ 1 & -1+\epsilon \\ \end{matrix}|||\\ \end{align}

How to prove that $|||\begin{matrix} \epsilon-1 & 1 \\ -1 & 1 \\ \end{matrix}|||\cdot|||\begin{matrix} 1 & -1 \\ 1 & -1+\epsilon \\ \end{matrix}||| $ contain some constant term？

Answer 1

All norms on finite dimensional spaces are equivalent, so pick any convenient norm.

One simple one is the Frobenius norm which gives $\kappa (B) = \sqrt{3+(1-\epsilon)^2} \sqrt{{3 \over \epsilon^2} + {(1-\epsilon)^2 \over \epsilon^2}}$.

We can write $\kappa (B) = {1 \over |\epsilon|} (3+ (1-\epsilon)^2)$.