# Does the norm of the matrix inverse alone say anything about the condition number

Is the fact that $$\| A^{-1} \|$$ is large, enough to conclude that the $cond(M)$ is large, where $cond()$ is the condition number of the matrix to determine if it is ill-conditioned. $$cond(M) = \| A^{-1} \| \|A\|$$ Since $cond()$ depends also on $\|A\|$, can it be that $$\| A^{-1} \|$$ is large and $$\| A \|$$ is small, producing a small condition number?

For $$\| A^{-1} \|$$ and $$\| A \|$$, I am considering only the infinity norm.

• you can scale the identity matrix to get arbitrarily large norm of the inverse. However, a scalar multiple of the identity has the same condition as the identity. Oct 2, 2015 at 1:02
• @user251257: You should probably post that as an answer :) Oct 2, 2015 at 8:21

The condition number is scaling invariant. That is, for each non-singular $A$ and $\alpha > 0$ we have $$\operatorname{cond} (\alpha A) = \alpha\alpha^{-1} \| A\| \|A^{-1}\| = \operatorname{cond}(A).$$ So, by multiplication with a scalar you can make the norm of the inverse arbitrarily large (or small) without changing the condition number.
However, if you fix $\|A\|$, say for example $=1$, then $\operatorname{cond}(A)$ is trivially proportional to $\|A^{-1}\|$.