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.

  • 3
    $\begingroup$ 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. $\endgroup$
    – user251257
    Oct 2, 2015 at 1:02
  • $\begingroup$ @user251257: You should probably post that as an answer :) $\endgroup$
    – PhoemueX
    Oct 2, 2015 at 8:21

1 Answer 1


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 $\|A^{-1}\|$.


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