If the norm is induced by an inner product, then:


where the second equality holds from the fact that if the norm is induced from an inner product, then $||Ax||=||x||$ for any orthogonal matrix $A$. Similarly, since $A^{-1}$ is also orthogonal we get $||A^{-1}||=1$ and thus $cond(A)=||A||\cdot||A^{-1}|| = 1 \cdot 1=1$.

Is this true that $cond(A) = 1$ for any orthgonal matrix $A$ for general norms too?

  • 2
    $\begingroup$ No, the Frobenius condition number of the identity is $2$. $\endgroup$
    – copper.hat
    Dec 19, 2019 at 19:07

1 Answer 1


No. Two counterexamples:

  • The Frobenius norm is submultiplicative and unitarily invariant. We have $\|I_2\|=\sqrt{2}$ and hence $\|I_2\|\|I_2^{-1}\|=2>1$.
  • Let $P=\pmatrix{1&1\\ 0&1}$ and $A=\pmatrix{0&-1\\ 1&0}$. Then $\|M\|=\max_{x\ne0}\frac{\|PMx\|}{\|Px\|}=\sigma_\max(PMP^{-1})$ is an induced norm and $\|A\|=\frac{3+\sqrt{5}}{2}>1$. Since $A^{-1}=-A$, we have $\|A\|\|A^{-1}\|=\|A\|^2>1$.

However, if $\|\cdot\|$ is a unitarily invariant induced norm, then $\|A\|=\|I\|=\max_{x\ne0}\frac{\|Ix\|}{\|x\|}=1$ for every real orthogonal matrix $A$. Hence the condition number of $A$ is $1$ in this case.


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