Show that for an $n$ x $n$ orthogonal matrix $A$ that $\operatorname{Cond}(A) \leq n$.

I need to use: $$\|x\|_1 \leq \sqrt n$$

I know that $\operatorname{Cond}(A)=1$ for $A$ orthogonal matrix. Also given that: $\operatorname{Cond}(A)= \|A\|_1 \cdot \|A^{-1}\|_1$

  • $\begingroup$ This question makes no sense. You know the condition number is 1 and you want to show that it is $\le n$? $\endgroup$
    – copper.hat
    Dec 4, 2017 at 18:47
  • $\begingroup$ Yeah, that's the only hint I have $\endgroup$ Dec 4, 2017 at 18:48
  • 2
    $\begingroup$ If you know it is 1 then it must be $\le n$, so there is nothing to show. $\endgroup$
    – copper.hat
    Dec 4, 2017 at 18:48
  • $\begingroup$ @JeanMarie: You are probably right, but I was hoping for input from the OP. $\endgroup$
    – copper.hat
    Dec 4, 2017 at 19:40

1 Answer 1


Since $A$ is orthogonal, $A^{-1}= A^T$

given $|| x||_1≤\sqrt n$

$Cond(A)= ||A||_1 * ||A^{-1}||_1 = ||A||_1 *||A^T||_1 ≤ \sqrt n * \sqrt n = n$

  • $\begingroup$ May I know why you multiply matrix 1-norm with matrix infinity-norm? $\endgroup$ Dec 4, 2017 at 23:44
  • $\begingroup$ @dembrownies I edited it, hope it is more clear now $\endgroup$
    – superman
    Dec 4, 2017 at 23:46
  • $\begingroup$ I know this might sound stupid but why did you replace ||A^T|| 1 norm with ||A|| infinity-norm? $\endgroup$ Dec 4, 2017 at 23:50
  • $\begingroup$ @dembrownies because infinity norm is the maximum of sum of rows, 1-norm is the maximum of sum of column , when you take transpose, rows become columns. and do you have the the complete questions? I realized a flaw in the solution: infinity norm is not bounded by $\sqrt n$, $\endgroup$
    – superman
    Dec 4, 2017 at 23:56
  • $\begingroup$ I edited the question already (with all the hints and complete question) $\endgroup$ Dec 4, 2017 at 23:57

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