# Condition number is less than n

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: $\operatorname{Cond}(A)= \|A\|_1 \cdot \|A^{-1}\|_1$

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

Since $A$ is orthogonal, $A^{-1}= A^T$
given $|| x||_1≤\sqrt n$
$Cond(A)= ||A||_1 * ||A^{-1}||_1 = ||A||_1 *||A^T||_1 ≤ \sqrt n * \sqrt n = n$
• @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 $\sqrt n$, Dec 4, 2017 at 23:56