$1\le p,q \le \infty$ with $\frac{1}{p}+\frac{1}{q} = 1$ Let $K_p(A) =\|A\|\|A^{-1}\|_p$ then prove that $ K_p(A)=K_q(A^t)$

Question

A be invertible matrix such that $1\le p,q \le \infty$ with $\frac{1}{p}+\frac{1}{q} = 1$ Let $K_p(A) =\|A\|\|A^{-1}\|_p$ then prove that $ K_p(A)=K_q(A^t)$

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One hint is given here https://math.stackexchange.com/a/3349025/750552 . $$\|A\|_p = \sup_{\|x\|_p = 1} \|A x\|_p = \sup_{\|x\|_p = 1} \sup_{\|y\|_q = 1} |y^T A x|$$ and $$\|A^t\|_q = \sup_{\|x\|_q = 1} \|A^t x\|_q = \sup_{\|x\|_q = 1} \sup_{\|y\|_p = 1} |y^T A^t x|$$

Then we have to prove that $$\sup_{\|x\|_q = 1} \sup_{\|y\|_p = 1} |y^T A^t x| =\sup_{\|x\|_p = 1} \sup_{\|y\|_q = 1} |y^T A x| $$

I don't understand how prove it ??

Answer 1

First transpose and then rename:$$\begin{align}\sup_{\|x\|_q = 1} \sup_{\|y\|_p = 1} |y^T A^t x|&= \sup_{\|x\|_q = 1} \sup_{\|y\|_p = 1} |x^TAy|\\ &=\sup_{\|y\|_q = 1} \sup_{\|x\|_p = 1} |y^TAx|. \end{align}$$