Let $A \in \mathbb{R}^{m\times n}$ and $p,q \in \mathbb{R}^{+}$ such that $\frac{1}{p}+\frac{1}{q}=1$. I am interested to prove the following:
$$ \|A\|_{p}=\|A^T\|_q$$
I have tried using Holder Inequality for vectors $Ax$ and $A^Ty$ and thereby mapping back to the original matrix norm using basic properties but I am unable to proceed further because the vector sizes are different.
Is there any other way to prove the same?
We have $$ \begin{eqnarray} LHS&=& \sup_{x\in \mathbb{R}^m\backslash \{0\},y\in \mathbb{R}^n\backslash \{0\}} \frac{|x^TAy|}{\|x\|_q\|y\|_p}\\ &\leq& \sup_{x\in \mathbb{R}^m\backslash \{0\},y\in \mathbb{R}^n\backslash \{0\}} \frac{ \|x\|_q\|Ay\|_p}{\|x\|_q\|y\|_p}\\ &=& \sup_{x\in \mathbb{R}^m\backslash \{0\},y\in \mathbb{R}^n\backslash \{0\}} \frac{\|Ay\|_p}{\|y\|_p}\\ &=&\|A\|_p. \end{eqnarray} $$ Similarly $LHS\leq \|A^T\|_q$. Note that by compactness argument there exist $y\in \mathbb{R}^m$ s.t. $\|Ay\|_p=\|A\|_p\|y\|_p$. We then have equality by choosing $x=sign(Ay).*|Ay|.^{p/q}$ where $.$ denotes element wise operation. Hence we have $$ \|A\|_p=LHS=\|A^T\|_q. $$
