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?



Holder inequality for a general rectangular matrix
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
$$
\|A\|\_p=LHS=\|A^T\|\_q.
$$


