Infininty norm and 1-norm on a matrix It is known that if $X,Y$ Banach spaces and $T\in \mathcal{L}(X,Y)$, then $T^*\in \mathcal{L}(Y^*,X^*)$ where $T^*(f):=f \circ T \in \mathcal{L}(X,\mathbb{R})$, and $\|T^*\|_{\mathcal{L}({Y^{*},X^*})}=\|T\|_{\mathcal{L}(X,Y)}$, where $$\mathcal{L}(A,B):=\{T:A\rightarrow B;\ \text{such that  $T$  is  linear and  continuous} \}$$
I have a simple question, let
$$A=\begin{pmatrix}
  1 &2 &3 \\
  4&5&6 \\
  7&8&9
 \end{pmatrix}, \qquad A^*=\begin{pmatrix}
  1 &4 &7 \\
  2&5&8 \\
  3&6&9
 \end{pmatrix}$$
Then, $\|A\|_1=18, \qquad \|A\|_\infty=24$, and
$\qquad$ $\|A^*\|_1=24, \qquad \|A\|_\infty=18.$
Why $\|A\|_1 \neq \|A^*\|_1$ and similarly for $\|\cdot \|_{\infty}$?
What is going wrong with my calculation here?
 A: An operator and its adjoint are related as that, operators. A norm defined on the entries has no reason to see the relation between $A$ and $A^*$. That is, there is no reason to expect $\|A\|_1=\|A^*\|_1$ are you seem to expect.
What is true is that if you put the operator norm on $A$ and use the induced operator norms on the duals, then $\|A^*\|=\|A\|$.
In this case, $X=Y=\mathbb R^3$.

*

*What you called $\|\cdot\|_1$, is the operator norm when you put the norm $\|(x,y,z)\|_1=|x|+|y|+|z|$ on both $X$ and $Y$. In that case the induced norm on $X^*=Y^*$ is the norm $\|(x,y,z)\|_\infty=\max\{|x|,|y|,|z|\}$.


*What you called $\|\cdot\|_\infty$ is the operator norm when you put the norm $\|(x,y,z)\|_\infty=\max\{|x|,|y|,|z|\}$ on both $X$ and $Y$. In this case the induced norm on $X^*=Y^*$ is the norm $\|(x,y,z)\|_1=|x|+|y|+|z|$.
In summary, if you consider $\|A\|_1=\|A\|_{1,1}$ as you defined it, you have
$$
\|A^*\|_\infty=\|A\|_1
$$
because the second norm is precisely the natural norm on $L(Y^*,X^*)$.
The same happens if you switch the roles of the norm (since we are in finite-dimension) and hence the two equalities you found are not a feature of your example but a general property.
