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suppose we have a matrix operator $A=(a_{ij})_{n\times n}:(R^n,\|\cdot\|_{\infty})\to(R^n,\|\cdot\|_{1})$, how to calculate $\|A\|$? or if there really is an exactly closed form for $\|A\|$?

if $x\in (R^n,\|\cdot\|_{\infty})$, then $\|x\|_{\infty}=\max \{|x_n|\}$, and by the definition $$\|A\|=\sup_{{x\in R^n}\atop{\|x\|_{\infty}=1}}\frac{\|Ax\|_1}{\|x\|_{\infty}} =\sup_{{x\in R^n}\atop{\|x\|_{\infty}=1}}\frac{\sum_{i=1}^n\left|\sum_{j=1}^na_{ij}x_j\right|}{\max\{|x_n|\}}.$$

before then, I have thought about the orthogonal transform, such that $P^{-1}AP=J$, $J$ is the Jordan form of $A$. but unfortunately, the norm of $J$ is not the same with $A$, because there are examples with the same Jordan form that have different norm, i.e. $A_1=\left(\begin{array}{cc}1 & n \\ & 2\end{array}\right)$ or $A_2=\left(\begin{array}{ccc}1 & x & y \\ & 2 & z\\ & & 3\end{array}\right)$, the calculation gives $\|A_1\|=n+3$ and $\|A_2\|=6+x+y+z$ for positive $x,y,z$.

So I have to go back to the definition of the norm. But the supremum $$\sup_{{x\in R^n}\atop{\|x\|_{\infty}=1}}\frac{\sum_{i=1}^n\left|\sum_{j=1}^na_{ij}x_j\right|}{\max\{|x_n|\}}$$ is not easy to calculate as in $(R^n,\|\cdot\|_{1})\to(R^n,\|\cdot\|_{1})$: $$\sup_{{x\in R^n}\atop{\|x\|_{\infty}=1}}\frac{\sum_{i=1}^n\left|\sum_{j=1}^na_{ij}x_j\right|}{\sum_{i=1}^n|x_i|} \le \sup_{{x\in R^n}\atop{\|x\|_{\infty}=1}}\frac{\sum_{j=1}^n\sup_j\sum_{i=1}^n|a_{ij}|\cdot |x_j|}{\sum_{i=1}^n|x_i|} = \sup_j\sum_{i=1}^n|a_{ij}|.$$

finally, I've tried some other examples such as $A=\left(\begin{array}{cc}1 & 2 \\ -3 & 4\end{array}\right)$, $x=(x_1,x_2)^T$ with $|x_i|\le 1$. then $\|Ax\|_1=|x_1+2x_2|+|3x_1-4x_2|\le8$, the equality achieved at $x_1=-1$ and $x_2=1$; for $A=\left(\begin{array}{cc}1 & 2 \\ -3 & -4\end{array}\right)$, $\|A\|=10$.

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If the columns of the matrix $A$ are such that all elements of the column have the same sign then in that case the norm is simple to calculate because

$\frac{||Ax||_1}{||x||_\infty}\leq \sum_{i,j}^n |a_{ij}|$

and if you consider the vector

$x_0=(sign(a_{11}),\dots , sign(a_{1,n}))$

then you have that

$\frac{||Ax_0||_1}{||x_0||_\infty}=\sum_{i=1}^n(|\sum_{j=1}^n a_{ij}sign(a_{1j})|=$

$=\sum_{i=1}^n(|\sum_{j=1}^n a_{ij}sign(a_{ij})|=$

$= \sum_{i,j}^n |a_{ij}|$

so you have that

$||A||= \sum_{i,j}^n |a_{ij}|$

for example if you consider

$\left[\begin{matrix}\frac{1}{9} & -\frac{1}{3} & \frac{1}{2} \\ \frac{1}{2} & -3 & \frac{1}{2} \\ \frac{1}{2} &- \frac{1}{2} & \frac{1}{2} \end{matrix}\right]$

then its norm will be

$||A||=\sum_{i,j}^3|a_{ij}|=6+\frac{1}{9}+\frac{1}{3}$

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    $\begingroup$ thanks for your answer, yes, this is a simple situation for my question, so the result for the general one is hard to imagine. $\endgroup$ – Larry Eppes Jun 15 at 8:51
  • $\begingroup$ Exactly, infact it is difficult think to a closed formula in which if the column of the matrix have all the same sign then you have that that formula becomes my simple formula. I don’t know how to generalize better. $\endgroup$ – Federico Fallucca Jun 15 at 8:53

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