# How to Calculate the norm of a matrix transform from $(R^n,\|\cdot\|_{\infty})$ to $(R^n,\|\cdot\|_{1})$

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 $$A=\left(\begin{array}{cc}1 & 2 \\ -3 & -4\end{array}\right)$$, $$\|A\|=10$$.

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

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

• thanks for your answer, yes, this is a simple situation for my question, so the result for the general one is hard to imagine. – Larry Eppes Jun 15 at 8:51
• 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. – Federico Fallucca Jun 15 at 8:53