# Operator norm of a matrix in terms of its coefficients

Let $$M:\mathbb{C}\to \mathbb{C}$$ be a matrix and equip $$\mathbb{C}$$ with the norm $$\|x\|_\infty=\max_{1\le j\le n}|x_j|.$$ If the operator norm is given by $$\|M\|=\sup_{\|x\|=1}|Mx|,$$ is it possible to compute the operator norm exactly in terms of the matrix entries?

Since $$(Mx)_i=\sum_{j=1}^nm_{ij}x_j$$, we have $$\|M\|=\sup_{\|x\|_\infty=1}\max_{1\le i\le n}\bigg|\sum_{j=1}^nm_{ij}x_j\bigg|.$$ From here, it is clear how one might bound this norm, but it is not clear to me how to compute it exactly without knowledge of the matrix.

• You have written $(Mx)_j$ incorrectly. We have $$(Mx)_i = \sum_{j=1}^n m_{ij}x_j.$$ Commented Jun 16, 2020 at 23:48

Hint: Using the triangle inequality, show that if $$\|x\|_\infty = 1$$, then $$|(Mx)_i| \leq \sum_{j=1}^n |m_{ij}|.$$ This gives you an upper bound for $$\|M\|$$, i.e. a value $$C$$ that depends on the entries of $$M$$ for which $$\|M\| \leq C$$. Using the entries of $$M$$, find a vector $$x$$ for which $$\|x\|_\infty = 1$$ and $$\|Mx\|_\infty = C$$.
• So the upper bound for $\|M\|$ that you mentioned is $C=\max_{1\le i\le n}\sum_{j=1}^n|m_{ij}|$? In that case, we can just take x to be the n-tuple with 1 in every entry. Then its infinity norm is 1 and $\|Mx\|=C$. Commented Jun 17, 2020 at 0:17
• Your upper bound is correct. As for finding the $x$, fix your solution by taking $x$ to have a $-1$ in the $j$'s corresponding to negative entries $m_{ij}$. Commented Jun 17, 2020 at 0:50
• That was my next thought, but it seems like there aren't enough entries in $x$ to record the sign of every $m_{ij}$. In other words, the sign of $x$ would depend on a choice of $i$. Commented Jun 17, 2020 at 1:01
• @zbrads2 It only needs to work for the sign of the $i$th row, where $i$ is such that $\sum_{j=1}^n |m_{ij}|$ is maximized Commented Jun 17, 2020 at 1:38