I wonder if there exits a general method to find an n by n non-symmetric matrix M that has the maximum $L_2$ norm possible under some interesting constraints. Recall $L_2$ norm of a matrix M can be defined as $$\sup_{\|v\|=1} \|Mv\|$$, or you can think it as the largest singular value of $M$.
In particular, I'm interested in imposing the following 3 constraints all at once to M
(a) The Diagonal of M are all 0's
(b) The sum of each column is 1. (not necessarily rows)
(C) All the other entries in M are within the range $[0,1]$
If M is symmetric, its $L_2$ norm would be largest absolute value of its eigenvalues. I don't think making M symmetric would help since matrix norm is usually larger than its eigenvalue as discussed here: Why is the norm of a matrix larger than its eigenvalue?
If you think these constraints are too specific, please feel free to share some general strategies to find such a matrix M.
Best Answer
First of all, you can equally assume that each row sums up to $1$ because $\|M^T\| = \|M\|$. Let $M$ be such a matrix (with your additional constraints). Then for $\|x\|=1$ we have \begin{align} \|Mx\|^2 &= \sum_{i=1}^n|(Mx)_i|^2 = \sum_{i=1}^n\left(\sum_{j\neq i}m_{ij}x_j\right)^2\le\sum_{i=1}^n\left(\sum_{j\neq i}m_{ij}^2\right)\left(\sum_{j\neq i}x_j^2\right)\\ &= \sum_{i=1}^n(1-x_i^2)\left(\sum_{j\neq i}m_{ij}^2\right)\,\le\,\sum_{i=1}^n(1-x_i^2) = n-1. \end{align} Hence, any such matrix has norm at most $\sqrt{n-1}$.
Now, choose $M_0$ to be the matrix with rows $m_1 = e_2^T$ and $m_i = e_1^T$ for all $i=2,\ldots,n$ (where $e_j$ is the $j$-th standard basis vector). Then $$ \|M_0e_1\|^2 = \|e_2+e_3+\ldots+e_n\|^2 = n-1. $$ Thus, $\|M_0\|=\sqrt{n-1}$ so that $M_0$ is a maximizer.