Consider $$\left\|2\sum\_{i<j}L\_{ij}+4\sum\_i \operatorname{diag}e\_i \right\|,$$ where


(1) $L\_{ij}=\operatorname{diag}e\_i+\operatorname{diag}e\_j-e\_ie\_j^T-e\_je\_i^T$


(2) $e\_i$ denotes $n$-by-$1$ vector with only $i$-th element equals to $1$ and others are $0$


(3) $\operatorname{diag}e\_i$ is a $n$-by-$n$ diagonal matrix with only $i$-th diagonal element equals to $1$ and others are $0.$


I am wondering is it possible to obtain the spectral norm of this matrix, or if it is not possible, can we obtain the upper bound of it?



How to compute the spectral norm of this matrix
Making through the terrible notations, we see that
$$M\_n:=2\sum\_{i<j}L\_{ij}+4\sum\_{i}\text{diag}\,e\_i=(2n+4)I\_n-2\,1\_n1\_n^\top,$$
where $I\_n$ is the $n\times n$ identity matrix and $1\_n$ is the $n\times1$ column matrix of $1$'s.


The eigenvectors of the symmetric matrix $M\_n$ are the nonzero multiples of $1\_n$ and the nonzero vectors orthogonal to $1\_n$. The eigenvalue corresponding to the eigenvector $1\_n$ is $4$, and eigenvalue corresponding to the eigenvectors orthogonal to $1\_n$ is $2n+4>4\ge0$. Thus, $\|M\_n\|=2n+4$.


