Lowest upper bound on matrix norm Let $A \in \mathbb{R}^{d \times d}$ be an invertible real matrix and $A'$ the matrix obtained from $A$ by setting all diagonal elements to $0$, namely
$$A'_{ij} = \begin{cases} A_{ij} & \text{if } i \neq j \\
0 & \text{otherwise.}
\end{cases}$$
I can prove that $\lVert A' \rVert_2 \leq \min(2, \sqrt{d})\lVert A \rVert_2$ where $\lVert \cdot \rVert_2$ is the $2$-norm (operator norm), but I don't think the bound is tight. I ran $20$ million samples with entries of $A$ generated both uniformly across integers from $-10$ to $10$ and from a standard normal distribution, and got
$$\lVert A' \rVert_2 \leq \begin{cases} \lVert A \rVert_2 & \text{for } d=2 \\
\approx 1.29 \lVert A \rVert_2 & \text{for } d=3 \\
\approx 1.339 \lVert A \rVert_2 & \text{for } d=4 \\
\approx 1.346 \lVert A \rVert_2 & \text{for } d=5 \\
\approx 1.28 \lVert A \rVert_2 & \text{for } d=6.
\end{cases} $$
Any ideas on what the lowest upper bound would be as a function of $d$?
 A: We consider the results for $d=3,4,5$. The bounds $k_d$ are (at least, I think so) 
$k_3=4/3,k_4=3/2,k_5=1.6$ and are reached for
 
The matrices $A_d$ are symmetric and $spectrum(A_d)=\{1,\cdots,1,-1\}$. Moreover, the entries of $A_d$ are fractions in $]-1,1[$ with denominators $d$.
I did not look for a proof, but I did your job. That is important 


*

*It's  not the value of the bound, but the form of the matrices that perform the best during the random tests. 

*To have a minimum of intuition (or experience); after random tests, we feel that the matrices are not very far from being symmetrical and not very far from having eigenvalues ​​with same modulus. Then we randomly test these special matrices and we approach the correct bound much faster...
EDIT. Using the above results, we can formulate 
$\textbf{Conjecture}$. Let $U$ be the $n\times n$  matrix of ones. For each $n$, the considered bound is reached for $A=I_n-\dfrac{2}{n}U$, that is $B=\dfrac{2}{n}(I_n-U)$.
It is easy to see that $||A||_2=1,||B||_2=2\dfrac{n-1}{n}$.
