Let $R\in\mathbb{R}^{d\times d}$ be an upper triangular matrix.
The eigenvalues of $R$ is then its diagonal entries, that is, $(R_{ii}, e_i)$ is an eigenpair of $R$ since $Re_i=R_{ii}e_i$.
Is it possible to bound the largest singular value $\sigma_\max (R)=\sqrt{\lambda_\max (RR^T)}$ by the diagonal entries of $R$?
Weyl's inequality tells us $\lambda_\max \le \sigma_\max$ which is the wrong direction. But maybe we can get something like $\sigma_\max < \lambda_\max^2$ assuming $\lambda_\max>1$? Or can the largest singular value be arbitrarily larger?
Is it possible to bound the largest singular value $\sigma_\max (R)=\sqrt{\lambda_\max (RR^T)}$ by the diagonal entries of $R$?
If you look at the unbounded nature of the Frobenius norm, the answer of course is no.
with standard basis vectors $\mathbf e_k$, consider
$R' := R +\alpha \mathbf e_1\mathbf e_d^T$
$R$ and $R'$ have the same diagonal entries and hence would have the same purported fixed upper bound $M$. Yet by selecting large enough $\alpha$ we have
$M\lt \frac{1}{d}\big \Vert R'\big\Vert_F \leq \sigma_1$