Are the eigenvalues of a matrix and just its diagonal related?

Question

I have a matrix $\textbf{A}$ and form the diagonal matrix $\bar{\textbf{A}}$ from the diagonal entries of $\textbf{A}$. Is there a relationship between the eigenvalues of $\textbf{A}$ and $\bar{\textbf{A}}$, or at least between their spectral norms $||\textbf{A}||_2$ and $||\bar{\textbf{A}}||_2$?

Answer 1

The Gershgorin circle theorem bounds the distance of the eigenvalues to the diagonal elements

Answer 2

There actually is a relationship between their spectral norms. In general, the spectral norm of $\textbf{A}$ is greater than any of the entries of $\textbf{A}$, i.e. $$||\textbf{A}||_2\ge\textbf{A}_{ij}, \ \forall i,j$$ The proof for this is as follows: \begin{align*} ||\textbf{A}||_2&=\sqrt{\underset{\textbf{x}}{\text{sup}}\frac{\textbf{x}^T\textbf{A}\textbf{A}^T\textbf{x}}{||\textbf{x}||_2^2}}\\ &=\underset{||\textbf{y}||_2=1}{\text{sup}}\underset{||\textbf{x}||_2=1}{\text{sup}}\textbf{x}^T\textbf{A}\textbf{y}\\ &\ge\textbf{e}_i^T\textbf{A}\textbf{e}_j\\ &=\textbf{A}_{ij} \end{align*} where $\textbf{e}_i$ is a zero vector with a single $1$ in the $i$th entry. It follows that $$||\bar{\textbf{A}}||_2=\max\limits_{i}|A_{ii}|\le||\textbf{A}||_2$$