Gershgorin discs and norm of a matrix

Question

Find a matrix, where the estimation of eigenvalues with the help of Gershgorin discs is

a, the same as

b, worse as

the estimation with the help of the norm of the matrix ($||A||_\infty$)

So, yes, it is a homework. And I'm not asking you to solve it for me (I know you guys hate that anyway) I just don't get the problem.

For a, e.g. the matrix

\begin{equation} A=\begin{bmatrix} 0 & 3 & 4 \\ 4 & 0 & 3 \\ 3 & 4 & 0% \end{bmatrix} \end{equation}

has a norm of $||A||_\infty=7$ and there are $3$ discs with a $[0,0]$ centre and with a $7$ radius. So it's basically the same as the estimation with the help of the norm.

About the second one, I'm clueless.

Answer 1

The point which is furthest away from $(0,0)$ in a Gershgorin's disk with center in $a_{ii}$ is the point $$\frac{a_{ii}}{|a_{ii}|}\left(|a_{ii}|+\sum_{j\ne i}|a_{ij}|\right)$$(easy to prove by drawing this disc). The distance is therefore $\sum_{j}|a_{ij}|$, which immediately gives us that Gersgorin's estimation can't be worse than $\|\cdot\|_\infty$.

As a matter of fact, the proof of Gershgorin's lemma involves the reasoning around $\|x\|_\infty$ where $x$ is the eigenvector.

Answer 2

We can certainly find an example where the Gershgorin estimate is superior. For example, we can take $$ B = \pmatrix{1&0\\0&2} $$ (or any non-zero diagonal matrix, for that matter). As for the other way around: the estimation with Gershgorin disks can ever be worse than the estimation using the matrix norm. In general, $\|A\|_\infty$ gives the radius of the smallest circle centered at the origin containing all Gershgorin disks.