Gershgorin disk bound Suppose $\mathbf{A}=\left[
 \begin{array}{ccc}
     2 & 2 & 0  \\
     2 & 3 & 4  \\
     0 & 4 & 5
 \end{array}\right]$.   
Give eigenvalue bounds based on Gershgorin disks for $\mathbf{A}$. 
   Note that the disk associated with the $(2,2)$ entry dominates the bounds.
   Consider a diagonal similarity $\mathbf{D}=\mathsf{diag}[1,\, d,\, 1]$ and the family of 
   Gershgorin disks for $\mathbf{D}^{-1}\mathbf{A}\mathbf{D}$ parameterized by $d>0$.
    Find the best eigenvalue bounds for $\mathbf{A}$ that you can derive from
   Gershgorin disks for $\mathbf{D}^{-1}\mathbf{A}\mathbf{D}$ by varying $d$.
Can someone please help me? 
 A: The disks are centered at the diagonal entries, and the radius is the sum of the absolute values of the non-diagonal entries in each row.
The currently the disks are $[0,4], [-3,9], [1,9]$
$D^{-1}AD = \begin{bmatrix} 2 & 2d & 0 \\ \frac {2}{d}&3&\frac {4}{d}\\ 0& 4d& 5\end{bmatrix}$
$D^{-1}AD$ has the same eigenvalues as $DAD$
The disks of $D^{-1}AD$ are $[2-2d, 2+2d], [3-\frac {6}{d},3+\frac {6}{d}, [5-4d, 5+4d]$
We are not going to find a $d$ (other than 1) that brings in the upper bound, as we already have 9 as an upper bound on two disks.
But we can find a $d$ that brings in the lower bound.  Is there a $d$ such that
$2-2d =  3-\frac {6}{d}$ or $5-4d = 3-\frac {6}{d}$
$0 = 2d^2 + d - 6\\
(2d-3)(d+2)\\
d=1.5$
$0= 4d^2 -2d -6\\
2(2d-3)(d+1)\\
d = 1.5$
1.5 is a winner
and the lower bound is $-1$
A: Of course, the matrix is symmetric so it has real eigenvalues.  It suffices to consider real intervals as opposed to disks.
Given your parameterization, we find
$$
D^{-1}AD = \pmatrix{2&2d&0\\2/d & 3& 4/d\\0&4d&5}
$$
The resulting intervals from the rows are
$$
[2 -2d,2+2d]\\
[3 - 6/d,3 + 6/d]\\
[5-4d,4+4d]
$$
So, we need to find the $d$ such that $\max\{2+2d,3+6/d,4+4d\}$ is as small as possible in order to get the tightest possible upper bound for the eigenvalues.  Similarly (but separately), find the $d$ such that $\min\{2-2d,3-6/d,5-4d\}$ is as large as possible in order to get the tightest possible lower bound for the eigenvalues.
