Gershgorin Circles Theorem: A counterexample?

Question

I have the following matrix: $\begin{bmatrix}8 \ 7 \ 7\\ 0 \ 2 \ \frac14\\ 0 \ 3 \ 1 \end{bmatrix}$

So my Gershgorin Circles are

$D(8,14)$

$D(2,0.25)$

$D(1,3)$

The Eigenvalues however are: $8,2.5,0.5$. Therefor the second eigenvalue is outside its gershgorin circle.

What am I missing?

Answer 1

According to the Gershgorin theorem the eigenvlues are within the union of Gershgorin disks.

Only if the disks are disjoint, we have each eignvalue located in its own disk.

In your problem the disks are not disjoint.

Thus there is no conflict here.

Answer 2

Gershgorin says that every eignvalue lies in $\cup D_i$, being $D_i$ the Gershgorin's Circles.

All the eignvalues you work out lies in this union, so there is no problem. I mean, there is no problem that one Circle does not have eignvalues, while the union of all such circles must have all values!