Given Ax=λx, take i for which |xᵢ| is largest. Look at the i'th equation: sum aᵢⱼxⱼ = λxᵢ, move the aᵢᵢxᵢ term to the rhs, take absolute values, divide by |xᵢ|, apply triangle inequality, and you have |aᵢᵢ - λ| ≤ sum |aᵢⱼ| over j≠i. So for every eigenvalue you can find such a disc.
That's by column, for row use Aᵀ.
If you had played around a bit with Laplacian matrices, like tri-diagonal matrices with stencil [-1, 2, -1], and found that its eigenvalues are within 2 ± 2, and if you also realized that A + τI has the same eigenvalues shifted by τ, then it's a small step to consider that the magnitude of the off-diagonal may have something to do with the spread of eigenvalues.
It's likely that Gerschgorin stumbled upon it like this.
Just trying to understand these terms. So is a 5x5 tridiagonal matrix with your stencil look like this?
2 -1 0 0 0
-1 2 -1 0 0
0 -1 2 -1 0
0 0 -1 2 -1
0 0 0 -1 2See https://en.wikipedia.org/wiki/Compact_stencil#Three_Point_St...
Math 101, simple is better.