Gershgorin Theorem

All eigenvalues of a matrix are contained
in the union of Gershgorin Disc in the complex plane.
i.e. any eigenvalue has:

for some .

Proof

Let be an eigen value of with eigenvector st .
Take the largest absolute coordinate in be
and divide through so that and for
(so WLOG take this at the start).
Now the -th row of gives

Rearrange to get

Finally, take absolute values, apply triangle inequality, and .