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 .