Let

be a linear operator and let

be a basis for

. Since

is diagnolizable if we form

the matrix is a diagnolizable, therefore

where

is a diagnol matrix whose entires consists of eigenvalues of

. Let

be the characheristic polynomial. If

are (repeated) eigenvalues of

then

. The problem asks us to show that

satisfies

. Note that

. Therefore

upon substituting this matrix into the equation. This becomes

. Behold, the middle expression. Computing

is simply raising each diagnol term to the

-th power. Therefore the middle expression consists of a diagnol matrix having its entries values to

. But

since they are eigenvalues. Thus, the middle matrix is the zero matrix. Thus,

. And so

is the zero operator.

Note this theorem (called Cayley-Hamilton) is true even if

is a non-diagnolizable linear operation but that is much harder to prove for we cannot diagnolize.