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**ThePerfectHacker** Let $\displaystyle T: V\to V$ be a linear operator and let $\displaystyle B$ be a basis for $\displaystyle V$. Since $\displaystyle T$ is diagnolizable if we form $\displaystyle [T]_B$ the matrix is a diagnolizable, therefore $\displaystyle [T]_B = ADA^{-1}$ where $\displaystyle D$ is a diagnol matrix whose entires consists of eigenvalues of $\displaystyle [T]_B$. Let $\displaystyle \det(xI - [T]_B) = f(x) = x + a_{n-1}x^{n-1}+...+a_1x+a_0$ be the characheristic polynomial. If $\displaystyle k_1,...,k_n$ are (repeated) eigenvalues of $\displaystyle [T]_B$ then $\displaystyle f(k_1) = ... = f(k_n) = 0$. The problem asks us to show that $\displaystyle [T]_B$ satisfies $\displaystyle X^n + a_{n-1}X^{n-1} + ... + a_1X + a_0I = \bold{0}$. Note that $\displaystyle [T]_B^k = (ADA^{-1})^k = AD^kA^{-1}$. Therefore $\displaystyle AD^nA^{-1} + a_{n-1}AD^{n-1}A^{-1} + ... + a_1ADA^{-1} + a_0AA^{-1}$ upon substituting this matrix into the equation. This becomes $\displaystyle A(D^n+a_{n-1}D^{n-1}+...+a_1D+a_0I)A^{-1}$. Behold, the middle expression. Computing $\displaystyle D^k$ is simply raising each diagnol term to the $\displaystyle k$-th power. Therefore the middle expression consists of a diagnol matrix having its entries values to $\displaystyle f(k_1),...,f(k_n)$. But $\displaystyle f(k_1)=...=f(k_n) = 0$ since they are eigenvalues. Thus, the middle matrix is the zero matrix. Thus, $\displaystyle A(D^n+a_{n-1}D^{n-1}+...+a_1D+a_0I)A^{-1} = A\bold{0}A^{-1} = \bold{0}$. And so $\displaystyle f(T)$ is the zero operator.

Note this theorem (called Cayley-Hamilton) is true even if $\displaystyle T$ is a non-diagnolizable linear operation but that is much harder to prove for we cannot diagnolize.