Question:
Let $V$ be a finite dimensional vector space and let $T:V\rightarrow V$ be a linear map. Suppose that $T$ is diagonalizable. Show that $Ch_T(T)=0$ without using the Cayley-Hamilton theorem. We are told as a hint to use: $f(T)\vec{v}=f(\lambda)\vec{v}$ where $\lambda\text{ and }\vec{v}$ are an eigenvalue and corresponding eigenvector of T. This was proved in the prior question. I would like to know if this is a valid proof.
Attempt:
Since $T$ is diagonalizable, $\exists ~~S\in\mathbb{R}^{n\times n}\text{ such that }D=S^{-1}TS$ Where $D$ is a diagonal matrix. Thus Matrices $D$ and $T$ are similar and they have the same characteristic equation and the same eigenvalues i.e. $Ch_T=Ch_D\text{ and } \lambda_T=\lambda_D$.
We have that $f(T)\vec{v}=f(\lambda)\vec{v}$, now let $f = Ch_T $.
$$Ch_T(T)\vec{v}=Ch_T(\lambda_T)\vec{v}$$
$$Ch_T(T)\vec{v}=Ch_D(\lambda_D)\vec{v}$$
Since $D$ is diagonal $Ch_D(\lambda_D)=0$ because there will be at least one multiple of zero in its factors. Thus:$$Ch_T(T)\vec{v}=0$$And since $\vec{v}$ is non-zero (its an eigenvector) this implies that $Ch_T(T)=0._{~~\square}$
Attempt 2:
Since $T$ is diagonalizable, there exists a basis $B$ consisting of eigenvectors of $T$ such that $[T]_B$ is diagonal. Let $B=\{b_1,b_2,\dots,b_n\}$. In the basis $B$ the characteristic equation of $T$ has the following factorisation: $Ch_T(x)=(x-\lambda_1)(x-\lambda_2)\dots(x-\lambda_n)$. Need to show that $Ch_T(T)\vec{v}=0$. Also note that $v=\sum_{k=1}^n a_kb_k$ where $a_k$ is an element of the field.
$$Ch_T(T)\vec{v}=\prod_{i=1}^n[T-\lambda_iI]\sum_{k=1}^n[ a_kb_k]$$
$$~~~~~~~~~~~~~~~~~~=\sum_{k=1}^n\prod_{i=1}^n[Ta_kb_k-\lambda_iIa_kb_k]$$
$$~~~~~~~~~~~~~~~~~~=\sum_{k=1}^n\prod_{i=1}^n[a_k\lambda_kb_k-a_k\lambda_ib_k]$$
$$~~~~~~~~~~~~=\sum_{k=1}^n\prod_{i=1}^n(\lambda_k-\lambda_i)a_kb_k$$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=0_{~~\square}$
That last step is because in every term in the sum, there will be a factor of zero in the product when k=i.
Best Answer
Your approach will lead to the right answer (but you may need to add the facts about an eigenbasis mentioned by @MorganRodgers)
An alternate approach is to observe that $A^n=SD^nS^{-1}$ where $A$ is the matrix of the linear transformation $T$ (after choosing a basis). Then, plugging in $A$ into the characteristic polynomial, you have $$ Ch_T(A)=S\begin{bmatrix}Ch_T(\lambda_1)\\&\ddots\\&&Ch_T(\lambda_n)\end{bmatrix}S^{-1} $$ But, since the roots of the characteristic polynomial are the eigenvalues, the diagonal matrix is actually the zero matrix. Therefore, $Ch_T(A)=0$.