$T^n=1$ then the vector space has $T-$eigenvectors. I have been working on this problem:-
Let $T:V\rightarrow V$ be a linear transformation on an $n$-dimensional vector space $V$ over $\mathbb{C}$ such that $T^{n}=1$. Prove that $V$ has a basis of $T-$eigenvectors. 
I feel I need to use idea of diagonalizable of linear transformation but not sure how wisely to used. any idea will appreciated.
 A: The minimal polynomial of $T$ divides $x^n - 1$. Over $\mathbb{C}$, the polynomial $x^n - 1$ splits into $n$ distinct linear factors and thus the minimal polynomial of $T$ also splits into distinct linear factors implying that $T$ is diagonalizable.
A: Whenever you have a polynomial $p(\lambda)=\prod_{j=1}^{k}(\lambda-\lambda_j)$ with no repeated roots for which $p(T)=0$, then $T$ is diagonalizable, meaning that it has a basis of eigenvectors. The key observation is that
$$
               (A-\lambda_l)\prod_{j\ne l}(A-\lambda_j)=0.
$$
Therefore, if you define $p_{l}(\lambda)=\prod_{j\ne l}(\lambda-\lambda_j)$, you see that the range of $p_{l}(A)$ is in the null space of $A-\lambda_l I$; in other words, everything in the range of $p_{l}(A)$ is either $0$ or is an eigenvector of $A$ with eigenvalue $\lambda_l$. To show that you have a basis of such eigenvectors uses the fact that
$$
                 1 = \frac{p_{1}(\lambda)}{p_{1}(\lambda_l)}+\frac{p_2(\lambda)}{p_2(\lambda_2)}+\cdots+\frac{p_k(\lambda)}{p_k(\lambda_k)}.
$$
Therefore,
$$
              I = \frac{1}{p_1(\lambda_1)}p_1(A)+\frac{1}{p_2(\lambda_2)}p_2(A)+\cdots+\frac{1}{p_k(\lambda_k)}p_k(A),
$$
which means that every vector $x$ can be written terms of eigenvectors of $A$:
$$
           x = \frac{1}{p_1(\lambda_1)}p_1(A)x+\frac{1}{p_2(\lambda_2)}p_2(A)x+\cdots+\frac{1}{p_k(\lambda_k)}p_k(A)x.
$$
