Diagonalizable Matrix and Linear dependence. I am tried to solve an exercise but for this I need to show that if  $\hspace{0.2cm }T%$ is a Linear Operator of a vector space $V$ $\;(\dim V = n< \infty)$  and T is diagonalizable with the algebraic multiplicity of eigenvalues ​​1. Then the set {$T, T^2 ,...,T^{n-1}$} is linearment independent.
I know that  there is a matrix $A$ of $T$ and $A$ is diagonal with all diagonal entries distinct, and the matrix of $T^2$ is $A²$, but how to prove that the set {$I,T, T^2 ,...,T^{n-1}$} is L.I?
 A: If $a_0 I + a_1 A + a_2 A^2 + \cdots + a_{n-1} A^{n-1} = 0$ where $A$ is diagonal with diagonal entries $\lambda_1, \ldots, \lambda_n$,
then the vector $(a_0, \ldots, a_{n-1})$ is in the nullspace of a certain Vandermonde matrix, namely
$$\begin{bmatrix}1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \lambda_n & \lambda_n^2 & \cdots & \lambda_n^{n-1}\end{bmatrix}.$$
If this matrix is invertible, then we must have $a_0 = \cdots = a_{n-1} = 0$.
A: Let $v$ be an eigenvector, $Tv=\lambda v$ for some $\lambda $.  Then $\{v,Tv,T^2v,\dots,T^{n-1}v\}=\{v,\lambda v,\dots,\lambda ^{n-1} v\}$.  Thus, if we let $\{v_1,\dots,v_n\}$ be a basis of eigenvectors,  then for any linear combination $\alpha_0I+\alpha _1T+\dots+\alpha_{n-1}T^{n-1}$, we have the system  $V=\begin{pmatrix}1&\lambda_1&\dots&\lambda_1^{n-1}\\1&\lambda_2&\dots &\lambda _2^{n-1}\\\vdots&\vdots&\ddots\\1&\lambda_n&\dots&\lambda_n^{n-1}\end{pmatrix}\begin{pmatrix}\alpha_0\\\alpha_1\\\vdots\\\alpha_{n-1}\end{pmatrix}=0$. 
The only solution is the trivial solution,  since the determinant of the Vandermonde matrix is $\prod_{i\lt j}(\lambda_j-\lambda_i)\neq0$.
A: Since the algebraic multiplicity of every eigenvalue is $1$, each occurs precisely once as a root of the characteristic polynomial $\chi_T(x)$ of $T$; thus the eigenvalues of $T$ are distinct, and since
$\dim V = n < \infty, \tag 1$
$T$ is an $n \times n$ matrix, so the characteristic polynomial $\chi_T(x)$ satisfies
$\deg \chi_T(x) = n; \tag 2$
it now follows from the fact that $T$ is diagonalizable that $\chi_T(x)$ splits completely into linear factors, and thus $T$ has $n$ distinct eigenvalues.  
Now suppose the set
$\{I, T, T^2, \ldots, T^{n -1} \} \tag 3$
were linearly dependent; then there would exist $n$ scalars $\alpha_i$, $0 \le i \le n - 1$, such that
$\exists j, 0 \le j \le n - 1, \; \alpha_j \ne 0; \tag 4$
and
$p(T) = \displaystyle \sum_0^{n - 1} \alpha_i T^i = \alpha_0 I + \alpha_1 T + \alpha_2T^2 + \ldots + \alpha_{n - 1} T^{n - 1} = 0; \tag 5$
if $\mu_k$ is an eigenvalue of $T$ with eigenvector $\vec v_k \ne 0$, 
$T \vec v_k = \mu_k \vec v_k, \tag 6$
then it is easy to see that, for any natural $m \in \Bbb N$,
$T^m \vec v_k = \mu_k^m \vec v_k; \tag 7$
thus
$p(\mu_k) \vec v_k  = \left ( \displaystyle \sum_0^{n - 1} \alpha_i \mu_k^i \right ) \vec v_i = \left ( \displaystyle \sum_0^{n - 1} \alpha_i T^i \right ) \vec v_i = p(T) \vec v_i= 0, \tag 8$
by virtue of (5); since $\vec v_k \ne 0$, this forces
$p(\mu_k) = 0, \; 1 \le k \le n; \tag 9$
but
$\deg p(x) = \deg \left (\displaystyle \sum_0^{n - 1} \alpha_i x^i \right ) = n - 1, \tag{10}$
hence $p(x)$ may have at most $n - 1$ zeroes; but since the $\mu_i$ are distinct, this contradicts (9), and thus the assumption (5) cannot bind; the $T^i$ must be linearly independent.
