$T$ is diagonalizable with vector space of finite dimension Let $F$ a field , $V$ a vector space ove $F$ with finite dimension and $T$ a linear operator on $V$. If $T$ is diagonalizable and $c_1,c_2,\ldots,c_n$ are distinct eigenvalues of $T$ and   $\{id_V, T, T^2, \ldots, T^d\}$ is a linearly independent set, then $d<r$, where $r$ is the number of distinct eigenvalues of $T$.
First I note that this is false if $T$ is not diagonalizable. For instance, consider the matrix
$$T=\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right).$$
For prove this i tried by induction for $\{1, T\}$ is linearly independent, but $r=1$ (the only eigenvalue is $0$).
I think I might use the fact that the minimal polynomial of $T$ is of the form $p=(x-c_1)(x-c_2) \cdots (x-c_r)$, where $c_1, c_2, \ldots, c_r$ are the distinct eigenvalues of $T$. I'm not sure of how to use this fact, I'd appreciate any hint to know how to apply this,help me please.
 A: Note that the degree of the minimal polynomial is $r$. Since the set $\{T^i:0\leq i\leq d\}$ is linearly independent(every subset also linearly independent), the degree of any annihilating polynomial should be greater than $d$. In particular  $r>d$, because the minimal polynomial is an annih1lating polynomial.
A: You are right in that induction is not the way to go here. The problem just does not have the kind of structure amenable to induction. Getting rid of one of the eigenvalues of $T$ would amount to replacing $V$ with a carefully selected quotient space or a subspace $V'$, and then making deductions about a relation in $End(V')$ or something like that. I guess it may be possible, but my experience tells me to look for a more direct path.

Hint: If $1,T,T^2,\ldots,T^d$ were linearly dependent, then there exist constants $a_0,a_1,\ldots,a_d\in F$, not all zero, such that $q(T)=0$ for the polynomial
$$
q(T)=a_0+a_1T+a_2T^2+\cdots a_dT^d.
$$
Given that you can assume $T$ to be diagonal what does that tell you about $q(c_i)$, $i=1,2,\ldots,r$?
