Is my attempt at the following Proof Correct?
PRELIMINATRY NOTATION AND THEOREMS
- $T\in\mathcal{L}(V)$ denotes the set of all linear operator on the vector space $V$.
- $(5.10)$ Given that $T\in\mathcal{L}(V)$ such that $\lambda_1,\lambda_2,...,\lambda_n$ are distinct eignenvalues of $T$ then the corresponding list of eigenvectors is linearly independent.
Theorem. Given that $T\in\mathcal{L}(V)$ such that $\dim\operatorname{range}T = k$ then $T$ has at most $k+1$ distinct eigenvalues.
Proof. We first prove the weaker claim that $T$ has at most $k$ non-zero distinct eigenvalues.
Assume on the contrary that $T$ has $n$ eigenvalues where $n>k$ namely $\lambda_1,\lambda_2,...,\lambda_n$ then using theorem $5.10$ we may establish that the corresponing list of vectors $v_1,v_2,...,v_n$ is linearly independent.
We now show that the list of vectors $Tv_1 = \lambda_1v_1,Tv_2 = \lambda_2v_2,...,Tv_n = \lambda_nv_n$ is linearly independent assume that $$c_1\lambda_1v_1+c_2\lambda_2v_2+\cdot\cdot\cdot+c_n\lambda_nv_n = 0$$ for some $c_1,c_2,...,c_n\in\mathbf{F}$ the linear independece of $v_1,v_2,v...,v_n$ implies that $c_j\lambda_j = 0$ for all $j\in\{1,2,...,n\}$ and since $\lambda_j\neq 0$ it follows that $c_j = 0$, but this implies that we have a linearly independent list of vectors namely $Tv_1,Tv_2,...,Tv_n$ with length larger than $\dim\operatorname{range}T$ an obvious contradiction.
Now returning to our original claim $0$ may or may not be an eigenvalue of $T$ in the event that it is we have $k+1$ distinct eigenvalues if not then we have $k$ distinct eigenvalues in both instances the number of distinct eigenvalues never exceeds $k+1$.
$\blacksquare$