Proving that an $n\times n$ matrix has at most $n$ distinct eigenvalues  $A$ is a $n\times n$ matrix over the field $F$. How can I prove that there are at most $n$ distinct scalars $c$ in $F$ such that $\det(cI - A) = 0$?
Thank you!
 A: If there are $c_1,\ldots,c_{n+1}$ $n+1$ distinct elements of $F$ such that for all $1\leq j\leq n+1$ we have $\det(c_jI-A)=0$ then $c_jI-A$ is not invertible. Hence we can find a vector $v_j\in F^n$ such that $Av_j=c_jv_j$. The family $\{v_j,1\leq j\leq n+1\}$ is linearly independent, otherwise there would exists a $j_0$ such that $\{v_1,\ldots,v_{j_0}\}$ is linearly independent, but not $\{v_1,\ldots,v_{j_0+1}\}$. So $v_{j_0+1}=\sum_{k=1}^{j_0}\alpha_kv_k$ for $(\alpha_1,\ldots,\alpha_{j_0})\neq (0,\ldots,0)$ and 
$$Av_{j_0+1}=c_{j_0+1}v_{j_0+1}=\sum_{k=1}^{j_0}\alpha_kc_kv_k.$$
We get 
$$\sum_{k=1}^{j_0}\alpha_kc_kv_k=\sum_{k=1}^{j_0}\alpha_kc_{j_0+1}v_k$$
so by linear independence $\alpha_k(c_k-c_{j_0+1})=0$ for all $1\leq k\leq j_0$. Since $c_k-c_{j_0+1}\neq 0$ we get $\alpha_k=0$, a contradiction.
A: The charecteristic polynomial of $A\in M_{n x n}(\mathbb R)$ is a polynomial of degree n with leading coefficient $(-1)^n$ and eigenvalues are zeros of this charesterictic polynomial so by Fundamental Theorem of Algebra n degree polynomial can have at most n roots.
