$A$ 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!
$A$ 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!
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.