Return to Answer

Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.

If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.

Assume that $K=k(x)$

$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$g_x(v^{q^l}) =(g_x(v))^{q^l}=(xv)^{q^l}=x^{q^l}v^{q^l}$$ Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and we have diagonalized $g_x$ $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$

If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$, wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. If the latter is reducible then its degree divides $n$ so that $F\subset K$. As before we obtain the diagonalization from $$h(v)=av \implies h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$ Finally $a = f(x)$ for some $f\in k[T]$ and $$h = f(g_x)=g_a$$

Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.

If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.

Assume that $K=k(x)$

$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$g_x(v^{q^l}) =(g_x(v))^{q^l}=(xv)^{q^l}=x^{q^l}v^{q^l}$$ Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and we have diagonalized $g_x$ $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$

If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$, wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. As before we obtain the diagonalization from $$h(v)=av \implies h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$ $v^{q^n}= v$ implies that $a^{q^n}=a$ thus $a\in K$. Whence $a = f(x)$ for some $f\in k[T]$ and $$h = f(g_x)=g_a$$

Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.

If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.

Assume that $K=k(x)$

$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$g_x(v^{q^l}) =(g_x(v))^{q^l}=(xv)^{q^l}=x^{q^l}v^{q^l}$$ Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and we have diagonalized $g_x$ $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$

If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$, wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. If the latter is irreducible then its degree divides $n$ so that $F\subset K$. We obtain the diagonalization $$h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$ Finally $a = f(x)$ for some $f\in k[T]$ and $$h = f(g_x)=g_a$$

Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.

If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.

Assume that $K=k(x)$

$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$g_x(v^{q^l}) =(g_x(v))^{q^l}=(xv)^{q^l}=x^{q^l}v^{q^l}$$ Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and we have diagonalized $g_x$ $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$

If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$, wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. If the latter is reducible then its degree divides $n$ so that $F\subset K$. As before we obtain the diagonalization from $$h(v)=av \implies h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$ Finally $a = f(x)$ for some $f\in k[T]$ and $$h = f(g_x)=g_a$$

Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.

If $k(x)=\Bbb{F}_{q^d}$ with $d<n$ then every $\Bbb{F}_{q^d}$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.

Assume that $K=k(x)$

$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$x^{q^l}v^{q^l}=(xv)^{q^l}=(g_x(v))^{q^l}=g_x(v^{q^l})$$ Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and hence $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$

If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$, wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. If the latter is irreducible then its degree divides $n$ so that $F\subset K$. We obtain as before $$h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$ Finally $a = f(x)$ for some $f\in k[x]$ and $$h = f(g_x)=g_a$$

Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.

If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.

Assume that $K=k(x)$

$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$g_x(v^{q^l}) =(g_x(v))^{q^l}=(xv)^{q^l}=x^{q^l}v^{q^l}$$ Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and we have diagonalized $g_x$ $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$

If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$, wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. If the latter is irreducible then its degree divides $n$ so that $F\subset K$. We obtain the diagonalization $$h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$ Finally $a = f(x)$ for some $f\in k[T]$ and $$h = f(g_x)=g_a$$