$k$ be an algebraically closed field , $R=M(n,k)$ , then $k \cong End_R (k^n)$

Question

Let $k$ be an algebraically closed field. Let $n \ge 1$ be an integer and let $R=M(n,k)$ . Now every element $f \in R=M(n,k)$ is a $k$-linear map $f:k^n \to k^n$, hence $k^n$ forms a left $R$-module with the external multiplication defined as $f.v:=f(v)$. Let $End_R(k^n)$ be the set of all $R$-module maps on $k^n$, so that $End_R (k^n)$ forms a ring with unity w.r.t. usual pointwise addition and function composition. Then how to show that $k \cong End_R (k^n)$ as rings with unity ?

Answer 1

If you unpack the definition, then it becomes clear that such an endomorphism is a linear map that commutes with all other linear maps. Equivalently a matrix that commutes with all other matrices.

It's a common exercise to show that if a matrix commutes with all other matrices, then it is a multiple of the identity. Consider the elementary matrices $E_{ij}$ with 1 in position (i,j) and zero elsewhere. Then first, checking what it means that your matrix commutes with $E_{ii}$, you find that all off-diagonal entries are zero. Then, by checking what it means that the matrix commutes with $E_{ij}$, you find that all the diagonal entries are equal.

So now you have an obvious morphism $k \to End_R(k^n)$ given by $\lambda \mapsto \lambda \operatorname{id}$. It's clearly injective, and the paragraph above shows that it's surjective, so it's an isomorphism.