Does $\text{End}_K(K^n) \cong \text{Mat}(n\times n, K)$?

Question

Let $K$ be a field, $K^n$ a vector space over $K$.

Is the following true?

$\text{End}_K(K^n) \cong \text{Mat}(n\times n, K)$

Does this change if $K$ is a ring, and $K^n$ a module over $K$?

Answer 1

OK, this is perhaps a really brief answer, but on rschwieb's request...

Yes, it is true - the "obvious" isomorphism (i.e. matrix $A$ is sent to the endomorphism $v \mapsto Av$) works.

If $R$ is a general ring, it will be also true, however, if one considers $R^m$ as a left module over $R$, the multiplication will be reversed - that is, there is an isomorphism $\mathrm{End}_R(R^m)\simeq M_{n}(R^{op})$, not $M_{n}(R)$ in general. If one takes $R^m$ as a right $R$-module, the "op" will not be necessary.