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

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 $K$?

• Yes, it is true. No, it will not change if you stick with the free module, i.e. $R^n$ (or $K^n$ in your notation). May 4, 2015 at 16:20
• @PavelČoupek Please put solutions in the solutions section. Consider transferring it out of the comments. Thanks! May 4, 2015 at 17:25

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.
• Thanks! Is it also true that $\text{End}_K(K^n) \cong \text{Mat}(n \times n, \text{End}_K(K)$? I would guess its true for fields (and probably division rings). But why? May 4, 2015 at 17:32
• Yes, this is perhaps the more natural one if linear algebra wouldn't be so well-known. For fields it holds. for example for the reason that $\mathrm{End}_K(K) \simeq K$, see e.g. this post. In the case of division rings, one needs to be a bit careful about the order mutliplication (you can see the discussion in the same post as well). But, again, in principle it should hold for any ring (with the multiplication consideration). May 4, 2015 at 17:46
• Mm... Thats strange then. I have to show that for a division ring $D$ (and as left-modul over itself) $\text{End}_D(D^n) \cong \text{Mat}(n\times n, \text{End}_D(D)) \cong \text{Mat}(n\times n, D^{\text{op}})$. But wouldn't this imply that $D \cong D^{\text{op}}$, wich is (at least in my understanding) false? May 4, 2015 at 18:04
• Sorry, I made a (left-right) mistake in this answer - indeed, $\mathrm{End}_R(R^m)$ is isomorphic to $M_{n}(R^{op})$, and not to $M_{n}(R^{op})$, if $R^m$ is a left $R$-module. May 4, 2015 at 18:44
• "[...] to $M_n(R^{op})$, and not to $M_n(R^{op})$ [...]" Where is the difference? May 4, 2015 at 19:04