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Can anyone give me an example of an algebra of endomorphisms of a vector space which is non-commutative?

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    How many algebras did you come up with that you rejected because they were commutative? It seems really difficult to entirely miss the obvious solutions to this question. So, we are led to believe that you did not come up with many, if any. – 2017-02-06
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    @rschwieb : see my comment to your answer – 2017-02-06
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    and maybe try to realize that I learned what an algebra just this morning, unlike yourself who evidently has PhD in Ring and module theory. – 2017-02-06
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    I understand! But you had to tell me that first :) Providing context in your questions is important: in the future, just mention something about your level of familiarity with algebras (or whatever topic is at hand) and it will help me and other posters meet you on the right level. (Otherwise I'm forced to ask stuff like "well what is going on? why didn't you try anything? etc") I did not intend to put you off (apologies if I did.) – 2017-02-06
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    ^point taken. thanks for your help. – 2017-02-06
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    And if I can offer any more help with any more particulars of this question, just let me know! – 2017-02-06

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For any vector space of dimension greater than $1$, the full ring of endomorphisms is noncommutative. This is because it is isomorphic to a matrix ring of the form $M_n(F)$ with $n>1$ which is always noncommutative.

Any sort of noncommutative subalgebra of these is also fair game. For example, the ring of upper triangular matrices.

Another thing: if you have ever seen two matrices $A,B$ such that $AB\neq BA$, then of course you can just generate a subalgebra of $M_n(F)$ using them, and that subalgebra is not commutative.

If your definition for vector spaces (like mine) allows noncommutative fields (= division rings), then there are even $1$-dimensional vector spaces whose endomorphism rings aren't commutative!

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    I guess I'm not getting the whole picture here (this $\mathbb{K}$-algebra stuff is very new to me). Suppose we take the following vector space $V$: $\mathbb{R}^2$ over the field $\mathbb{K}$. Let $f,g$ be two endomorphisms of $V$ such that $f((x,y))=(f_1(x),f_2(y))$ and $g((x,y))=(g_1(x),g_2(y))$. Then $f(g)=g(f)$ iff $f_1(g_1)=g_1(f_1)$ and $f_2(g_2)=g_2(f_2)$. But $f_1$ and $g_1$ are endomorphisms of the $\mathbb{R}$-vector space over $\mathbb{K}$, which is of dimmension one, and thus their composition is commutative. So the endomorphism algebra of $V$ should also be commutative. – 2017-02-06
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    @JoshuaBenabou Why would you think that every endomorphism $f$ of $\mathbb R^2$ can be written as $f((x,y))=(f_1(x),f_2(y))$? What about, for example $f(x,y)=(y,x)$? – 2017-02-06
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    @ZoranLoncarevic : seems to follow directly from the definition...can you give me a example of an endomorphism which can't be written in this way. – 2017-02-06
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    @JoshuaBenabou The user provided one (just saying something in case he edited it into his comment after your question and it passed you by.) – 2017-02-06
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    @JoshuaBenabou The bottom line is this: the (entire) ring of endomorphisms of $F^n$ is isomorphic to the matrix ring $M_n(F)$. This should give you a complete picture about what it looks like. Every ring of endomorphisms of $F^n$ is going to be a subring of $M_n(F)$. – 2017-02-06
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    @JoshuaBenabou If you'd like to learn more about this isomorphism, start with [the wiki article](https://en.wikipedia.org/wiki/Linear_map#Matrices) and maybe a linear algebra text that defines rings/algebras. – 2017-02-06
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$ \textrm{M}_2(\mathbb R) $ is the $\mathbb R $-algebra of endomorphisms of the $ \mathbb R $-vector space $ \mathbb R^2 $, and it is not commutative.