If $n$ is given and $A$ is a subalgebra of $M_n(\mathbb C)$, the algebra of $n \times n$ matrices with entries in the field of complex numbers, then what are the possible values of dimension of $A$ as a vector space over $\mathbb C$?

$\begingroup$ Ok, I just found out that there is an interesting result due to Schur which gives a partial answer to my question. Here it is for those who are interested: If F is a field, then there exists a "commutative" subalgebra A of M_n(F) with dim_F A = k if and only if k \leq [n^2/4] + 1, where [ ] is the floor function. I'm starting to think that there exists a subalgebra of M_n(F) of any dimension! $\endgroup$ – abcba Mar 28 '10 at 8:02

1$\begingroup$ @abcba: here's a hint for constructing that commutative subalgebra: write an n x n matrix as (A B;C D) with A,B,C,D n/2 x n/2 matrices, and then consider the space with A=C=D=0. To get further start eating into B. Add scalar multiples of the identity if you're the sort of person whose algebras have to contain 1. $\endgroup$ – Kevin Buzzard Mar 28 '10 at 8:05

4$\begingroup$ Is there an 8 dimensional subalgebra of M_3? $\endgroup$ – Jonas Meyer Mar 28 '10 at 8:35

3$\begingroup$ One can get all dimensions up to $n(n+1)/2$ by using subalgebras of upper triangular matrices. We can alo get some larger examples by the construction $(A\ B;0\ D)$ where $A$ and $D$ run through given subalgebras of $M_k$ and $M_{nk}$ and $B$ is arbitrary. Some dimensions are not accessible by these constructions, e.g., dimension $8$ when $n=3$. Are there any subalgebras with these dimensions? $\endgroup$ – Robin Chapman Mar 28 '10 at 8:49

3$\begingroup$ A nice proof of Schur's theorem is at M. Mirzakhani `A simple proof of a theorem of Schur' Amer. Math. Monthly 105 (1998), 260262. $\endgroup$ – Robin Chapman Mar 28 '10 at 8:51
Rough answer : almost all small dims can appear, there are some restrictions to large dims.
For example, considering 1 matrix all dims between 1 and n appear. Taking centralizers of these all numbers of the form sum a_i^2 where a is a partition of n appear.
In general, consider ktuples of positive integers a and b such that their scalar product a.b=n (a should be thought of as the Morita setting, b as the matrixsizes of the semisimple part of the subalgebra), then any number of the form
sum b_i^2 + subsum b_ib_j
is possible (here 'subsum' means that one takes all terms b_xb_y for all x,y in a substring
1 <= i_1 < i_2 < ... < i_l <=k for any 0<=l<=k)
Edit : the subsum gives the dimension of the Jacobson radical. This answer cannot be the final one, as it only detects the subalgebras of global dimension 1. For example any ndiml algebra can be embedded in nxn matrices.
There are some obvious restriction wrt large dimensions. For example, there cannot be an 8dml subalgebra of 3x3 matrices as its semisimple part can be at most C x M_2(C) and so its dimension must be smaller or equal to 7.
For general n there cannot be subalgebras with dimensions between the dim of the largest parabolic subgroup of GL(n) and n^2.
Edit : a closely related question can be found here : problems concerning subspaces of mxm matrices.
Soit $E$ un $\mathbb C$espace vectoriel de dimension $n$. J'ai démontré entre autres les deux résultats suivants dans un article à paraître dans la revue française Quadrature :
On suppose que $k$ vérifie les inégalités $k \ge 2$ et $k^{2}\le n$. Soit $\mathcal{A}$ une sousalgèbre de $\mathcal{L}(E)$ qui vérifie la relation $n^{2}kn+k^{2}k+1 < \dim \mathcal{A} < n^{2}kn+n.$ Alors, $\mathcal{A}$ vérifie la relation $\dim \mathcal{A}=n^{2}kn+k^{2}.$
Soient $n$ un entier naturel et $p$ un entier de l'intervalle $[0,n^{2}].$ On suppose $p$ écrit sous la forme $p=n(nk)+t,\ 0\le t \le n1$. Alors il existe une sousalgèbre de dimension $p$ dans $\mathcal M_n (\mathbb C )$ si et seulement s'il existe une sousalgèbre de dimension $t$ dans $\mathcal M_k(\mathbb C)$.

1$\begingroup$ Is there a copy of this paper available on the internet? $\endgroup$ – S. Carnahan♦ Jul 4 '11 at 2:42

$\begingroup$ On peut consulter mon article à l'adresse : logique.jussieu.fr/~chalons/z2009/articleabou.pdf Bonne lecture $\endgroup$ – abou Jul 8 '11 at 0:01
I think that the fact that every proper subalgebra is contained in am maximal parabollic follows immediately from Jacobson's density theorem because if a subalgebra does not preserve any subspace, then $C^n$ is a simple module for it. This is of course true over any field.
In the case of Lie algebras rather than associative algebras, then a classification of maximal subalgebras of finite dimensional simple Lie algebras over the complex numbers was obtained by Dynkin. In the positive characteristic case a classiifcation can probably be obtained using arguments which were used for the classifcation of maximal subgroups of finite simple groups. This is at least what I understood talking to Liebeck and Seitz, but I am not an expert on these matters.
However, in the Lie case an elementary argument that the maximal dimension of a proper subalgebra of $sl_n(F)$ is $n^2n$, assuming $F$ has characteristic different than 2 can be found in Y. Barnea and A. Shalev, Hausdorff dimension, prop groups, and KacMoody algebras, Trans. Amer. Math. Soc. 349 (1997), 50735091 (Theorem 1.7). Other related stuff (related to possible dimensions) but more on the group theoretic side can be found in the same paper. A generalization of this to other classical Lie algebras can be found in Abért, Miklós; Nikolov, Nikolay; Szegedy, Balázs Congruence subgroup growth of arithmetic groups in positive characteristic. Duke Math. J. 117 (2003), no. 2, 367383 (Theorem 4).