TITLE: How big can a commutative algebra of $n \times n$ matrices be?
QUESTION [6 upvotes]: What's the maximum possible dimension of a commutative subalgebra of the algebra of $n \times n$ complex matrices?  
There's a theorem of Burnside saying that any commutative subalgebra of a matrix algebra can be upper triangularized.  My friend Bruce Smith pointed out that for $n$ even we can get a commutative subalgebra of dimension $(\frac{n}{2})^2 + 1$.  For $n = 4$ its elements look like this:
$$\left( \begin{array}{cccc}
a & 0 & b & c \\
0 & a & d & e \\
0 & 0 & a & 0 \\
0 & 0 & 0 & a 
\end{array}\right)$$
and the same trick works in any even dimension.  For $n$ odd we can get dimension $\frac{(n-1)(n+1)}{2} + 1$ using a similar idea, with a rectangle rather than a square of nonzero entries in the upper right corner.
Can one do better?  Someone must have figured this out.

REPLY [2 votes]: Thanks, Christian Remling!  There's a much harder question on MathOverflow to which Robin Chapman gave an answer to my question here.  
In short: Bruce's guess is indeed the best we can do.  This fact was proved by Schur, and there's also a proof here:

Maryam Mirzakhani, A simple proof of a theorem of Schur, Amer. Math. Monthly 105 (1998), 260–262.