How big can a commutative algebra of $n \times n$ matrices be? [duplicate]

Question

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.

Answer 1

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.