# A Bound on the Dimensions of Certain Types of Subspaces

Let $$V$$ be a $$4$$-dimensional vector space over the complex numbers, and let $$S$$ be a subspace of the endomorphisms of $$V$$ such that the elements of $$S$$ commute.

If there exists an element in $$S$$ that has at least two distinct eigenvalues, is the dimension of $$S$$ at most $$4$$? If so (or if not), why?

An example of such a subspace of dimension $$5$$ that does not have an element with at least two distinct eigenvalues is the set of matrices of the form $$\left(\begin{matrix} a & 0 & c & d \\ 0 & a & e & f \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a \end{matrix}\right) .$$

• I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though. – Matt Samuel Dec 10 '18 at 1:55

By an old theorem of Schur (see this simple proof in an early paper by the late great Maryam Mirzakhani), the maximal number of linearly independent linear endomorphisms of $$\mathbb{C}^n$$ is $$N(n) = \lfloor n^2/4 \rfloor + 1$$. Your example gives the maximum dimension $$N(4)=5$$. However, if some matrix $$A \in S$$ has two different eigenvalues, then $$S$$ will be reducible since every matrix in $$S$$ has these different eigenspaces of $$A$$ as invariant subspaces. For $$n<4$$, one always has $$N(n) = n$$, so the maximum dimension in that case is $$4$$.