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 $$\left(\begin{matrix} a & 0 & c & d \\ 0 & a & e & f \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a \end{matrix}\right) .$$

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    $\begingroup$ I see where this is coming from, but I would guess not. Can't find a counterexample at the moment though. $\endgroup$ – 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 $4$.

  • $\begingroup$ +1 Thanks for the theorem and for this nice application. $\endgroup$ – Tengu Dec 12 '18 at 12:30

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