Orthogonal idempotents in $M_{4\times 4}(\mathbb{C})$. If $e_1,\cdots,e_n \in M_{4\times 4}(\mathbb{C})$ are $n$ distinct, nonzero, $4\times 4$ matrices with complex entries that satisfy $e_ie_j = e_je_i = 0$ and $e_i^2 = e_i$ for all $1\leq i \leq n$, then I want to show that $n \leq 4$.
My initial instinct was to look at the Jordan Canonical forms of the matrices $e_i$, and note that the minimal polynomial of $e_i$ must divide $x^2-x$, and hence must be $x,(x-1)$ or $x(x-1)$. The minimal polynomial can't be $x$ (since the $e_i$'s were assumed nonzero) and it can't be $x-1$, since then $e_i$ is the identity matrix which is only orthogonal to the zero matrix.
Thus, then minimal polynomial of $e_i$ is $x(x-1)$, and the invariant factor decomposition of $e_i$ must be one of
\begin{align*}
&(x,x,x(x-1)) &(x-1,x-1,x(x-1))& &(x(x-1),x(x-1)).
\end{align*}
However, at this point I realized that this method does not seem to be on the right track, since  there is no reason why the $e_i$'s shouldn't be similar, so knowing their JCF doesn't seem to help me reason why the total number of such matrices should be less than $4$.
Any thoughts or hints would be greatly appreciated.
 A: By this minimal polynomial condition we see that $e_1$ is diagonalizable, so there's a basis in which $$e_1=\begin{bmatrix}I_d & 0 \\ 0 & 0\end{bmatrix}$$ for some $1\leq d\leq 4$. Now we want to use trivial product relations and this basis to analyse the remaining matrices.
Continue for spoiler:

 Since $e_1e_i=0$ for $i\geq 2$, the columns of $e_i$ are contained in the kernel of $e_1$. So in this basis, the columns of $e_i$ must have the first $d$ entries equal to $0$. Similarly, since $e_ie_1=0$ for $i\geq 2$, the rows of $e_i$ must have the first $d$ entries equal to $0$ (in the given basis). So the $e_i$ really live in the space of $(n-d)\times(n-d)$ matrices. Then this problem reduces to a lower dimension and you just keep going until you hit dimension $1$ where the problem is trivial.

A: Consider $I_i=\sum_{j\ne i}e_jR$, where $R=M_{4\times4}(\mathbb{C})$. If $x\in e_iR\cap I_i$, then
$$
x=\sum_{j\ne i}e_jr_j
$$
and also $x=e_ir$, from which $e_ix=e_ie_ir=e_ir=x$, so
$$
x=e_ix=\sum_{j\ne i}e_ie_jr_j=0
$$
We conclude that
$$
R=\bigoplus_{1\le i\le n}e_iR
$$
as a right $R$-module.
Now note that $R$ is a simple artinian ring, so it has a unique minimal right ideal up to isomorphism. Since the right ideal formed by matrices of the form $[u\ 0\ 0\ 0]$ (only the first column being possibly nonzero) is minimal, it has dimension $4$ over $\mathbb{C}$.
Each $e_iR$ is a nonzero right ideal of $R$, so it contains a minimal right ideal. Thus the number of direct summands cannot exceed $4$ or dimension $16$ of $R$ would be exceeded as well.
