Is every $A \in \mathrm{SL}_n(\mathbb C)$ a product of four unipotent matrices?

In response to Qiaochu's question in the comments, Fong and Sourour prove in their paper The group generated by unipotent operators that every element of $\mathrm{SL}_n(\mathbb C)$ is a product of three unipotent matrices.

Edit: Sourour proves in the paper A factorization theorem for matrices that a nonscalar matrix in $\mathrm{SL}_n(\mathbb C)$ is indeed a product of two unipotent matrices, confirming YCor's generalization.

If you only need four factors then you can have much more than that. Namely, you can take as the unipotent matrices upper and lower uni-triangular matrices. It was noted by G. Strang in "Every unit matrix is a LULU" that for any field $K$ anfy $A\in M(n,K)$ can be presented as $A=U_1L_1U_2L_2$, where $U_i$ are upper uni-triangular and $L_i$ are lower uni-triangular.

In fact, this is true not only for fields, but also for rings of stable rank $1$ (in particular, for semilocal rings, for boolean rings, for the ring of all entire functions and for the ring of algebraic integers), see this paper (the result was known to the experts long before that, but mostly in terms of triangular decompositions).

As noted in the answer by Gjergji Zaimi, there is a paper by A. Sourour, where he proved that over a field $F$ there is a presentation as a product of three unipotent matrices (two for non-central elements). This can not be carried to the decompositions in terms of lower and upper uni-tringular marices. Namely, every matrix $A\in \operatorname{SL}(n, R)$ is a product $U_1LU_2$ if and only if $R$ is a boolean ring (reference). However, T. Toffoli proved that almost all (w.r.t. Lebesgue measure) elements of $\operatorname{SL}(n,\mathbb{R})$ are ULU, and that for any $A$ there exists a permutation matrix $P$ such that either $PA$ or $-PA$ is ULU.

I have not seen this result before, but here is a proof. Let $e_{ij}$ be the matrix with a $1$ in position $(i,j)$ and all other entries $0$. Let $x_i(t)$ and $y_i(t)$ be the Chevalley generators $\mathrm{Id} + t e_{i (i+1)}$ and $\mathrm{Id} + t e_{(i+1) i}$.

Since being unipotent is unchanged by conjugacy, and every matrix in $\mathrm{SL}_n$ is conjugate to an upper triangular matrix, we may assume that our matrix is upper triangular. So let $A$ be an upper triangular matrix with determinant $1$.

We can choose constants $t_1$, $t_2$, ..., $t_n$ such that $B:= A x_1(t_1) x_2(t_2) \cdots x_{n-1}(t_{n-1})$ has $B_{12} = B_{23} = \cdots = B_{(n-1)n}=1$.

We can then inductively choose $u_1$, $u_2$, ..., $u_{n-1}$ such that the upper left $k \times k$ minor of $C:=B y_1(u_1) y_2(u_2) \cdots y_{n-1}(u_{n-1})$ is $1$. Namely, we choose $u_k$ to make the $k \times k$ minor of $B y_1(u_1) y_2(u_2) \cdots y_k(u_k)$ correct; the later factors won't effect this minor. Since our original matrix is in $\mathrm{SL}_n$, the $n \times n$ minor will also be $1$.

Now, since the upper left minors of $C$ are $1$, the matrix $C$ has an $LU$ factorization where $L$ is lower triangular unipotent and $U$ is upper triangular unipotent. Thus, we have written $$A = L*U* \left( y_{n-1}(-u_{n-1}) \cdots y_2(-u_2) y_1(-u_1) \right)* \left( x_{n-1}(-t_{n-1}) \cdots x_2(-t_2) x_1(-t_1) \right).$$ This is a product of $4$ unipotent factors.

There was a lot of slack in this argument! I wouldn't be surprised if you could prove something much better, either in the direction of non-algebraically closed fields, or in the direction of a more limited class of factors, but this is what you asked for.