Is every square matrix a sum of a diagonal matrix and a nilpotent matrix The question is in the title. I am working with complex matrices. This is true for $2\times 2$ matrices, but becomes complicated already for $3\times 3$ matrices if we try to brute force it. By brute force, I mean taking an arbitrary $3\times 3$ matrix and imposing conditions on the diagonal entries so that the matrix is nilpotent.
A follow up question for matrices which can be expressed as a sum of diagonal and nilpotent matrices is whether there are infinitely many ways of doing this? This feels unlikely because the final answer seems to depend on solving $n$ independent polynomial equations in $n$ variables, but I might be wrong.
 A: This is always possible. Given $A \in M_n(\mathbb{C})$, we need to find a diagonal matrix $D$ such that $A - D$ is nilpotent. By a well-known result, this is equivalent to requiring that $\operatorname{tr}((D - A)^k) = 0$ for $1 \leq k \leq n$. Denote the diagonal entries of $D$ by $x_1, \dots, x_n$. Written explicitly, the conditions
$$ \operatorname{tr}(D-A) = \operatorname{tr}((D-A)^2) = \dots = \operatorname{tr}((D-A)^n) = 0 $$
gives us a system of $n$ polynomial equations for $x_1,\dots,x_n$ of degrees $1,\dots,n$:
$$ \operatorname{tr}(D) - \operatorname{tr}(A) = 0, \\
\operatorname{tr}(D^2) - 2\operatorname{tr}(DA) + \operatorname{tr}(A^2) = 0, \\
\operatorname{tr}(D^3) - 3\operatorname{tr}(D^2A) + 3\operatorname{tr}(DA^2) - \operatorname{tr}(A^3) = 0, \\
\vdots \\
\operatorname{tr}(D^n) + \textrm{ lower order terms } + (-1)^n \operatorname{tr}(A^n) = 0.
$$
We can homogenize the equations by adding an extra variable $w$ and then they will have the form:
$$ \operatorname{tr}(D) - w\operatorname{tr}(A) = 0, \\
\operatorname{tr}(D^2) - 2w\operatorname{tr}(DA) + w^2\operatorname{tr}(A^2) = 0, \\
\operatorname{tr}(D^3) - 3w\operatorname{tr}(D^2A) + 3w^2\operatorname{tr}(DA^2) - w^3\operatorname{tr}(A^3) = 0, \\
\vdots \\
\operatorname{tr}(D^n) + w \cdot \left( \textrm{lower order terms} \right) = 0.
$$
Now, Bézout's theorem guarantees that the homogenized system has either infinitely many solutions in
$$\mathbb{P}^n = \{ [x_1 : \dots : x_n : w] \, | \, (x_1, \dots, x_n,w) \in \mathbb{C}^{n+1} \setminus \{ \vec{0} \} \},$$
or, if it has finitely many solutions, then the number of solutions is $1 \cdot 2 \cdot \dots \cdot n = n!$, counted with multiplicities. Any solution $[x_1 : \dots : x_n : w]$ of the homogenized system for which $w \neq 0$ gives us a solution $\left( \frac{x_1}{w},  \dots, \frac{x_n}{w} \right)$ to the original system of equations while solutions of the form $[x_1 \dots : x_n : 0]$ correspond to extra solutions we have possibly added by moving to the projective space. However, if $w = 0$, the homogenized system of equations becomes
$$ \operatorname{tr}(D) = \operatorname{tr}(D^2) = \dots = \operatorname{tr}(D^n) = 0. $$
This implies that $D$ is both nilpotent and diagonal so $D = 0$ which means that the homogenized system has no solutions of the form $[x_1 : \dots : x_n : 0]$. Hence, we haven't added any solutions by moving to the projective space and the bottom line is that the original system has either infinitely many solutions or $n!$ solutions, counted with multiplicities. In any case, such a $D$ always exist.

A few comments:

*

*If you only care about the existence of $D$, you only need to know that the zero set of $n$ homogeneous equations in $\mathbb{P}^n$ is always non-empty. This is certainly weaker than Bézout's theorem but I'm not an expert and have no idea if this result by itself has an "elementary" proof.

*The argument depends heavily on the fact that $\mathbb{C}$ is algebraically closed. You can verify by a direct calculation that the result is false for $2 \times 2$ real matrices.

*Generically, one expects the number of different ways to write $A$ as a sum of nilpotent and a diagonal matrix to be $n!$. It can certainly be less (this already happens for $2 \times 2$ matrices and corresponds to solutions with multiplicities) but I suspect that it will always be finite. Geometrically, this should correspond to the fact that the $n$ different hypersurfaces defined by the equations have "no common components" but I don't know how to justify this rigourously.

