This a follow-on of this question "Q" asked some days ago on Maths SE to which I had given an answer that will be denoted "A".
This question "Q" was dealing with the mapping :
$$f_{A_0}(X)=A_0^2X+A_0XA_0+XA_0^2$$
In fact (it wasn't mentionned in "Q"), this mapping is, in the 9-dimensional space $\frak{M}_{3 \times 3}$ of $3 \times 3$ matrices, the differential of function $f:A \to A^3$ in "point" $A_0$.
As such $f_{A_0}$ is a linear function having a certain rank $r(f_{A_0})$ which can be maximal, i.e., equal to $9$ (for most $A_0$) or have a rank deficiency (equivalent naming : "rank drop"), like $8$ for some $A_0$ or even less in very rare cases (see statistical study below).
It will be easier to work on the complementary value of the rank, i.e., the dimension of the kernel :
$$\underbrace{\dim(\ker(f_{A_0}))}_k = 9 - \underbrace{\operatorname{rank}(f_{A_0})}_r$$
Taking a pragmatic approach, I have done a large scale simulation on matrices $A$ having integer entries between $-3$ and $3$ which has given these approximate proportion of values for the dimension $k$ of the kernel of $f_{A_0}$ :
$$\begin{array}{c|ccccccc}k&0&1&2&3&4&5&6 \\\hline &97 \% &2.6 \%&0.2\%&0.1 \%&0.0 \%&0.0 \%&0.1 \%\end{array}$$
Side remark : In fact the relative percentages aren't important ; the important fact is the "holes" in the spectrum : no matrices for $k=4,5,7,8$. I know that this approach doesn't meet the approval of some readers : this question has been downvoted probably for this reason, but I don't bother, I think that all the tools we now have, in particular simulation methods, are good to use in order to establish conjectures.
Question 1 : Set apart the fact that in a large majority of cases this differential is injective, how is it possible to explain the presence of these particular cases ($k=1, 2,3,6$) and the absence of other cases ?
Let us enter into the details.
- Case $k=1$ : for matrices $A$ having a one-dimensional kernel, I observe on my simulations that this kernel is always with rank-1 matrices $X$. Why ? Moreover, one doesn't see the rationale existing between $A$ and $X$ (btw, it was the main question in question "Q").
Two examples of a matrix $A$ and an element $X$ of the kernel of $f_A$ :
$$A=\begin{pmatrix}0&-1&1\\1&-2&2\\-1&2&2\end{pmatrix} \text{ and } X=\begin{pmatrix}0&0&0\\0&1&1\\0&1&1\end{pmatrix}$$
$$A=\begin{pmatrix}1&2&0\\2&1&0\\0&-2&0\end{pmatrix} \text{ and } X=\begin{pmatrix}0&0&0\\0&0&0\\4&-2&3\end{pmatrix}$$
Let us turn now to other cases $k=2,3,6$ where $X$ has (necessarily) rank $1$ :
- Case k=2 : examples
$$A=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix} \text{ and } X_1=\begin{pmatrix}0&-1&0\\0&0&1\\0&0&0\end{pmatrix}, X_2=\begin{pmatrix}-2&2&0\\0&1&2\\-4&0&1\end{pmatrix}$$
- Case k=3 : example
$$A=\begin{pmatrix}0&1&1\\1&0&0\\-1&1&0\end{pmatrix}$$
We are here in a particular case of a matrix where rank$(A)=2$, rank$(A^2)=1$ and $A^3=A^2$.
- Case k=6 : example
$$A=\begin{pmatrix}-1&1&-2\\2&0&1\\2&1&1\end{pmatrix}.$$
This matrix has a very specific property $A^3=3I$. In fact, all matrices of this category ($k=6$) are such that $A^3=\mu A$ for a certain real number $A$. Why that ? (Another example of this special case $k=6$ can be seen in my former answer "A").
Question 2 (the main question in fact) How can the disparate observations seen above be understood and proved under a common globalizing explanation ?