In the very beginning, I'm going to refer to a similar post dealing with theory:Why is dimension of solution space of homogeneous equations n-r?
Let $A,S\in M_n, n\in\mathbb N,$ where $S$ is regular.
$(a)$ Prove that the solution spaces of homogeneous systems $AX=0\;\&\;(SA)X=0$ coincide.
$(b)$ Prove that dimensions of the solution spaces of the homogeneous systems $AX=0\;\&\;(AS)X=0$ are equal.
$(c)$ Let $\operatorname{rank}(A)<\frac{n}2$ Prove that systems $AX=0\;\&\;(AS)X=0$ have at least $1$ non-trivial common solution.
My attempt:
$(a)$ I took the regularity of $S$ into account and saw that left multiplication by $S$ would affect neither the result in the second equation nor change the dimension (what is to be proven in $(b)$). I first thought of proving $$A,S\in M_n\;\&\;\det(S)\ne0\implies[\{A\}]=[\{SA\}]$$ and use the fact that the coefficient matrices $A\;\&\;SA$ are of the same type just like $AX\;\&\;(SA)X$. However, I wasn't sure if $\text{associativity}$ was sufficient. What I know for sure is that $X$ is either a$\text{zero-divisor}$ (depending on $\operatorname{rank}(A)=\operatorname{rank}(SA)$, which is to be proven in $(c)$) or a $\text{null-matrix}$. My notation got complicated, but I should be able to transform the matrices into canonical ones to get the basis.
$A,S\in M_n\implies X\in M_{nk}$
Let $N,P\leq M_{nk}$ $$N:=\{X_{nk}:AX=0, A\in M_n\}, P:=\{X_{nk}: (SA)X=0, A,S\in M_n\}$$ then I have to prove: $$\begin{aligned}N=P\iff& N\leq P\;\&\; P\leq N\\\iff&[\{N\}]=[\{P\}]\\\iff& [\{N\}]\subseteq [\{P\}]\;\&\; [\{P\}]\subseteq [\{N\}]\\\iff& (\forall X'\in N) X'\in P\;\&\;(\forall X''\in P) X''\in N\end{aligned}$$ I decided to express the elements of the following matrices according to properties of the matrix product in order to prove the statement above is true for every $\text{column (row) vector}$ in the vector space-$\text{union of linear spans of}$ $\operatorname{rank}(AX)=\operatorname{rank}((SA)X)$ $\text{vectors}$.
\begin{aligned}(AX)_{im}&=\sum_{j=1}^n a_{ij}x_{jm},\ i\in\{1,\ldots,n\},m\in\{1,\ldots,k\}&\\(AS)_{ij}&=\sum_{l=1}^n a_{il}s_{lj},&\\((AS)X)_{im}&=\sum_{t=1}^n(AS)_{it}x_{tm}&\\&=\sum_{t=1}^n\left(\sum_{j=1}^na_{il}s_{lj}\right)_{it}x_{tm}\forall C=((AX)_{1m},(AX)_{2m},\ldots,(AX)_{im}),j\in\{1,\ldots,n\}\\ &C\in\bigcup_{m=1}^{m=k}[\{(((SA)X)_{1m},((SA)X)_{2m},\ldots,((SA)X)_{im})\}]\end{aligned} And vice-versa for $D=(((SA)X)_{1m},((SA)X)_{2m},\ldots,((SA)X)_{im}),\ m\in\{1,\ldots,k\}$. I don't think I've proven anything so far.
In part $(c)$, $\operatorname{rank}(A)\leq \left\lfloor\frac{n}2\right\rfloor,n=2k-1, k\in\Bbb N\;$ or $\;r(A)\leq \frac{n}2-1\;,n=2k,k\in\Bbb N.$ There should be, as mentioned in the answer to the linked question, more than $\frac{n}{2}\;\text{linearly independent zero-divisors}$?