Let $A_{3x3},B_{3x3}$, and $AB=0, BA \neq 0$ the span of the homogeneous for $Ax=0$ is $Sp\{(1,1,1)\}$ find $\ rank(B)$

Question

I have this problem :

Let $A_{3x3},B_{3x3}$, and $AB=0, BA \neq 0$ the span of the homogeneous for $Ax=0$ is $Sp\{(1,1,1)\}$ find $rank(B)$.

Since $BA \neq 0$ and $B_{3x3} \implies 0 < rank(B) \leq 3$

Since $A_{\rightarrow}B_{\downarrow}=0$ Meaning every column in B is a solution for the $A_{i \rightarrow} (i=0,1,2)$ homogeneous equation

Since we know that the solution of $Ax=0$ is $Sp\{(1,1,1)\}$ So we can conclude that the dimension for the homogeneous equation is $1$, and since every column in $B$ is a solution for the homogeneous equation, we can conclude $rank(B)=1$ Since every colum in $B$ is a $Sp\{1,1,1\}$.

Is this solution correct?

Thanks!

Answer 1

Yes, the solution is correct. Here's a simpler version.

I assume you mean that the solution set of the homogeneous system $Ax=0$ is spanned by $(1,1,1)$. This means that $Ax=0$ implies $x=\alpha(1,1,1)$ for some $\alpha$.

Thus, since $AB=0$, every column of $B$ is of that form and therefore $$ B=\begin{bmatrix} \alpha & \beta & \gamma \\ \alpha & \beta & \gamma \\ \alpha & \beta & \gamma \end{bmatrix} $$ so its rank is at most $1$.

Since $BA\ne0$, the rank of $B$ is at least $1$.