The question was that whether $\operatorname{rank}(ABC)$ is equal to $1$ or not. The matrices are $3\times3$. So I wanted someone to help me understand this question without using formulas but giving me the intuition behind this.
My attempt: let $T_c:\mathbb{R}^3\to\mathbb{R}^3$. The rank is $2$ which means that the image space of $T_c$ will have two vectors which are linearly independent. Now $T_b:\mathbb{R}^3\to\mathbb{R}^3$ So the domain of $T_b$ will have two linearly independent vectors and as $\operatorname{rank}(B)=1$, the nullspace of $T_b$ will have one vector and the domain of $T_a$ will have one linearly independent vector. Is this making sense because I am stuck over here.
Best Answer
We have $\text{rank}(ABC) \leq 1$, because: $$\text{rank}(ABC) \leq \text{rank}(BC) \leq \text{rank}(B) = 1.$$
So is $\text{rank}(ABC)$ equal to $0$ or $1$? It could be either. Consider $A = \begin{bmatrix}0 & e_2 & e_3\end{bmatrix}, B = \begin{bmatrix}0 & e_2 & 0\end{bmatrix}, C = \begin{bmatrix}e_1 & e_2 & 0\end{bmatrix}$. Then $ABC = \begin{bmatrix}0 & e_2 & 0\end{bmatrix}$, which has rank 1. On the other hand, now consider $A = \begin{bmatrix}e_1 & 0 & e_3\end{bmatrix}$, and let $B, C$ be as before. Then $ABC = 0$, which has rank $0$.