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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.

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  • $\begingroup$ See math.stackexchange.com/questions/48989/… $\endgroup$ – Minus One-Twelfth May 13 '20 at 18:32
  • $\begingroup$ @MinusOne-Twelfth from the formula I can conclude that rank(ABC)$\le$1 .However this is a true or false question asking whether it is 1 or not $\endgroup$ – smita May 13 '20 at 18:39
  • $\begingroup$ I see. In that case you can try and show that $BC$ might equal $O$ (the zero matrix). $\endgroup$ – Minus One-Twelfth May 13 '20 at 19:04
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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 $0$.

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  • $\begingroup$ If you want some intuition, as you asked for: the reason that $\text{rank}(AB) \leq \text{min}\{\text{rank}(A), \text{rank}(B)\}$ is answered well in various answers on this site. As for why $\text{rank}(ABC)$ can be $0$ or $1$, we can see in the examples I gave that $BC$ can destroy the $x$ and $z$ components while leaving the $y$ component unchanged, i.e. it sends $(x, y, z)$ to $(0, y, 0)$. So the final step of applying $A$ will depend on what $A$ is. In the first case, it preserves the $y$ component, so nothing changes. In the second case, it destroys it, so we get $0$. $\endgroup$ – twosigma May 13 '20 at 19:32
  • $\begingroup$ However, I've only described the intuition for this particular example I gave. In general, the rank might be $0$ or $1$ due to various reasons. Another example is if we have $BC = 0$ so that $ABC = 0$, then we get rank($ABC$) = $0$; but if we had different $B, C$ then maybe $BC$ isn't $0$ and $ABC$ might not be $0$. $\endgroup$ – twosigma May 13 '20 at 20:03
  • $\begingroup$ So just to understand that in terms of linear transformations $T_c(e_1)=e_1$ ,$T_c(e_2)=e_2$,$T_c(e_3)=0$ .Now $T_b(e_2)=e_2$,$T_b(e_1)=0$ and $T_a(e_2)=0$.is that how it is working ? @twosigma $\endgroup$ – smita May 14 '20 at 4:12
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    $\begingroup$ For those examples, the linear transformation induced by those matrices do result in those, yes. However, in general, a linear transformation can correspond to different matrices, depending on the choice of basis. As for why those matrices in my examples do what they do (e.g. $B$ applied to a vector $(x, y, z)$ will give you $(0, y, 0)$), you can see my answer here about the column space and matrix-vector multiplication. $\endgroup$ – twosigma May 14 '20 at 4:21

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