# $\operatorname{rank}(A)=2$, $\operatorname{rank}(B)=1$ and $\operatorname{rank}(C)=2$. Find $\operatorname{rank}(ABC)$.

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.

• – Minus One-Twelfth May 13 '20 at 18:32
• @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 – smita May 13 '20 at 18:39
• I see. In that case you can try and show that $BC$ might equal $O$ (the zero matrix). – Minus One-Twelfth May 13 '20 at 19:04

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 $$0$$.
• 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 $0$. – twosigma May 13 '20 at 19:32
• 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 $0$. – twosigma May 13 '20 at 20:03
• 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 – smita May 14 '20 at 4:12
• 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. – twosigma May 14 '20 at 4:21