For $n\times n$ matrices $A,B,$ and $C,$ is it always true that $\mathrm{rank}(ABC)\leq\mathrm{rank}(AC)$?

Question

I'd appreciate any thoughts on the correctness of the following proof:

Using the generalized Frobenius inequality:

$\mathrm{rank}(ABC)\geq\mathrm{rank}(AB)+\mathrm{rank}(BC)-\mathrm{rank}(B)$ and

$\mathrm{rank}(AC)\geq\mathrm{rank}(A)+\mathrm{rank}(C)-n,$ we have:

$$\mathrm{rank}(AC)-\mathrm{rank}(ABC)\geq \mathrm{rank}(A)+\mathrm{rank}(B)+\mathrm{rank}(C)-\mathrm{rank}(AB)-\mathrm{rank}(BC)-n.$$

Using $\mathrm{rank}(AB)\leq\mathrm{rank}(A)\Rightarrow -\mathrm{rank}(AB)\geq-\mathrm{rank}(A)$ and similarly $-\mathrm{rank}(BC)\geq-\mathrm{rank}(B),$ we get:

$$\mathrm{rank}(AC)-\mathrm{rank}(ABC)\geq\mathrm{rank}(A)+\mathrm{rank}(B)+\mathrm{rank}(C)-\mathrm{rank}(A)-\mathrm{rank}(B)-n$$

$$\Rightarrow \mathrm{rank}(AC)-\mathrm{rank}(ABC)\geq\mathrm{rank}(C)-n.$$

If $\mathrm{rank}(C)=n,$ then $\mathrm{rank}(ABC)\leq\mathrm{rank}(AC).$ Instead of using $-\mathrm{rank}(BC)\geq-\mathrm{rank}(B),$ if we use $-\mathrm{rank}(BC)\geq-\mathrm{rank}(C),$ then we can show that if $\mathrm{rank}(B)=n,$ then $\mathrm{rank}(ABC)\leq\mathrm{rank}(AC).$

Answer 1

It is not always true under the given hypotheses that $\mathrm{rank}(ABC) \le \mathrm{rank}(AC).$

Consider

$A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $C = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$.

Then $AC = 0$ has rank 0, but $ABC = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ has rank 1.