# Frobenius Inequality Rank

I was looking for an answer for this problem in terms of matrices, but I really don't know how to prove this result. The proposition says that:

Let $A\in M_{m\times k}(\mathbb{C})$, $B\in M_{k\times p}(\mathbb{C})$ and $C\in M_{p\times n}(\mathbb{C})$, then $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$

Notation:

• $\rho(\cdot)$ stands for rank,
• $\ker(\cdot)$ for null space (aka kernel),
• $\text{im}(\cdot)$ for column space (aka image).

All we need is the following well known identity (see this answer for a proof): $$\rho(AB)=\rho(B)−\dim(\text{im}(B) \cap \ker(A)) \tag{1}$$ and the following observation: $$\text{im}(BC) \cap \ker(A) \subseteq \text{im}(B)\cap \ker(A)\tag{2}$$ which holds since $\text{im}(BC)\subseteq \text{im}(B)$.

Now we want to write $\rho(ABC)$ in such a way that $\text{im}(BC)\cap \ker(A)$ pops up, so we could make use of $(2)$. Analogously to $(1)$:

$$\rho(ABC)=\rho(BC)−\dim(\text{im}(BC) \cap \ker(A)) \tag{3}.$$

From $(1)$ and $(3)$:

$$\rho(AB)+ \rho(BC) = \rho(B) + \rho(ABC) + \underbrace{\dim(\text{im}(BC) \cap \ker(A))−\dim(\text{im}(B) \cap \ker(A))}_{\leq 0 \text{ due to } (2)}$$ which implies the desired inequality.

• i just wanna start studying linear algebra , do you recommend a self-study text book? i wanna learn everything for example it was awesome when you used (1) to solve the problem. – Arsh Gh Sep 20 '16 at 4:53

If you know the Sylvester inequality then it's a two lines proof.

## Sylvester Inequality:

Consider $$A_{m\times r}$$ and $$B_{r\times n}$$. Then $$r(AB)\geq r(A)+r(B)-r \tag{1}$$ where $$r(\cdot)$$ is the rank.

Let $$B=U_{k\times r} V_{r\times p}$$ be a full-rank factorization of $$B$$.

Then by $$(1)$$,

\begin{align} r(ABC)&\geq r(AU)+r(VC)-r\\ &=r(AB)+r(BC)-r(B). \end{align}

## For a proof of Sylvester inequality refer the following:

• Please provide a link or name the book from where the screenshot of Sylvester Theorem has been taken. – Shatabdi Sinha May 17 at 12:56