Inequality about rank of matrix

Question

$A,B$ are $n\times n $ matrix over a field, and $AB=BA$. Let $C=(A\mid B)$ (an $n\times 2n $ matrix). How to prove the following inequality about rank. $$\text{rank}(A)+\text{rank}(B)\ge \text{rank}(C)+\text{rank}(AB)$$ The difficulty is how I should interpret $AB=BA$. I tried to imitate the proof of Sylvester Rank Inequality but failed.

Answer 1

Let $\text{Ker}(M)$ denote the left kernel of a matrix $M$. Note that if $M$ has $n$ rows, then $\dim (\text{Ker}(M)) = n - \text{rk}(M)$. Thus, we can rewrite the inequality we need to prove as \begin{equation} \dim(\text{Ker}(A)) + \dim(\text{Ker}(B)) \leq \dim(\text{Ker}(C)) + \dim(\text{Ker}(AB)) \end{equation} It follows immediately that the left kernel of $C$ is $\text{Ker}(A) \cap \text{Ker}(B)$, hence \begin{align} \dim(\text{Ker}(C)) &= \dim(\text{Ker}(A) \cap \text{Ker}(B)) \\ &= \dim(\text{Ker}(A)) + \dim(\text{Ker}(B)) - \dim(\langle \text{Ker}(A), \text{Ker}(B) \rangle), \end{align} by Grassmann's identity. Thus, we need to prove that \begin{equation} \dim(\text{Ker}(AB)) \geq \dim(\langle \text{Ker}(A), \text{Ker}(B) \rangle). \end{equation} We will do this by proving that $\langle \text{Ker}(A), \text{Ker}(B) \rangle$ is a subspace of $\text{Ker}(AB)$. Indeed, take a vector $v \in \langle \text{Ker}(A), \text{Ker}(B) \rangle$, then $v = v_A + v_B$ for some vectors $v_A \in \text{Ker}(A)$ and $v_B \in \text{Ker}(v_B)$. Then, using that $A$ and $B$ commute, \begin{equation} ABv = AB(v_A + v_B) = A(B v_A + B v_B) = A (B v_A + \vec 0) = AB v_A = BA v_A = B \vec 0 = \vec 0. \end{equation} This concludes the proof.