Sylvester rank inequality: $\operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n$ [duplicate]

Subtract $2n$ from both sides and then multiply by $-1$; this gives the equivalent inequality $\def\rk{\operatorname{rank}}(n-\rk A)+(n-\rk B)\geq n-\rk(AB)$. In other words (by the rank-nullity theorem) we must show that the sums of the dimensions of the kernels of $A$ and of $B$ gives at least the dimension of the kernel of $AB$. Intuititively vectors that get killed by $AB$ either get killed by $B$ or (later) by $A$, and the dimension of $\ker(AB)$ therefore cannot exceed $\dim\ker A+\dim\ker B$. Here is how to make that precise.

One has $\ker B\subseteq \ker(AB)$, so if one restricts (the linear map with matrix) $B$ to $\ker(AB)$, giving a map $\tilde B:\ker(AB)\to K^n$, one sees that $\ker(\tilde B)=\ker(B)$ (both inclusions are obvious). Since the image of $\tilde B$ is contained in $\ker A$ by the definition of $\ker(AB)$, one has $\rk\tilde B\leq\dim\ker A$. Now rank-nullity applied to $\tilde B$ gives $$\dim\ker(AB)=\dim\ker\tilde B+\rk\tilde B\leq\dim\ker B+\dim\ker A,$$ as desired.

As left or right multiplications by invertible matrices (and in particular, elementary matrices) don't change the rank of a matrix, we may assume WLOG that $A=\pmatrix{I_r&0\\ 0&0}$. Therefore, \begin{align*} \operatorname{rank}(A)+\operatorname{rank}(B) &=\operatorname{rank}(A)+\operatorname{rank}(AB+(I-A)B)\\ &\le\operatorname{rank}(A)+\operatorname{rank}(AB)+\operatorname{rank}((I-A)B)\\ &\le r+\operatorname{rank}(AB)+(n-r)\\ &=\operatorname{rank}(AB)+n. \end{align*} The last but one line is due to the fact that the first $r$ rows of $(I-A)B$ are zero.

$$n+r(AB)=r(M)=r \begin{pmatrix} I_n & 0 \\ 0 & AB \end{pmatrix}$$ Using generalized elementary transformation to $M$: $$M=\begin{pmatrix} I_n & 0 \\ 0 & AB \end{pmatrix} \to \begin{pmatrix} I_n & 0 \\ A & AB \end{pmatrix} \to \begin{pmatrix} I_n & -B \\ A & 0 \end{pmatrix} \to \begin{pmatrix} B & I_n \\ 0 & A \end{pmatrix},$$ hence $$n+r(AB)=r(M)= r\begin{pmatrix} B & I_n \\ 0 & A \end{pmatrix}\geq r(A)+r(B):$$

This solution is from here.

Let $T:V\to W,\ S:W\to V$ be linear transformations which correspond to $A,B$. Equivalently we want prove that $r(T)+r(S)\leq r(TS)+n$ also similarly show that $$\dim \ker(T) + \dim \ker(S) \gt \dim \ker(TS).$$ I know that $\ker (S) \subset \ker (TS)$. Let {$a_1,a_2,\ldots,a_r$} be a base for $\ker (S)$; expand this base for $\ker(ST)$ to {$a_1,a_2,\ldots,a_r,a_{r+1},a_{r+2},\ldots,a_s$} and then show that {$S(a_{r+1}), S(a_{r+2}),\ldots,S(a_s)$} are a set of independent elements s.t. $S(a_i)\in \ker (T) \, \forall i:r+1,\ldots,s$ and indicate $\ker (T) \ge s-r,$ so $\dim \ker T+ \dim \ker S\ge s-r+r=\dim \ker TS$. $$ 1.\dim \ker T+ \dim \ker S\ge \dim \ker TS$$ $$2.\dim \ker T+r(T)=n$$ $$3.\dim \ker S T+r(S)=n$$ $$4.\dim \ker TS+r(TS)=n.$$ By 1, 2, 3, 4 easily conclude $r(T)+r(S)\leq r(TS)+n$.

Denote by $r(A)$ the rank of $A$ and by $n(A)$ the nullity of $A$, i.e. the dimension of the nullspace of $A$. The rank plus nullity theorem says that $r(A)+n(A)=n$ and $r(B)+n(B)=n$ and so $2n= r(A)+r(B)+n(A)+n(B)$.

Let $N(A)$ be the nullspace of $A$ and $R(B)$ the range-space of $B$. Then $r(AB)=n - n(B)- \dim(N(A) \cap R(B))$ (try proving this, hint: what is the nullspace of $AB$?). This gives $n = r(AB) +n(B)+ n(N(A) \cap R(B))$ and so $n + r(AB) = r(A)+r(B) +n(A) - n(N(A) \cap R(B)) \ge r(A)+r(B)$.