An inequality about rank of matrices Suppose $A$ and $B$ are $m \times n$ matrices, prove that:
$$\operatorname{rank}(A)+\operatorname{rank}(B)+\operatorname{rank}(A+B)
\ge \operatorname{rank}\pmatrix{A & B}+\operatorname{rank}\pmatrix{A \\ B},$$
where the two matrices on the right hand side are block matrices.  
 A: We can consider $A, B$ to be mappings $K^n \to K^m$.
Let $\varphi \colon K^n \to K^m \times K^m$ be defined by $\varphi(X) = (AX, BX)$. Then $\varphi$ is, in essence, the mapping with matrix
$\begin{bmatrix} A \\ B \end{bmatrix}$. Let $U = \textrm{im} \, \varphi$, let $\Delta = \{ (u,v) \in K^m \times K^m \mid u + v = 0 \}$, and let $\pi_1$ and $\pi_2$ be the first and second projections $K^m \times K^m \to K^m$.
Now we have
$$ \textrm{rk}(A) = \dim \textrm{im}(A)  = \dim \pi_1(U)$$ 
$$ \textrm{rk}(B) = \dim \textrm{im}(B)  = \dim \pi_2(U)$$ 
$$ \textrm{rk} \begin{bmatrix} A & B \end{bmatrix}= \dim (\textrm{im}(A) + \textrm{im}(B))  = \dim (\pi_1(U) + \pi_2(U))$$
$$ \textrm{rk} \begin{bmatrix} A \\ B \end{bmatrix}= \dim \textrm{im}(\varphi)  = \dim U$$
$$\begin{align} \textrm{rk}(A + B) &= n - \dim \ker(A + B) \\
& = n - \dim \varphi^{-1}(\Delta) \\ &= n - \dim \ker \varphi - \dim(\Delta \cap \textrm{im}(\varphi)) \\ &= \dim \textrm{im}(\varphi) - \dim(\Delta \cap \textrm{im}(\varphi)) \\ &=\dim U - \dim (\Delta \cap U)
\end{align}$$
We can now rewrite the inequality to be proved as 
$$\dim (\Delta \cap U) \leq \dim \pi_1(U) + \dim \pi_2(U) - \dim (\pi_1(U) + \pi_2(U)),$$
which is equivalent to
$$\dim(\Delta \cap U) \leq \dim(\pi_1(U) \cap \pi_2(U)).$$
But this last inequality results from the fact that the restriction of $\pi_1$ to $\Delta \cap U$ realizes an injection of this space into $\pi_1(U) \cap \pi_2(U)$.
