Rank of the concatenation of two matrices

Question

Let $A, B \in \Bbb R^{n_1 \times n_2}$ and let $C := [A\,\,\,B]$ be a $n_1 \times 2n_2$ matrix whose first $n_2$ columns are $A$ and whose remaining columns are $B$. Is it true in general that $\mbox{rank} (C) \le \mbox{rank} (A) + \mbox{rank} (B)$?

Answer 1

Yes. Let $$ A= \begin{pmatrix} a_1&\cdots&a_{n_2} \end{pmatrix} $$ and $$ B= \begin{pmatrix} b_1&\cdots&b_{n_2} \end{pmatrix}. $$ Then $$ \operatorname{rank} A = \dim \operatorname{span}(a_1,\dots,a_{n_2})\operatorname{\hat{=}}\dim V_1, $$ $$ \operatorname{rank} B=\dim\operatorname{span}(b_1,\dots,b_{n_2})\operatorname{\hat{=}}\dim V_2, $$ $$\operatorname{rank} C = \operatorname{dim}\operatorname{span} (a_1,\dots,a_{n_1},b_1,\dots,b_{n_2})\operatorname{\hat{=}}\dim (V_1+V_2). $$ As $$ \dim (V_1+V_2)=\dim V_1+\dim V_2-\dim (V_1\cap V_2)\leq \dim V_1+\dim V_2, $$ we have $\operatorname{rank} C\leq \operatorname{rank} A+ \operatorname{rank}B$.