How to prove the following relation for rank of matrices?

Question

If $\mathbb{A}$ and $\mathbb{B}$ have same rows then how to show that the following relation holds $$\operatorname{rank}([\mathbb{A~B}])=\operatorname{rank}(\mathbb{A})+\dim(\text{Proj}_{\mathcal{A^C}}\mathcal{B})$$ where $\text{Proj}_{\mathcal{A^C}}\mathcal{B}$ is the projection of span of $\mathbb{B}$, denoted by $\mathcal{B}$, on the orthogonal complement of the span of $\mathbb{A}$, denoted by $\mathcal{A^c}$. $\dim$ denotes the number of dimension.

Answer 1

If I understood your question correctly you have the space $\mathcal A=[\vec a_1,\dots,\vec a_k]$ generated by the columns of $\mathbb A$ and $\mathcal B=[\vec b_1,\dots,\vec b_l]$ generated by the columns of $\mathbb B$. And you are asking about the dimension of $$\mathcal A+\mathcal B=[\vec a_1,\dots,\vec a_k,b_1,\dots,\vec b_l].$$ You can write each vector $\vec b_i$ as $$\vec b_i=\underset{\in\mathcal A}{\underbrace{\vec b'_i}}+\underset{\in\operatorname{Proj}_{\mathcal A^c}\mathcal B}{\underbrace{\vec b''_i}}.$$ From this you get $$\mathcal A+\mathcal B=[\vec a_1,\dots,\vec a_k,b''_1,\dots,\vec b''_l]=\mathcal A \oplus \operatorname{Proj}_{\mathcal A^c}\mathcal B$$ and this implies the claim about dimensions.