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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.

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  • $\begingroup$ Deos $[\mathbb A \mathbb B]$ mean the matrix which contains column of $\mathbb A$ and then columns of $\mathbb B$? (Or is it product of the two matrices,) When you say "span of a matrix" do you mean the spaces generated by columns? (Or the space generated by rows?) $\endgroup$ Apr 20 '19 at 4:48
  • $\begingroup$ Did you mean to say that the two matrices have "the same number of rows" rather than just "the same rows"? (If you say that they have the same rows, it seems that $\mathbb A=\mathbb B$, or maybe they are not exactly the same but the order of rows might be changed.) $\endgroup$ Apr 20 '19 at 4:51
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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.

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