How to prove the following relation for rank of matrices?

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If $mathbbA$ and $mathbbB$ have same rows then how to show that the following relation holds $$rank([mathbbA~B])=rank(mathbbA)+dim(textProj_mathcalA^CmathcalB)$$ where $textProj_mathcalA^CmathcalB$ is the projection of span of $mathbbB$, denoted by $mathcalB$, on the orthogonal complement of the span of $mathbbA$, denoted by $mathcalA^c$. $dim$ denotes the number of dimension.










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    If $mathbbA$ and $mathbbB$ have same rows then how to show that the following relation holds $$rank([mathbbA~B])=rank(mathbbA)+dim(textProj_mathcalA^CmathcalB)$$ where $textProj_mathcalA^CmathcalB$ is the projection of span of $mathbbB$, denoted by $mathcalB$, on the orthogonal complement of the span of $mathbbA$, denoted by $mathcalA^c$. $dim$ denotes the number of dimension.










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      If $mathbbA$ and $mathbbB$ have same rows then how to show that the following relation holds $$rank([mathbbA~B])=rank(mathbbA)+dim(textProj_mathcalA^CmathcalB)$$ where $textProj_mathcalA^CmathcalB$ is the projection of span of $mathbbB$, denoted by $mathcalB$, on the orthogonal complement of the span of $mathbbA$, denoted by $mathcalA^c$. $dim$ denotes the number of dimension.










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      If $mathbbA$ and $mathbbB$ have same rows then how to show that the following relation holds $$rank([mathbbA~B])=rank(mathbbA)+dim(textProj_mathcalA^CmathcalB)$$ where $textProj_mathcalA^CmathcalB$ is the projection of span of $mathbbB$, denoted by $mathcalB$, on the orthogonal complement of the span of $mathbbA$, denoted by $mathcalA^c$. $dim$ denotes the number of dimension.







      matrices matrix-rank dimension-theory






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      asked Sep 10 at 8:53









      Frank Moses

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