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Consider an $n \times n_1$ matrix $A_1$ and an $n \times n_2$ matrix $A_2$ with the following properties:

  • $\mathrm{Rank} (A_1)=n_1$,
  • $\mathrm{Rank} (A_2)=n_2$,
  • $\mathrm{Rank} (A_1 : A_2)=n_1+n_2$

Let $B=(B_1:B_2)$ be a nonsingular $n \times n$ matrix such that $A_1^TB_1=I_{n_1}$ and $A_1^TB_2=0$.

The paper I'm reading says that

  • $\mathrm{Rank} (A_2^TB)=n_2$,
  • $\mathrm{Rank} (A_2^TB_2)=n_2$,
  • $\mathrm{Rank} (A_1^TB:A_2^TB)=n_1+n_2$

I'm not sure why the above are true. Can someone help me understand? I think part of the reason is that $B$ is nonsingular but I'm not sure.

Davide Giraudo
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user103828
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  • I think the first point is true by these posts http://math.stackexchange.com/questions/745071/if-a-in-mathbbcm-times-n-is-full-column-rank-matrix-then-is-rankab?rq=1 and http://math.stackexchange.com/questions/801250/rm-rankba-rm-rankb-if-a-in-mathbbrn-times-n-is-invertible?rq=1 – user103828 Jul 27 '15 at 08:22
  • what is $A_1^{\shortmid}$? – Ofir Schnabel Jul 27 '15 at 08:23
  • transpose of $A_1$ (replaced with $A_1^T$ to make it clearer). – user103828 Jul 27 '15 at 08:24
  • Do you know that if $A$ is invertible then $rank(AB)=rank(B)$ for any matrix $B$? – Ofir Schnabel Jul 27 '15 at 08:27
  • $A_j$ can be a rectangle matrix but it has full rank... but $B$ is nonsingular so $rank(A_j^TB)=rank(A_j)$ so the first point is true.... but I'm not sure about the last two points. – user103828 Jul 27 '15 at 08:29
  • Okay, I think the last point is also true because $B$ is invertible so $Rank(A_1^TB:A_2^TB)=Rank([A_1:A_2]^TB)=Rank(A_1:A_2)$. – user103828 Jul 27 '15 at 08:32
  • For the second multiply $(A_1:A_2)^T$ with $B$ and use $A_1^TB_2=0$. – Ofir Schnabel Jul 27 '15 at 08:38
  • @OfirSchnabel Okay, I think I get it now. Do you want to write up a solution and I can mark it correct?.... I can also do it. – user103828 Jul 27 '15 at 08:53

1 Answers1

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Hints:

(i) For the first use the fact that $B$ is invertible.

(ii) For the second multiply $(A_1:A_2)^T$ with $B$ and use $A_1^TB_2=0$.

(iii) The last follows from the argument you write in your above comment.

user103828
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Ofir Schnabel
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  • Just to make sure I understand the second point: $$n_1+n_2=Rank(A_1:A_2)=Rank(A_1^TB:A_1^TB)=Rank \left(\begin{array}{cc} I_{n_1} & A_2^TB_1 \ 0 & A_2^TB_2 \end{array}\right)$$ so $Rank(A_2^TB_2)=n_2$. – user103828 Jul 27 '15 at 09:06