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Suppose both $A$ and $B$ are matrices with dimensions $n \times m$ and $n \times p$ respectively. Assume $m + p < n $ and $A \neq 0$ and $B \neq 0$. Further, assume Im($B$) $\cap$ Im($A$) = {0}. Then, it means that columns of $B$ are independent of columns of $A$. Hence, I can claim the following: \begin{equation} rank([A\ \ B]) = rank(A) + rank(B) \end{equation}

I believe this should be correct (if not correct me please). However, I was hoping to find a reference for it. It would be nice if someone could can suggest me a good book that contains many useful identities related to rank of matrices (like the one I just showed). Alternatively, I would like to get a formal proof. Thank you in advance!

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  • $\begingroup$ what is the identity? I don't see any identity in your question. $\endgroup$ – john May 28 '13 at 14:48
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    $\begingroup$ Let $A = 0$ and $B \not = 0$ to see a counterexample. $\endgroup$ – Steve Huntsman May 28 '13 at 14:52
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    $\begingroup$ I think by $[A,B]$ he means what I would call $[A B]$, where the two matrices (of the same height $n$) are placed side by side. To the OP: everybody else uses $[A,B]$ to mean AB-BA, so I'm changing your writeup. $\endgroup$ – Allen Knutson May 28 '13 at 15:53
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    $\begingroup$ If the columns of B are independent of the columns of A, then your identity holds. (Assuming that "independent" in this context means that only the zero vector is the common vector in the spaces the images of the individual matrices span.) You don't really need to find a reference for this, since this is basic linear algebra. $\endgroup$ – Per Alexandersson May 28 '13 at 16:57
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    $\begingroup$ This question seems clearly to be off-topic in MO. $\endgroup$ – Benoît Kloeckner May 28 '13 at 18:02
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I conjecture that you didn't ask what you meant to ask. The reason for my conjecture is that you wrote "assume Im$(B)\not\subseteq$ Im$(A)$ ... it means that columns of $B$ are independent of columns of $A$." In fact, it doesn't mean that at all. The assumption Im$(B)\not\subseteq$ Im$(A)$ allows some (though not all) of the columns of $B$ to be linear combinations of columns of $A$ (or, indeed, to be columns of $A$). If you really intended the hypothesis to be Im$(B)\not\subseteq$ Im$(A)$, then there are lots of counterexamples, and Felix has given one of them. But if you intended the hypothesis to be that the columns of $B$ are independent of those of $A$, in the sense that no non-zero linear combination of the columns of $B$ equals a linear combination of the columns of $A$ (in other words, Im$(B)\cap$ Im$(A)=\{0\}$), then your proposed equation about ranks is OK, simply because all columns of $[A\ \ B]$ are linearly independent.

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  • $\begingroup$ Well, not all columns of $[A\:B]$ are linearly independent, for there may well be linear relations among the columns of, say, $A$. :-) $\endgroup$ – Mariano Suárez-Álvarez May 28 '13 at 17:28
  • $\begingroup$ So, you are right that I actually made a mistake with my question. I went to Felix's suggestion - Marsaglia-Styan paper and saw similar statements. Geometrically, it make sense too. $\endgroup$ – Yao Hong Kok May 28 '13 at 19:38
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Try $A=\begin{bmatrix}0 & 1\\\\ 1 & 0\\\\ 0 & 0\\\\ 0 & 0\\\\ 0 & 0\end{bmatrix}$ and $B=\begin{bmatrix}0 & 1\\\\ 0 & 0\\\\ 1 & 1\\\\ 0 & 0\\\\ 0 & 0\end{bmatrix}$. Then $rank([A,B])=3$ but $rank(A)=rank(B)=2$.

What is true however, is $rank([A, B])=rank(A)+rank(E_{A}B)$, where $E_{A}$ is the projection defined by $E_{A}=I-AA^{\dagger}$. See p.4 in this paper (the results are much older, going at least back to Marsaglia-Styan in the 70s, but Tian's paper gives a good recap and is easily accesible).

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  • $\begingroup$ Your answer to my previous question was correct. But after studying the comment from Andreas, I decided to change my question because that was actually what I wanted. $\endgroup$ – Yao Hong Kok May 28 '13 at 19:30

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