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I know rank AB is not same as rank BA.

I take here particular case. Is this particular case true?

Pradip
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1 Answers1

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Yes, provided we're in the real case. First off, rank is invariant under transposition i.e. $\text{rank}\ B=\text{rank}\ (B^T).$ One can show that \begin{align*}\text{rank}\ A&=\text{rank}\ (A^TA)\\ \text{rank}\ A^T&=\text{rank}\ (AA^T). \end{align*} Since $\text{rank}\ A=\text{rank}\ A^T,$ we get that $$\text{rank}\ (A^TA)=\text{rank}\ (AA^T)$$ You can find proofs of the basic facts that I've used all over this site, although I'd suggest for you to try to show them yourself before looking.

cmk
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  • But note that it's only true for real matrices! – mathphys Jul 01 '19 at 15:12
  • Of course, which is why I used transpose, instead of Hermitian adjoint. – cmk Jul 01 '19 at 15:13
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    But you could use the transpose on a complex matrix and then the equality wouldn't hold, for example $\begin{bmatrix} 1 & i \ 0 &0 \end{bmatrix}\begin{bmatrix} 1 & 0 \ i &0 \end{bmatrix}$; I think it would work for complex matrices when you use the conjugate transpose/Hermitian adjoint. – mathphys Jul 01 '19 at 15:14
  • Sure, but typically one uses the standard transpose when talking about real matrices and the conjugate transpose for complex matrices. – cmk Jul 01 '19 at 15:19
  • @mathphys Okay, I've edited because apparently that's worthy of a down vote from somone. – cmk Jul 01 '19 at 15:21
  • @cmk Ok I'll +1 (though the -1 wasn't from me) – mathphys Jul 01 '19 at 15:24
  • The downvote was mine. I cast it because I didn't understand how you got $rank(A^TA)=rank(AA^T)$ from the identity $rank(B)=rank(B^T)$. The transpose of $A^TA$ is $A^TA$ itself, not $AA^T$. Did I miss something? – user1551 Jul 01 '19 at 18:08
  • @user1551 That's actually fair. I actually edited my post a little bit ago to explain that further. What you can show, without too much effort, is that the rank of $A$ is the same as $A^TA$, and that the rank of $A^T$ is the rank as $AA^T$. Now, we use the identity for $A$ and $A^T$. – cmk Jul 01 '19 at 18:17
  • It makes sense now. Thanks for the explanation. – user1551 Jul 01 '19 at 18:20