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I am pondering over the question:

Whether $\mathrm{rank}\mathbf{AA^T}\stackrel{?}{=}\mathrm{rank}\mathbf{A^TA}\stackrel{?}{=}\mathrm{rank}\mathbf{A}$ for any matrix $\mathbf{A}\in M_{m\times n}$

I looked through these two answers: Gram Matrix Rank and Prove rank$(A^TA)$=rank$(A)$ for any $A\in M_{m\times n}$.

In the second proof of Gram Matrix Rank, and this proof if we replace $A$ with the $A^T$, we get $$\mathrm{rank}\mathbf{A^T}=\mathrm{rank}AA^T$$ But $\mathrm{rank }A^T=\mathrm{rank}A$. So, we have the following:$$\mathrm{rank}A=\mathrm{rank}A^T=\mathrm{rank}AA^T=\mathrm{rank}A^TA\enspace\enspace\cdots (1)$$

Since the proofs were done in the above are independent of the entries of the matrix. So, consider$$B^T=\begin{bmatrix}1& i\end{bmatrix}$$ where $i^2=-1$.
For the matrix $B$ the result in $(1)$ fails to hold.

I am unable to understand is there a problem with those proofs or I am missing something? Thank you in advance.

Kumar
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    Did you notice this comment ? – Arnaud D. Oct 06 '20 at 09:09
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    Both proofs make use of the fact that $A$ is real. Your $B$ is not real. – Christoph Oct 06 '20 at 09:10
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    For complex-valued matrices, the four matrices $A$, $A^$, $A^A$, and $AA^$ have the same rank, where $A^$ is the conjugate transpose of $A$. –  Oct 06 '20 at 09:11
  • @Christoph But I don't see that fact. – Kumar Oct 06 '20 at 09:12
  • @ArnaudD. Yes. But I wasn't able to comprehend the comment. – Kumar Oct 06 '20 at 09:13
  • @Bungo Can you please generalize the result for any Field? – Kumar Oct 06 '20 at 09:15
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    @Kumar The result can be false for finite fields. Over $\mathbb{F}_2$ (or any field of characteristic $2$), if $A=(1,1)$ then $A^TA=\left(\begin{smallmatrix}1 & 1\ 1 & 1 \end{smallmatrix}\right)$ and $AA^T=0$. – Arnaud D. Oct 06 '20 at 09:15
  • And the same counterexample works for any prime characteristic $p$ if you put $A = (1, 1, \ldots, 1)$ (with $p$ elements). –  Oct 06 '20 at 09:21
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    @Kumar The first proof uses $|\mathbf x|^2 = \mathbf x^T, \mathbf x$, which only holds for real vectors $\mathbf x$. For complex vectors $|\mathbf x|^2 = \mathbf x^, \mathbf x$, where ${}^$ denotes the Hermitian conjugate. The second proof also uses $\mathbf y^T, \mathbf y = 0\ \Longrightarrow\ \mathbf y=0$ for $\mathbf y = A\mathbf x$, which also only works for real $\mathbf y$ for the same reason. – Christoph Oct 06 '20 at 09:47

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Well, in my eyes, the most useful tool to analyze this type of questions, in the non-square matrix case, is the SVD decomposition.

Let us decompose our matrix: $A=U\Sigma V^\star$, then the rank of $A$ is equal to the amount of non zero entries in $\Sigma$. However, each non zero entry is the square of an eigenvalue of $A A^\star$. Thus yielding a perfect correspondence between non zero singular values of $A$ to non zero eigenvalues of $AA^\star$ and $A^\star A$ (Which are actually the same). Hence, the ranks are indeed equal.

Alon Yariv
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  • Here $\star$ means complex conjugate transpose – Alon Yariv Oct 06 '20 at 09:16
  • Well, is there some similar result for finite fields? – Kumar Oct 06 '20 at 09:17
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    The case of a finite field is infinitely more complex, mostly in my eyes due to the absolute Galois group of a finite field being much more complex than that of $R$. I red a paper a few months ago which describes the equivalent result, but it relies on orbits in the algebraic group of matrices over a finite field.

    https://arxiv.org/pdf/1805.06999.pdf

     – Alon Yariv Oct 06 '20 at 09:25
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    The SVD is a pretty big hammer to use. A simpler argument is to show that $A$ and $A^A$ have the same null space. This is obvious in one direction: if $Ax = 0$ then $A^Ax = 0$. Conversely, if $A^Ax = 0$ then $x^A^Ax = 0$, hence $|Ax| = 0$, hence $Ax = 0$. Now observe $A$ and $A^A$ have the same number of columns, and use rank-nullity to conclude that they have the same rank. –  Oct 06 '20 at 09:27
  • The SVD is a big hammer for a reason. I tried to emphasise here on how useful this tool is. @Bungo, I don't see a way to generalize your argument to the finite field case. – Alon Yariv Oct 06 '20 at 09:30
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    @AlonYariv Does the SVD work over fields other than $\mathbb R$ or $\mathbb C$, e.g. finite fields? –  Oct 06 '20 at 21:37
  • I attached in a previous comment a paper by Guaralnik describing a way to generalize the SVD over finite fields. The way he interprets it is by orbits in the algebraic group, but that made me think, it should be possible to replicate the result using Galois twist. For any Algebraically closed field, it is straight forward, by using the same ideas in the standard SVD. – Alon Yariv Oct 07 '20 at 09:36