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I saw in a proof somewhere that a square matrix $AA^T$ is similar to $A^T A$, so I thought about it and I don't know why (or whether) it's true.

I tried using the fact that every matrix is similar to its transpose and maybe transpose the entire expression $AA^T$ but what I get is $(AA^T)^T=A^{T^T} A^T=AA^T$ which is obvious because $AA^T$ is symmetric.

I tried to run some examples like $$ A = \begin{bmatrix} 1 & 4 \\ 3 & 2 \end{bmatrix} \qquad \qquad \qquad A^T = \begin{bmatrix} 1 & 3 \\ 4 & 2 \end{bmatrix} $$ And I get that $AA^T$ and $A^TA$ have the same characteristic polynomial so obviously they have the same trace, eigenvalues and determinant.

But is it true for the general case?

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    $\begingroup$ If $\mathrm A \in \mathbb R^{m \times n}$, where $m \neq n$, then $$\mathrm A \mathrm A^T \in \mathbb R^{m \times m}$$ and $$\mathrm A^T \mathrm A \in \mathbb R^{n \times n}$$How can two square matrices of different dimensions be similar? $\endgroup$ – Rodrigo de Azevedo Jul 26 '16 at 10:15
  • $\begingroup$ That's a good point. I assumed it's not always true but really wondered when it is. $\endgroup$ – PanthersFan92 Jul 27 '16 at 9:04
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If $A$ is a square real matrix and let $A=U D V^T$ be the SVD decomposition.

$$A^TA=VD^2V^T$$

$$AA^T=UD^2U^T$$

Notice that $$(UV^T)A^TA(VU^T)=AA^T$$

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  • $\begingroup$ Can you elaborate the last step please? $\endgroup$ – PanthersFan92 Jul 26 '16 at 8:31
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    $\begingroup$ First let's check that the equality holds: $$(UV^T)A^TA(VU^T)=(UV^T)VD^2V^T(VU^T)=UD^2U^T=AA^T$$ We have used the fact that $V^TV=I$. Also, note that $(UV^T)^{-1}=VU^T$ since $U$ and $V$ are orthogonal matrices. $\endgroup$ – Siong Thye Goh Jul 26 '16 at 8:40
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In general the statement is false. E.g. consider the complex matrix $A=\pmatrix{1&i\\ 0&0}$, for which $AA^T=0\ne A^TA$. It is for real square matrices, though. A proof was given for in the other answer here, but that proof can actually be made simpler if you are allowed to use polar decomposition: let $A=PU$, where $P$ is symmetric positive semidefinite and $U$ is real orthogonal (so that $U^T=U^{-1}$). Then $A^TA=U^TP^2U$ is similar to $AA^T=P^2$.

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  • $\begingroup$ But if the statement is false in general, how come the proof above works? and why only for reals? I thought that SVD works for every matrix without exception. $\endgroup$ – PanthersFan92 Jul 26 '16 at 10:14
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    $\begingroup$ @PanthersFan92 For complex matrices, the correct generalisation is $AA^\ast\sim A^\ast A$. Transpose is different from conjugate transpose and in general, $UU^T\ne I=UU^\ast$ for complex unitary matrices. $\endgroup$ – user1551 Jul 26 '16 at 10:16

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