Questions on factor analysis and linear algebra Is it possible that $A*A^T=B*B^T$ while A and B have different ranks?
Thanks in advance!
 A: (My answer assumes that your matrices are $\in M(m \text{ x }n, \mathbb{R})$.)
No, it is not. The key here is to prove that $\text{rank}(A) = \text{rank}(AA^T)$. For this, prove that $$A^TAx = 0 \iff Ax = 0\ \forall x \in \mathbb{R}^n$$  
(then they have the same kernel, thus dimension formula provides that $\text{rank}(A) = \text{rank}(A^TA)$ and thus $\text{rank}(A) = \text{rank}(A^T) = \text{rank}(AA^T)$)
"$\Leftarrow$" is obvious. For "$\Rightarrow$" assume $A^TAx = 0$ for $x \in \mathbb{R}^n$. Then $$x^TA^TAx = 0 \implies (Ax)^TAx = 0 \implies \langle Ax,Ax \rangle = 0 \implies Ax = 0$$, where "$\langle\ , \rangle$" is the Euclidian scalar product (this doesn't hold for $\mathbb{C}$, for example, thus my assumption that the matrices are elements of $M(m \text{ x }n, \mathbb{R})$).  
So now we have $\text{rank}(A) = \text{rank}(AA^T)$, thus $AA^T \neq BB^T$ if $A$ and $B$ have different ranks.
A: The rank of $A*A^T$ = $m*m$ Where $m$ is the no. of rows in $A$. So, as long as $A$ and $B$ have same no. of rows, it is possible 
