Let $T \in \mathbb{R}^{n\times n}$ be invertible and $A \in \mathbb{R}^{m\times n}$ have independent rows. Does the following hold?
$$T (A T)^+ = A^+,$$
where $(\cdot)^+$ gives the M-P inverse.
My proof is given as follows:
Set $B = T (A T)^+$, and we get $A B = (A T) (A T)^+ = I$, i.e., $B = A^+$ which implies $T (A T)^+ = A^+$.
Am I right? Thanks!
PS: What if the condition that $A$ has independent rows is removed?
Using this property (case 3), we have $$ T(AT)^+ = T T^+A^+ = T T^{-1} A^+ = A^+, $$ which is what we wanted. This uses the fact that $A$ has linearly independent rows, so I am not sure about the more general case.
The assumption of independent of rows cannot be dropped. For unit vectors $u,v$ and $A = uv^\top$, we find that $A^+ = vu^\top$ and $$ T(AT)^+ = \frac{1}{\|T^\top v\|^2}TT^\top vu^\top, $$ so that the desired equality holds iff we have $v = \frac{TT^\top}{\|T^\top v\|^2} v$; in particular, we see that $v$ must be an eiegnvector of $TT^\top$. This fails to hold, for instance, with $$ v = (1,1)^\top, \quad T= \pmatrix{1&0\\0&2}, $$ and any choice of $u$.