How to make $ABA^T$ as diagonal as possible by designing symmetric pd $B$? Given $A\in\mathbb{R}^{n\times m}$, where $n\geq m$, assume $\mathrm{rank}(A)=m.$

How to make $ABA^T$ as diagonal as possible by designing $m\times m$
symmetric positive definite $B$?

By "as diagonal as possible", I mean the followings:

*

*$ABA^T$ equals to diagonal matrix, with possible zeros in diagonal elements, or

*$ABA^T$ is block diagonal with each block being rank $1$, or

*$ABA^T$ is block diagonal.


My attempt: When $n=m$, we can choose $B=(A^TA)^{-1}$, then $ABA^T=I_n$, where $I_n$ is an $n\times n$ identity matrix.
When $n>m$, then $\mathrm{rank}(ABA^T)=m$, thus we cannot achieve identity matrix.
First attempt was to split $B=CC^T$ and choose $C=A^+$, (pseudoinverse of $A$), however it turns out for tall matrix $A$, pseudoinverse must be multiplied from the left in order to get an identity.
Second attempt was to do SVD of $A=U\Sigma V^T$, and choose $B=V\Sigma_m^{-2}V^T$, where $\Sigma=\begin{bmatrix} 
\Sigma_m\\ 
0_{n-m}
\end{bmatrix}$. However, $ABA^T=U\begin{bmatrix} 
I_m&\\ 
&0_{n-m}
\end{bmatrix}U^*$, I couldn't get rid of $U$.
If there is no way to diagonalize, I am really interested in case 2: block diagonalize with each block being rank $1$.

This isn't always possible. Suppose $v\in\mathbb R^2$ is entrywise nonzero and
$$
A=\pmatrix{I_2\\ v^T}
\ \Rightarrow\ C:=ABA^T=\pmatrix{B&Bv\\ v^TB&v^TBv}.
$$
Then $Bv\ne0$. Hence $C$ is not a block-diagonal matrix in the form of $M\oplus \lambda$, where $M$ is $2\times2$. It is not in the form of $\lambda\oplus M$ either, for,

*

*if $B$ is not a diagonal matrix, then $c_{12}=b_{12}$ is nonzero;

*if $B$ is diagonal, then $c_{13}$, which is equal to the first entry of $Bv$, is nonzero.

So, in this example, $ABA^T$ is never block-diagonal (and in particular, it is never diagonal).
I don't know in general how often $B$ can be chosen to make $ABA^T$ block-diagonal, but I suspect that this is rarely possible (unless $A$ is a square nonsingular matrix, of course).
