To find the rank of $T_A$ Let $A$ be an $n \times n$ matrix and let $$T_A : M_{n \times n}(F) \to M_{n \times n}(F)$$ be the linear transformation $$X \to AXA.$$ We want to find the rank of $T_A$. Also we have to prove that there exists a matrix $B$ such that $ABA = A$.
I tried to find the Ker of $T_A$. Can someone give any hints or direction to proceed?
 A: Let $r$ be the rank of $A$. Then $A=USV$ for some invertible matrices $U$ and $V$ with $S=I_r\oplus0$. Now define two linear maps $\alpha$ and $\beta$ on $M_n(\mathbb R)$ by
$$
\alpha(X)=VXU,\ \beta(X)=UXV.
$$
Then $\alpha,\beta$ are invertible linear maps and $T_A=\beta\circ T_S\circ\alpha$. Therefore,

*

*$T_A$ has the same rank as $T_S$;

*if $SCS=S$ for some matrix $C$, then $ABA=A$ when $B=V^{-1}CU^{-1}$.

Hence the problem boils down to the special case where $A=S$. You may continue from here.
A: For the second thing you want to prove, the idea is that $A$ can have elementary column operatortions performed on it to turn it into a block matrix where the upper left block is an invertible $(n-k) \times (n-k) $ matrix and all the other blocks are $0$.
Let $C$ be the product of all elementary matrices used to perform these operations. Then take $D$ to be the a block matrix of the same type where the top left is the inverse to the top left block of $A$ and the rest are also $0$ blocks. Then $B = CD$.
