# How to prove $CAA^t=BAA^t$ implies $CA=BA$?

Suppose $A$, $B$, $C$ are $n\times n$ matrices. $A'$ denotes the transpose of $A$. $CAA'=BAA'$. How to prove $CA=BA$?

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Show that the kernel on the left of AA' equals the kernel on the left of A. –  Phira Nov 16 '11 at 15:50
thanks. I got it. –  Acky Nov 16 '11 at 15:56
May I suggest answering your own question, if you've truly gotten it? –  Ｊ. Ｍ. Nov 16 '11 at 16:05
@Phira would you mind answering? I would be interested. –  draks ... Jul 17 '12 at 19:06

The given equation $(C-B)AA'=0$ means that each row of the matrix $C-B$ is in the left kernel of $AA'$. The desired equation $(C-B)A=0$ means that each row of the matrix $C-B$ is in the left kernel of $A$. So, I want to show that the (left) kernel of the matrix $AA'$ is already the (left) kernel of the matrix $A$.
Suppose the vector $v$ is in the left kernel of $AA'$, i.e. $$vAA'=0$$ which implies $$vAA'v'=0=(vA)\cdot (vA)'=\|vA\|^2$$ (where $v'$ is the transpose of the vector $v$).
So, $vA$ is already the zero vector and therefore, $v$ is in the left kernel of $A$.