Suppose $A$, $B$, $C$ are $n\times n$ matrices. A' denotes the transpose of $A$. CAA'=BAA'. How to prove $CA=BA$?
How to prove $CAA^t=BAA^t$ implies $CA=BA$?
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0@Phira would you m$in$d answering? I would be interested. – 2012-07-17
1 Answers
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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 $A$.