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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$\begingroup$ Show that the kernel on the left of AA' equals the kernel on the left of A. $\endgroup$– PhiraNov 16, 2011 at 15:50
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5$\begingroup$ May I suggest answering your own question, if you've truly gotten it? $\endgroup$– J. M. ain't a mathematicianNov 16, 2011 at 16:05
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$\begingroup$ @Phira would you mind answering? I would be interested. $\endgroup$– draks ...Jul 17, 2012 at 19:06
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1 Answer
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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 $A$.
