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If matrix multiplication $B*A=C*A$, does it mean $B=C$?

If A is invertible, then I guess this should work. If not, then?

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    $\begingroup$ Hint: try $A=0$. If $A$ is invertible, then you can multiply on the right hand side by $A^{-1}$. $\endgroup$ – mdp Oct 13 '14 at 10:05
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For a given matrix $A$, we can say that $BA=CA \Rightarrow B=C$ iff $DA={\bf{0}}\Rightarrow D=\bf{0}$.

If the rows of $A$ are linearly independent, then the second condition will hold, since for any (row) vector $\vec{v}$, $\vec{v}A$ gives a linear combination of the rows of $A$, and only the trivial linear combination gives the zero-vector. In particular this will hold for each row of $D$.

Conversely if the rows of $A$ are linearly dependent, we can find a nonzero row vector $\vec{v}$ such that $\vec{v}A=\vec{0}$.

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If $A$ is invertible, you get: $$BA = CA \implies BAA^{-1} = CAA^{-1} \implies BI = CI \implies B = C.$$ If $A$ is not invertible, it is false.

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    $\begingroup$ In fact, if $A$ is not invertible, then there always exist two matrices $B\neq C$ such that $BA=CA$. $\endgroup$ – 5xum Oct 13 '14 at 10:13
  • $\begingroup$ @5xum Why " there always exist two matrices ..."? Can you give an answer and explain why? Thanks $\endgroup$ – Tyler 十三将士归玉门 Oct 13 '14 at 10:31
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    $\begingroup$ @5xum: Actually be careful here, your comment is true if $A$ is square, but not necessarily otherwise. More accurate would be to say that if the rows of $A$ are not linearly independent, then there exist two matrices $B\ne C$ such that $BA=CA$. $\endgroup$ – paw88789 Oct 13 '14 at 10:33
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    $\begingroup$ @Tyler If $A$ is not invertible and square, then its rows are not linearly independent, meaning there exists a nonzero vector $v$ such that $v^TA=[0,0,\dots, 0]$, so if I make a matrix $B$ in which every row is equal to $v$, you have $BA=0=0\cdot A$, so $BA=CA$ for $C=0$, however, $B\neq 0 = C$ $\endgroup$ – 5xum Oct 13 '14 at 10:53

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