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What I try is: if $\;A\;$ is not invertible then I can do elementary row operations and get $\;\overline A\;$ with one row all of zeros (which I can put at the bottom), and then I can put

$$\begin{pmatrix}0&\ldots&0\\0&\ldots&1\end{pmatrix}\overline A=0\rlap{\;\;\;\;/}\implies\begin{pmatrix}*&\ldots&*\\0&\ldots&1\end{pmatrix}=0$$

and get contradiction.

But I have the problem that $\;\overline A\neq A\;$, and I thought it is fine because $\;\overline A\;$ is similar to $\;A\;$ , but elementary row operations don't really keep the similarity.

Thus, I'm stuck in proving that if $\;BA=0\implies B=0\;$ , for all matrices $\;B\;$ , and these are squared matrices, then it must be that $\;A\;$ is invertible.

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Since $\overline{A}$ comes from $A$ using elementary row operations, there is an invertible matrix $E$ with $\overline{A}=EA$.

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  • $\begingroup$ Thank you, but I still doubt a little: how can I use this if $\;BA=0\;$ ? This isn't as $\;B(EA)=E(BA)=0\;$, right? $\endgroup$ – user177692 Feb 17 '15 at 17:26
  • $\begingroup$ Use your $B$, $B(EA)=(BE)A$ The bottom row of $E$ is not all zeros because it is invertible, so $BE\neq0$. But $B(EA)=0$. $\endgroup$ – Empy2 Feb 17 '15 at 17:29
  • $\begingroup$ Oh, I see. Very nice. Thank you very much. $\endgroup$ – user177692 Feb 17 '15 at 17:31

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