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Let $F$ be a subfield of the complex numbers. Let $A,B\in F^{n\times n}$ and $X\in F^{n\times 1}.$ If $A\ne B,$ then $AX\ne BX$ for every non-zero $X.$

I am NOT sure if the above statement is right. I did some computation with random matrices and it seems true to me. This is the argument I tried :

Let $A=[A_1,A_2,\dots,A_n]$ and $B=[B_1,B_2,\dots,B_n]$ where $A_i,B_i \in F^{n\times 1}$ are columns of $A$ and $B$ respectively. Let $X=\begin{bmatrix}x_1\\\vdots \\x_n\end{bmatrix}.$

Since $A\ne B,$ choose $j$ such that $A_j\ne B_j.$ Let $J_1$ be the set of all such $j$'s and $J_2=I$ \ $ J_1$ where $I=\{1,2,\dots ,n\}.$ Then, $AX=\sum_{i\in J_1} x_iA_i +\sum_{i\in J_2} x_iA_i \ne \sum_{i\in J_1} x_iB_i + \sum_{i\in J_2} x_iB_i= BX.$

(1) Is my argument correct?
(2) Also, can $F$ be generalized?

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It's not correct, as it would imply that, as soon as $A\ne B$, $A-B$ is non-singular. Why should it be true?

Actually, $A\ne B$ only implies that for some $X$, one has $AX\ne BX$.

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  • $\begingroup$ Thanks! I just noticed that if $X$ is such that $x_i=0$ for $i\in J_1$ then $AX=BX.$ $\endgroup$
    – Bijesh K.S
    Jul 7 '17 at 10:07
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    $\begingroup$ That's absolutely true. For an assertion that sounds too good to be true, it's often useful to check what it would imply. $\endgroup$
    – Bernard
    Jul 7 '17 at 10:13
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In an integral domain $R$, we have $ax=bx\Rightarrow a=b$ for all $a,b,x\in R$ with $x\ne 0$. To see this, write $(a-b)x=0$. Since $R$ has no zero divisors and $x\ne 0$ we must have $a-b=0$, i.e., $a=b$. But an $n\times n$ matrix ring is not an integral domain for $n\geq 2$.

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Let $A = \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}, B= \begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}$ and $X=\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$ and look what happens !

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  • $\begingroup$ $X$ is supposed to be in $F^{n\times 1.}$ In fact I just noticed that if $X$ is such that $x_i=0$ for $i\in J_1$ then $AX=BX.$ So my argument is not right. $\endgroup$
    – Bijesh K.S
    Jul 7 '17 at 10:14

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