# $Av_1=Av_2 \neq 0$ implies $v_1=v_2$.

This is a seemingly simple question, but had me thinking quite a bit...maybe someone has an elegant proof.

For an arbitrary field $\mathscr{F}$ let $A$ be an $n \times n$ matrix over $\mathscr{F}$ and $v_1, v_2$ vectors in $\mathscr{F}^n$ such that $Av_1=Av_2 \neq 0$ and suppose $v_2 \neq v_1 +n$ where $n$ is some vector in the null space of $A$ and $n\neq 0$. Note that by the given conditions $A$ is not the zero matrix, and $v_1,v_2$ are not zero vectors. Prove or disprove $v_1=v_2$.

• $A(v_2-v_1)=0$ so that $v_2-v_1$ is in the null space of A, no? – Paul Sep 6 '16 at 9:12
• correct, but you have to show $v_2-v_1=0$, if the result is true – Christiaan Hattingh Sep 6 '16 at 9:13
• By assumption, $v_2-v_1\neq n$, so it cannot be in the nullspace of $A$, contradiction. – Dietrich Burde Sep 6 '16 at 9:14
• You need to change the line "$v_2\neq v_1+n$ for some $n\in\ker(A)$" to "for some $n\neq 0$", because $A(v_2-v_1)=0$, and $0$ is always in the nullspace of $A$. Then the correct condition gives $v_2=v_1+0$, which you want. – Dietrich Burde Sep 6 '16 at 9:22
• What quantification is on $n$? If it says $\forall n\neq 0$, then implication is true. If it says $\exists n$, then implication is not true, since $v_2-v_1\in\ker A$, and the condition then translates to "there is more than one element in $\ker A$". – Ennar Sep 6 '16 at 9:43

$Av_1=Av_2\Rightarrow A(v_1-v_2)=0\Rightarrow v_1-v_2\in N(A)\Rightarrow v_1=v_2.$
• just to be explicit: $v_1-v_2=n$ so that $v_2=v_1-n$, and since $-n$ cannot be nonzero, the result follows. – Christiaan Hattingh Sep 6 '16 at 10:20