-1
$\begingroup$

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 $v_1=v_2$.

$\endgroup$
  • 2
    $\begingroup$ $A(v_2-v_1)=0$ so that $v_2-v_1$ is in the null space of A, no? $\endgroup$ – Paul Sep 6 '16 at 9:12
  • $\begingroup$ correct, but you have to show $v_2-v_1=0$, if the result is true $\endgroup$ – Christiaan Hattingh Sep 6 '16 at 9:13
  • $\begingroup$ By assumption, $v_2-v_1\neq n$, so it cannot be in the nullspace of $A$, contradiction. $\endgroup$ – Dietrich Burde Sep 6 '16 at 9:14
  • 1
    $\begingroup$ 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. $\endgroup$ – Dietrich Burde Sep 6 '16 at 9:22
  • 1
    $\begingroup$ 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 $\ker A$". $\endgroup$ – Ennar Sep 6 '16 at 9:43
2
$\begingroup$

$Av_1=Av_2\Rightarrow A(v_1-v_2)=0\Rightarrow v_1-v_2\in N(A)\Rightarrow v_1=v_2.$

$\endgroup$
  • $\begingroup$ Thanks Rafael, yes it's actually really simple...can't believe I missed that...thanks for your patience $\endgroup$ – Christiaan Hattingh Sep 6 '16 at 10:14
  • 1
    $\begingroup$ 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. $\endgroup$ – Christiaan Hattingh Sep 6 '16 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.