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Assume that $A$ is an linear operator on a real vector spacev $V$.

I wish to prove that if for some $x,y \in V$ such that $x\neq 0$ or $y \neq 0$ and some $a,b \in \mathbb R$ and $b\neq 0$, the following conditions hold $$ Ax=ax-by, Ay=ay+bx $$ then $x,y$ are linearly independent.

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closed as off-topic by Semiclassical, Jyrki Lahtonen, Najib Idrissi, Vedran Šego, beep-boop Oct 11 '14 at 14:50

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    $\begingroup$ What did you try? A proof by contradiction would begin by assuming $x,y$ are not linearly independent. Combine the expressions resulting from that assumption with the conditions above involving $a,b,x,y$ and see if this gives a contradiction. $\endgroup$ – hardmath Oct 11 '14 at 12:44
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Suppose the contrary, i.e., that they are linearly dependent. Then $x=ky$, where $k \ne 0$, so

\begin{align*} Ax=kAy &\implies ax-by=k(ay+bx) \implies x(a-kb)=y(ak+b) \\ &\implies ky(a-kb)=y(ak+b) \implies ak-k^2b=ak+b \\ &\implies k^2=-1. \end{align*}

This is impossible as $V$ is a real vector space.

Hence they are linearly independent.

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