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**Stormey** Hi guys.

I need to prove the following:

Let $\displaystyle V$ be a vector space over $\displaystyle \mathbb{R}$, such that $\displaystyle dim(V)>1$.

Let $\displaystyle \varphi:V\rightarrow V$ be a linear transformation such that: $\displaystyle \varphi^2=-I$.

Prove that for every $\displaystyle v\neq 0$: $\displaystyle v,\varphi(v)$ are linear independent.

This is what I tried so far:

Let $\displaystyle v\in V$.

suppose $\displaystyle \alpha_1v+\alpha_2\varphi(v)=0$, I need to prove $\displaystyle \alpha_1=\alpha_2=0$.

since $\displaystyle \varphi$ is linear, and $\displaystyle \alpha_1v+\alpha_2\varphi(v)=0$, then also $\displaystyle \varphi(\alpha_1v+\alpha_2\varphi(v))=0$, and then:

$\displaystyle \varphi(\alpha_1v+\alpha_2\varphi(v))=0$

$\displaystyle \varphi(\alpha_1v)+\varphi(\alpha_2\varphi(v)))=0$

$\displaystyle \alpha_1\varphi(v)+\alpha_2\varphi^2(v)=0$ now since $\displaystyle \varphi^2=-I$:

$\displaystyle \alpha_1\varphi(v)-\alpha_2v=0$

but now I'm not sure what conclusion should I draw from this...

I can really use some guidance here.

Thanks in advanced!