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Thread: Help with linear transformation

  1. #1
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    Help with linear transformation

    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!
    Last edited by Stormey; Apr 6th 2013 at 01:23 AM.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Help with linear transformation

    Quote Originally Posted by Stormey View Post
    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!
    Hi Stormey!

    Suppose $\displaystyle v,\varphi(v)$ are linearly dependent.
    Then there is some $\displaystyle t \in \mathbb R$ such that $\displaystyle \varphi(v) = t v$.

    What do you get for $\displaystyle \varphi(\varphi(v))$?
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  3. #3
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    Re: Help with linear transformation

    Hi ILikeSerena!
    Thanks for the help.

    let me see if I got it right:

    if $\displaystyle \varphi(v)=tv$, then $\displaystyle \varphi(\varphi (v))=\varphi(tv)$, and then we'll get a contradiction since:

    $\displaystyle \varphi(\varphi (v))=\varphi(tv)$
    $\displaystyle t\varphi(v)=-v$
    $\displaystyle \varphi(v)=-\frac{1}{t}v$ (because according to the assumption $\displaystyle t\neq 0$)
    $\displaystyle t=-\frac{1}{t}$
    $\displaystyle t^2=-1$
    no solution in $\displaystyle \mathbb{R}$

    ?
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Help with linear transformation

    Yep!
    Thanks from Stormey
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