Help with linear transformation

• Apr 6th 2013, 01:20 AM
Stormey
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.
• Apr 6th 2013, 04:44 AM
ILikeSerena
Re: Help with linear transformation
Quote:

Originally Posted by 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.

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 $\displaystyle \varphi(\varphi(v))$?
• Apr 6th 2013, 05:19 AM
Stormey
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 $\displaystyle \mathbb{R}$

?
• Apr 6th 2013, 05:24 AM
ILikeSerena
Re: Help with linear transformation
Yep!