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!

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.

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 $\displaystyle \varphi(\varphi(v))$?

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 $\displaystyle \mathbb{R}$

?

Re: Help with linear transformation