# Thread: Help with linear transformation

1. ## Help with linear transformation

Hi guys.
I need to prove the following:

Let $V$ be a vector space over $\mathbb{R}$, such that $dim(V)>1$.
Let $\varphi:V\rightarrow V$ be a linear transformation such that: $\varphi^2=-I$.
Prove that for every $v\neq 0$: $v,\varphi(v)$ are linear independent.

This is what I tried so far:
Let $v\in V$.
suppose $\alpha_1v+\alpha_2\varphi(v)=0$, I need to prove $\alpha_1=\alpha_2=0$.
since $\varphi$ is linear, and $\alpha_1v+\alpha_2\varphi(v)=0$, then also $\varphi(\alpha_1v+\alpha_2\varphi(v))=0$, and then:
$\varphi(\alpha_1v+\alpha_2\varphi(v))=0$
$\varphi(\alpha_1v)+\varphi(\alpha_2\varphi(v)))=0$
$\alpha_1\varphi(v)+\alpha_2\varphi^2(v)=0$ now since $\varphi^2=-I$:
$\alpha_1\varphi(v)-\alpha_2v=0$

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

I can really use some guidance here.

2. ## Re: Help with linear transformation

Originally Posted by Stormey
Hi guys.
I need to prove the following:

Let $V$ be a vector space over $\mathbb{R}$, such that $dim(V)>1$.
Let $\varphi:V\rightarrow V$ be a linear transformation such that: $\varphi^2=-I$.
Prove that for every $v\neq 0$: $v,\varphi(v)$ are linear independent.

This is what I tried so far:
Let $v\in V$.
suppose $\alpha_1v+\alpha_2\varphi(v)=0$, I need to prove $\alpha_1=\alpha_2=0$.
since $\varphi$ is linear, and $\alpha_1v+\alpha_2\varphi(v)=0$, then also $\varphi(\alpha_1v+\alpha_2\varphi(v))=0$, and then:
$\varphi(\alpha_1v+\alpha_2\varphi(v))=0$
$\varphi(\alpha_1v)+\varphi(\alpha_2\varphi(v)))=0$
$\alpha_1\varphi(v)+\alpha_2\varphi^2(v)=0$ now since $\varphi^2=-I$:
$\alpha_1\varphi(v)-\alpha_2v=0$

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

I can really use some guidance here.
Hi Stormey!

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

What do you get for $\varphi(\varphi(v))$?

3. ## Re: Help with linear transformation

Hi ILikeSerena!
Thanks for the help.

let me see if I got it right:

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

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

?

Yep!