Vector Spaces and Correspondence

1. Oct 8, 2012

Ninty64

1. The problem statement, all variables and given/known data
This question came out of a section on Correspondence and Isomorphism Theorems

Let $V$ be a vector space and $U \neq V, \left\{ \vec{0} \right\}$ be a subspace of $V$. Assume $T \in L(V,V)$ satisfies the following:
a) $T(\vec{u} ) = \vec{u}$ for all $\vec{u} \in U$
b) $T(\vec{v} + U) = \vec{v} + U$ for all $\vec{v} \in V$
Set $S=T-I_{V}$. Prove that $S^{2}=\vec{0}_{V \rightarrow V}$

2. Relevant equations
$I_{V}$ is the identity map
$L(V,V)$ is the map of all linear operators on V

3. The attempt at a solution
I have trouble understanding the question.
Since $T \in L(V,V)$ then how is $T(\vec{v} + U) = \vec{v} + U$ for all $\vec{v} \in V$?
Wouldn't that mean $T:V/U \rightarrow V/U$?
I don't understand, what is $T(\vec{v})$ equal to?
Does $T(\vec{v})=\vec{v}$ or $T(\vec{v}) = [\vec{v}]_W$ or something else?

I'm sorry if this is a silly question.

2. Oct 8, 2012

micromass

Staff Emeritus
With $T(v+U)=v+U$, they mean that the set v+U is mapped to the set v+U.
So

$$T(\{v+u~\vert~u\in U\})=\{v+u~\vert~u\in U\}$$

Or in another way: for each $v\in V$ and $u\in U$, there exists $u^\prime\in U$ such that $T(v+u)=v+u^\prime$.

3. Oct 8, 2012

Ninty64

This is where I confuse myself. T is a linear transformation from V to V. If it maps the set v+U to the set v+U, then wouldn't that be mapping cosets of V mod U to cosets of V mod U instead of mapping V to V?

4. Oct 9, 2012

Ninty64

I think I get it now. The function maps the coset of V mod U to the same coset of V mod U by mapping each individual element to another element in that coset.

I'm sorry if it seemed I brushed over your post and didn't completely read it. I did. I just didn't understand it. I'm having trouble with my Linear Algebra 2 class, and I'm glad that you responded. Thank you a lot!

So then I get
Let $\vec{v} \in V$ be arbitrary
$S^2(\vec{v} + \vec{u}), \vec{u} \in U$
$=S(T(\vec{v} + \vec{u}) - I(\vec{v} + \vec{u}))$
$=S(\vec{v} + \vec{u} - ( \vec{v} + \vec{u}))$ where $\vec{u} \in U$
$=S(\vec{u} - \vec{u})$
$=S(\vec{y})$ where $\vec{y} = \vec{u}- \vec{u} \in U$
$=T(\vec{y}) - I(\vec{y})$
$=\vec{y} - \vec{y}=\vec{0}$