Vector Spaces and Correspondence

[Ninty64]

Homework Statement

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}

Homework Equations

I_{V} is the identity map
L(V,V) is the map of all linear operators on V

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.

 

[micromass]

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.

 

[Ninty64]

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

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?

 

[Ninty64]

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.

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}