Linear transformation proof

[trojansc82]

Homework Statement

Let T: V --> W be a linear transformation and let U be a subspace of W. Prove that the set T^-1 (U) = {v E V: T (v) E U} is a subspace of V. What is T^-1 if U = {0}?

Homework Equations

The Attempt at a Solution

Since U is a subspace, k(v) = ku. Also, if u and v are in U, u + v lies in U. Therefore, any inverse linear transformation of a vector within u will lie in the subspace U.

 

[micromass]

There are three things to prove about $$T^{-1}(U)$$, can you tell me what those things are?

 

[trojansc82]

1. T^-1 a one to one transformation.

2. Closure under addition

3. Closure under scalar multiplication

 

[micromass]

1. T^-1 a one to one transformation.

2. Closure under addition

3. Closure under scalar multiplication

1 isn't always true. So, that is incorrect. What I was looking for was:

1) It contains the zero vector

2&3 are correct.

So, start with (1), try to prove that $$0\in T^{-1}(U)$$.

 

[trojansc82]

For vector u in T^-1(U), 0u = 0. (zero vector property).

 

[micromass]

For vector u in T^-1(U), 0u = 0. (zero vector property).

Correct, but I don't see how that has to do with anything. By definition, we have that

$$T^{-1}(U)=\{v\in V~\vert~T(v)\in U\}$$

Thus, we have $$v\in T^{-1}(U)~\Leftrightarrow~T(v)\in U$$

So, to prove that $$0\in T^{-1}(U)$$, the statement above shows that it is sufficient to show that $$T(0)\in U$$. Do you see why this is the case?

 

[trojansc82]

How would I proceed with scalar multiplication and closure under addition? The same process?

 

[micromass]

Yes, it's exactly the same thing!

 

[HallsofIvy]

Homework Statement

Let T: V --> W be a linear transformation and let U be a subspace of W. Prove that the set T^-1 (U) = {v E V: T (v) E U} is a subspace of V. What is T^-1 if U = {0}?

The question is asking you to determine the set of all u such that T(u)= 0. There is a specific name for that set!

 

[trojansc82]

The question is asking you to determine the set of all u such that T(u)= 0. There is a specific name for that set!

Yeah, the kernel.