# Linera Transformation Questions

1. Jul 21, 2008

### JinM

Hey everyone,

Let T: W -> V be a linear transformation and T(w_1) = v_1 for some w_1 \in W and v_1 \in V. Now set S = {w \in W | T(w) = T(w_1) = v_1}. Prove that S = w_1 + Kernel(T) = {w_1 + a | a \in Ker(T)}.

Let w \in S. T(w_1) = T(w) = T(w) + T(a) = T(w_1) + T(a) = T(w_1 + a) = v_1, where a \in Ker(T).

Thus

S_1 = {w \in W | T(w) = T(w_1) = v_1}; S_2 = {w_1 + a | a \in Ker(t)}

S_1 = S_2 = S and we're done. Correct?

Last edited: Jul 21, 2008
2. Jul 21, 2008

### JinM

Hmm....

All elements of S = S_1 constitute elements whose images are v_1, and the elements of S_2 constitute elements whose images are v_1. Now how can I prove they're no preimages in S_1 that aren't present in S_2 and vice versa?

3. Jul 21, 2008

### Defennder

This looks ok, though the steps appear to be out of order.

You mean how do you prove there isn't any $$\vec{u} \in W \ | \ T(\vec{u}) \neq \vec{v_1}$$? Didn't you just do that with the above? Note that by definition all such $$\vec{u} \ \mbox{can be expressed as} \ \vec{w_1} + \vec{a}$$. What does that tell you?

4. Jul 21, 2008

### HallsofIvy

Staff Emeritus
If x is any member of S then T(x)= T(w_1)= w by definition of S. Now, let a= x- w_1 so that x= w_1+ a. It is easy to prove T(a)= 0 (do that yourself) and so x is in S_2.

Conversely, if x is any member of S_2, then x= w_1+ a by definition of S_2. But then T(x)= T(w_1+ a)= T(w_1)+ T(a)= ? (and so x is in S).