# I Proof that the general solution of a linear equation is....

1. Mar 19, 2016

### Woolyabyss

any particular solution plus the general solution to the homogeneous equation.

I'm having difficuilty understanding this proof from my lecture notes

Theorem
: Let T : V → W be a linear transformation. Let w ∈ W and suppose T(u0) = w

T(v) = 0. where v ∈ V (the kernel )

to prove:
T(u) = w, where u = u0 + v

proof: (from my lecture notes)

First, we show that every vector of the form u = u0 + v
where T(u0) = w and T(v) = 0, satisfies T(u) = w:
T(u) = T(u0 + v) = T(u0) + T(v) = w + 0 = w

Now we show that every solution looks like this.
Suppose that T(u) = w. Let v = u − u0.
Then u0 + v = u0 + (u − u0) = u and v is in the kernel of
T: T(v) = T(u − u0) = T(u) − T(u0) = w − w = 0

I can follow the first part of the proof. The second part is where I'm having difficulty. I'm not sure how the last three lines show that every solution looks like u = u0 + w, particularly the second last line. Is that not showing what we were already given?
Any help would be appreciated

2. Mar 19, 2016

### BvU

Replace
"Suppose that T(u) = w. Let v = u - u0"
by
"Suppose that T(y) = w. Let v = y - u0 "
and now it no longer shows what we were already given !

3. Mar 19, 2016

### WWGD

This is the way that makes sense to me:

Assume T(a)=w =T(b). Then T(a)-T(b)=T(a-b)=0 , so a-b $\in K$ ; K :=Kernel T.

Then a-b=k, so a =b+K. Then T(b)=w iff b=a-K . EDIT: BTW this is related to one of the
isomorphism theorems stating that if T: V-->W is a homeomorphism (a linear map, in our case)

with kernel K , then T: V/K --> T(W) is an isomorphism. Moding out by the kernel just collapses
elements with the same preimage, so you get a linear bijection between V/K and the image.

Last edited: Mar 19, 2016