# Thread: Prove a direct sum...

1. ## Prove a direct sum...

Here's the question as an image because it's easier to format:

Please note I'll just write + to mean the plus with the circle around it (direct sum). + is just a normal addition.

My work so far

V = im(T) + ker(S) means that im(T) ∩ ker(S) = {0} and that im(T) + ker(S) = V.

If ST = 1v, then TS = 1w (Identity transformations). Thus w = T(v) and v = S(w).
S[w - TS(w)] = S(w) - STS(w) = v - ST(v) = v - S(w) = v -v = 0, therefore it's in ker(S).

Now I'm stuck. I don't know how to use this to do the proof... I think showing the intersection might go:
im(T) ∩ ker(S) = T(v) ∩ w - TS(w) = w ∩ 0 = 0. But I'm not sure.

I have no idea about the im(T) + ker(S) part though.

2. edit: disregard me, I'm an idiot

Spoiler:

****preserved for reference****

Suppose we can write a vector $\displaystyle v\in V$ as the sum of $\displaystyle s+T(t)$ where S(s)=0 and $\displaystyle t\in V$ ie as the sum of an element of ker S and of im T. Apply S to the equation $\displaystyle v=s+T(t)$ to get $\displaystyle S(v)=S(s) + ST(t) = 0 + t$, so $\displaystyle t=S(v)$. Therefore $\displaystyle s = v - TS(v)$. So we have the form of the required decompositon, $\displaystyle v = (v - TS(v)) + T(S(v))$ . Does this help you?

3. Hmm, I'm not sure I got all of that. I'll repeat to make sure.

So we get to the point v = (v - TS(v)) + T(S(v)). The first element is in ker(S), and the second element is in im(T).

Then ker(S) + im(T) = v.

I'm confused though. Does this hold for all v in V? Part of what was tripping me up before was that w-TS(w) is a specific element, so aren't you just proving one case?

4. edit: disregard me, I'm a moron

Spoiler:

*****preserved for referene*****

Well the relation is only formal since s is in W. You need to find a bijection between formal sums of elements of ker S and im T, and vectors in V. You should see v = (v - TS(v)) + TS(v) as "suggestive" of this bijection.

edit: This was really vague. Will fix in a minute.
edit2: Note that this is formal. It doesn't really hold per se since v is not in the domain of S so the compositions don't really make sense. It was meant to suggest why the term w-TSw will be useful. To find the bijection I was talking about, break V into the ker T and its complement and look at how the complement relates to im S.

5. I don't really understand what you mean... am I supposed to show that it's one to one?

6. Thanks! It definitely shows how w-TS(w) will be useful.

I looked up "complement" in the index of my book and it's not there... but based on what I remember from stats, it would be everything that is NOT in ker(S). Which I'm guessing should turn out to be equal to im(T).

So I understand how this works in theory, just not sure how to apply it.

7. Wait, disregard all of that. If ST=1, don't V and W have to be isomorphic? I have no idea anymore. Disregard me! Sorry!