Direct Sum Proof Homework: Solve V = im(T) + ker(S)

jumbogala
Messages
414
Reaction score
4

Homework Statement


Here's the question... it was easier to format it in paint haha:
linalg.jpg


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

Homework Equations





The Attempt at a Solution


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. 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.
 
Physics news on Phys.org
Update: I'm thinking about using

w = (w - TS(w)) + TS(w).

Then w-TS(w) is in ker(S) and TS(w) is in im(T). Although I guess this is not really helpful because this doesn't show that it's equal to the space V...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top