
#1
Jan3104, 04:41 AM

P: 34

Hi! I'm in serious need of some help.
I am supposed to show that a transformation [tex]\mathcal{A} = \mathbb{R}^n \rightarrow \mathbb{R}^m[/tex] can be separated into [tex]\mathcal{A} = i \circ \mathcal{B} \circ p[/tex] where
I hardly know where to start! I would really like some help. I asked this question before, in a different topic, but got a response I didn't understand. I posted a followup, but got no response on that. Thanks in advance, Nille 



#2
Jan3104, 08:00 AM

Sci Advisor
HW Helper
P: 9,398

let K be the kernel of B. Then A is K direct sum K*, where we'll use * to denote the complementary vector space.
Let p be the map p(k) = 0 if k in K, and p(x)=x for x in K*, extended linearly. This means that any vector in A can be written as x+k for x in K* and k in K, and then p(x+k)=x. This is your projection. Notice that for all v in A that Bp(v)=v. The inclusion is the dual construction: Let I be the image of B. This is a subspace of of R^n. Pick a complementary subspace I* Then there is a natural map from I to Idirect sum I*, just the inclusion of the vector, call tis map i. Obviously the map iBp is the same as B. This is just the Isomorphism theorems glued together. 


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