Vector Transformation


by nille40
Tags: transformation, vector
nille40
nille40 is offline
#1
Jan31-04, 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
  • [tex]p[/tex] is the projection on the (orthogonal) complement of the kernel of [tex]\mathcal{A}[/tex].

    [tex]\mathcal{B}[/tex] is an invertible transformation from the complement to the kernel to the image of [tex]\mathcal{A}[/tex].

    [tex]i[/tex] is the inclusion of the image in [tex]\mathbb{R}^n[/tex]

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 follow-up, but got no response on that.

Thanks in advance,
Nille
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matt grime
matt grime is offline
#2
Jan31-04, 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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