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Vector Transformation

  1. Jan 31, 2004 #1
    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,
  2. jcsd
  3. Jan 31, 2004 #2

    matt grime

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    Homework Helper

    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


    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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