## Vector Transformation

Hi! I'm in serious need of some help.

I am supposed to show that a transformation $$\mathcal{A} = \mathbb{R}^n \rightarrow \mathbb{R}^m$$ can be separated into $$\mathcal{A} = i \circ \mathcal{B} \circ p$$ where
• $$p$$ is the projection on the (orthogonal) complement of the kernel of $$\mathcal{A}$$.

$$\mathcal{B}$$ is an invertible transformation from the complement to the kernel to the image of $$\mathcal{A}$$.

$$i$$ is the inclusion of the image in $$\mathbb{R}^n$$

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
 PhysOrg.com science news on PhysOrg.com >> Intel's Haswell to extend battery life, set for Taipei launch>> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens
 Recognitions: Homework Help Science Advisor 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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