# Vector Transformation

1. Jan 31, 2004

### nille40

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.

Nille

2. Jan 31, 2004

### matt grime

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.

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.