Vector Transformation in \mathbb{R}^n and \mathbb{R}^m with Separable Components

[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.

Thanks in advance,
Nille

 

[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.


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.

 

[M98Ranger]

Hi Nille,

I'd be happy to help you with this problem. Let's break it down step by step.

First, let's define our transformation \mathcal{A}: \mathbb{R}^n \rightarrow \mathbb{R}^m. This means that \mathcal{A} takes in a vector in \mathbb{R}^n and outputs a vector in \mathbb{R}^m. So we can represent \mathcal{A} as a matrix A with m rows and n columns.

Next, let's define the kernel of \mathcal{A}. The kernel of a transformation is the set of all vectors that get mapped to the zero vector in the output space. In other words, it's the set of all x \in \mathbb{R}^n such that \mathcal{A}(x) = 0.

Now, let's define the complement of the kernel. This is the set of all vectors in \mathbb{R}^n that are not in the kernel of \mathcal{A}. In other words, it's the set of all x \in \mathbb{R}^n such that \mathcal{A}(x) \neq 0.

The projection on the complement of the kernel of \mathcal{A} is a transformation p that takes in a vector x \in \mathbb{R}^n and outputs a vector p(x) \in \mathbb{R}^n, where p(x) is the projection of x onto the complement of the kernel of \mathcal{A}. This means that p(x) is the closest vector to x that is not in the kernel of \mathcal{A}. We can represent this as a matrix P with n rows and n columns.

Now, let's define \mathcal{B}. This is a transformation from the complement of the kernel of \mathcal{A} to the image of \mathcal{A}. This means that \mathcal{B} takes in a vector x \in \mathbb{R}^n and outputs a vector \mathcal{B}(x) \in \mathbb{R}^m, where \mathcal{B}(x) is the transformation of x by \mathcal{A}. We can represent this as a matrix B with m rows and n columns.