Linear transformation proof

I'm trying to prove if we have a linear transformation that maps to itself that every element in the domain can be written as the sum of elements in the kernal and image. Would this be a sufficient proof?

Suppose the domain and co-domain are both S.

Then if s is an element of S we can write s=s-T(s)+T(s)

so then by T(s) is an element of im(T) by definition. Also s-T(s) is in kerT since T(s-T(s))=T(s)-T(s)^2=T(s)-T(s)=0 [since].

Hence s=kerT+imT.

Reply 1

maths learner

I also thought that I could prove it using the rank nullity theorem possibly?

Reply 2

ghostwalker

Original post by maths learner

Consider

f:\mathbb{R}^2 \to \mathbb{R}^2

given by

f(x,y)=(y,0)

This is a linear transformation, and it contradicts what you're trying to prove.

Your next task is to find out what you've assumed in your proof that is incorrect.

(edited 10 years ago)

Reply 3

BlueSam3

Original post by maths learner

I'm trying to prove if we have a linear transformation that maps to itself

This is unclearly worded - I'm assuming that you mean the linear map in question maps a vector space to itself throughout this post.

that every element in the domain can be written as the sum of elements in the kernal and image. Would this be a sufficient proof?

Suppose the domain and co-domain are both S.

Then if s is an element of S we can write s=s-T(s)+T(s)

so then by T(s) is an element of im(T) by definition. Also s-T(s) is in kerT since T(s-T(s))=T(s)-T(s)^2=T(s)-T(s)=0 [since].

Hence s=kerT+imT.

It is not clear why you are assuming that

T \circ T = T

. This is not necessarily true. Instead, take a basis of the image and a basis of the kernel, and construct a basis of the whole space from them.

Original post by ghostwalker

Consider

f:\mathbb{R}^2 \to \mathbb{R}^2

given by

f(x,y)=(y,0)

This is a linear transformation, and it contradicts what you're trying to prove.

Your next task is to find out what you've assumed in your proof that is incorrect.

That can be written in the desired form:

(1,0)

is in the kernel, whilst

(0,1)

is in the image.

Reply 4

maths learner

Original post by BlueSam3

This is unclearly worded - I'm assuming that you mean the linear map in question maps a vector space to itself throughout this post.

It is not clear why you are assuming that

T \circ T = T

. This is not necessarily true. Instead, take a basis of the image and a basis of the kernel, and construct a basis of the whole space from them.

That can be written in the desired form:

(1,0)

is in the kernel, whilst

(0,1)

is in the image.

Sorry about the poor wording. I'm assuming that we have a Linear transformation from a vector space S to itself. And that it's an idempotent linear transformation I.E. ToT=T.

So I can do it, by kind of using the dimension theorem, and having a basis in kerT and expanding it to the whole basis, and then show the image is also a basis?

Reply 5

ghostwalker

Original post by BlueSam3

That can be written in the desired form:

(1,0)

is in the kernel, whilst

(0,1)

is in the image.

I beg to differ.

(0,1)

can't be in the image as it consists of elements of the form

(y,0)

in my example.

But the OP has now stated that ToT=T, so my example is no longer pertinent to the question. :sigh:

(edited 10 years ago)

Reply 6

BlueSam3

Original post by ghostwalker

I beg to differ.

(0,1)

can't be in the image as it consists of elements of the form

(y,0)

in my example.

But the OP has now stated that ToT=T, so my example is no longer pertinent to the question. :sigh:

EDIT: Oops, completely misread your example, sorry.

(edited 10 years ago)

Reply 7

BlueSam3

Original post by maths learner

Sorry about the poor wording. I'm assuming that we have a Linear transformation from a vector space S to itself. And that it's an idempotent linear transformation I.E. ToT=T.

Yes, in that case, your proof works, I think. It's not completely clear, but I think it's OK.

So I can do it, by kind of using the dimension theorem, and having a basis in kerT and expanding it to the whole basis, and then show the image is also a basis?

Well no, because the image isn't a basis. However, it does have a basis (by virtue of being a subspace of

S

), and hence a dimension.

Reply 8

maths learner

Original post by BlueSam3

Yes, in that case, your proof works, I think. It's not completely clear, but I think it's OK.

Well no, because the image isn't a basis. However, it does have a basis (by virtue of being a subspace of