Prove that dim(Ker(T)) is either n or n-1

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The discussion revolves around proving that the dimension of the kernel of a linear transformation T from a vector space V to the real numbers is either n or n-1, where the dimension of V is n. Participants explore the dimension theorem, noting that the dimension of the image of T must be less than or equal to 1. There is also a related problem regarding the equality of kernels for two transformations, T and S, and the conditions under which they can be scalar multiples of each other. The importance of rigor in mathematical proofs is emphasized, particularly in establishing the truth of statements about linear transformations and their kernels. The conversation highlights the need for careful reasoning and verification in mathematical arguments.
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Homework Statement


Having:

\displaystyle T:V\to \mathbb{R}
\displaystyle \dim\left( V \right)=n

Prove that \displaystyle \dim\left( \ker \left( T \right) \right)=n\text{ or }n-1

Homework Equations



The Attempt at a Solution



I would start stating the theorem of the dimensions:

\displaystyle \dim\left( V \right)=\dim\left( \ker \left( T \right) \right)+\dim\left( \operatorname{Im}\left( T \right) \right)

But I get nowhere with that theorem. Any clue?

Thanks!
 
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Hernaner28 said:

Homework Statement


Having:

\displaystyle T:V\to \mathbb{R}
\displaystyle \dim\left( V \right)=n

Prove that \displaystyle \dim\left( \ker \left( T \right) \right)=n\text{ or }n-1

Homework Equations



The Attempt at a Solution



I would start stating the theorem of the dimensions:

\displaystyle \dim\left( V \right)=\dim\left( \ker \left( T \right) \right)+\dim\left( \operatorname{Im}\left( T \right) \right)

But I get nowhere with that theorem. Any clue?

Thanks!

What is dim(V)? What are possibilities for dim(im(T))?
 
Hmm yes, I know how to determine the rank of a matrix. What do you mean by the dimension of T? Im(T)?

Thanks
 
Ahhhhmmm, dim(Im(T)) has to be less or equal than 1... or better: it either is 1 or zero Thanks!
 
Alright, then I'm asked the following:

Having:
\displaystyle T:V\to \mathbb{R}
\displaystyle S:V\to \mathbb{R}
\displaystyle \dim\left( V \right)=n

Prove that ker(T)=ker(S) iff exists a real number a (different to zero) such that T=aS

This looks obvious for me because if the transformations are equal, then not only their kernels but also their images and everything must be equal.. What is what I have to show then?

Thanks
 
Sure, if two transformations are equal then their kernels must be equal. But what does that have to do with the problem here??
 
Hernaner28 said:
Alright, then I'm asked the following:

Having:
\displaystyle T:V\to \mathbb{R}
\displaystyle S:V\to \mathbb{R}
\displaystyle \dim\left( V \right)=n

Prove that ker(T)=ker(S) iff exists a real number a (different to zero) such that T=aS

This looks obvious for me because if the transformations are equal, then not only their kernels but also their images and everything must be equal.. What is what I have to show then?

Thanks

I don't see why that is obvious. One direction looks fairly clear, but you should still show the details, because I think it will familiarize you more with the elements of vector spaces and linear maps. (By the way, why do you say equal, we don't have T=S, we have T=aS.)

The other direction looks less clear, if their kernels are equal, It's not obvious to me why they have to be scalar multiples of one another. It may require the use of the image being 1-d.
You may ask, "why should I be so careful?" You could easily lead yourself to believing something false. We need to establish whether a given statement is true. That takes some of what's called mathematical maturity. You can come to recognize the care you must take. Your belief in the statement does not use the fact that a is nonzero. In fact, you seem to read it as S=T. Your belief in the statement does not use the fact that the codomain is ℝ, 1-d. It may require this fact. It seems if somebody adjusted the statement a bit so that it was false, you would not notice, and and that you would not be prepared to verify.

Now these things come with lots of practice, but I just wanted to point out some possible goals. And let's keep trying to investigate the problem you stated above!
 

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