Linear transformation and Ker(T)

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Homework Help Overview

The discussion revolves around linear transformations and their properties, specifically focusing on the kernel of a transformation and the implications of injectivity and surjectivity on invertibility.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to clarify the nature of the kernel of a linear transformation, particularly regarding its dimension when it consists only of the zero vector. They also question the relationship between injectivity and surjectivity in the context of invertibility.

Discussion Status

Participants are exploring the definitions and implications of linear transformations, with some providing clarifications on the dimension of the kernel and the conditions under which a transformation can be considered invertible. Multiple interpretations regarding injectivity and surjectivity are being discussed.

Contextual Notes

There is a focus on technical definitions and properties of linear transformations, with an emphasis on the implications of these properties in different dimensional contexts.

Benny
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Hi, suppose I have a linear transformation T and Ker(T) consists of only the zero vector. Then is it true that a basis for Ker(T) consists of no vectors and is of dimension zero? I would like these technicalities to be clarified. Any help would be good thanks.
 
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Yes, the "subspace" consisting only of the 0 vector has dimension 0.
 
Ok thanks HallsofIvy.

Just one more thing, if a transformation is injective then it is not necessarily onto? I ask this because I would like to know if in determining whether a transformation is invertible, if it is sufficient to look at whether or not it is injective.
 
Just one more thing, if a transformation is injective then it is not necessarily onto?

Correct. It's very easy to find an example of an injection which is not surjective.

I ask this because I would like to know if in determining whether a transformation is invertible, if it is sufficient to look at whether or not it is injective.

Sometimes it is (such as when the transformation is between two spaces with the same (finite) dimension).
 
Thanks for quick reply Muzza.
 

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