Linear transformation and Ker(T)

[Benny]

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.

 

[HallsofIvy]

Yes, the "subspace" consisting only of the 0 vector has dimension 0.

 

[Benny]

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.

 

[Muzza]

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).

 

[Benny]

Thanks for quick reply Muzza.