Finite dimensional

  • #1
terryfields
44
0
problem here on a kernal and range question. on the first part of the question it asks me to define what is meant by kernal and image of a mapping T:U>V
answers being kerT={uEU:T(u)=0} and imT=T(U)={T(u):uEU}
then there's a second part to the question asking me to state the definitions of rank and nullity if this is a finite dimensional space, what does finite dimensional mean? is this just the normal definitions of rank and nullity i.e the dimension of the kernal and the dimension of the image?
 

Answers and Replies

  • #2
radou
Homework Helper
3,134
7
then there's a second part to the question asking me to state the definitions of rank and nullity if this is a finite dimensional space, what does finite dimensional mean? is this just the normal definitions of rank and nullity i.e the dimension of the kernal and the dimension of the image?

Yes, this should simply be the definition of rank and nullity of the operator, i.e. the dimensions of its image and kernel, respectively.

I assume you know what it means for a vector space to be finite dimensional.
 

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