Special Subspaces Proof

Question

Let L: Rⁿ -> Rⁿ be  linear mapping. Assume that {v₁,...,v_n} is a basis for Rⁿ such that {v₁,...,v_k} is a basis for Ker(L). Prove that {L(v_k+1),...,L(v_n)} spans Range(L).

 

Not sure how to go about this proof. Any help is appreciated!

Answer 1

Ryan,

 

I am not sure what you know at this point, so I am sending you a simple proof. If this doesn't work for you, you will have to reply or try something else.

 

Recall the following definitions:

 

If L is a linear transformation (same as your linear mapping), then:

 

rank(T) is the dimension of the range of T as a subspace of the codomain or T, and

nullity(T) is the dimension of the kernel of T as a subspace of the domain of T.

 

A nice theorem from your subject states that:

 

rank(T) + nullity(T) = n,

 

where n is the dimension of the domain (given by you).

 

Based upon what you were given, nullity(T) = k (the number of basis vectors is the dimension of the subspace). Using this formula, rank(T) = n - k.

 

Since L maps every vector in the kernel to 0 (the zero vector), the images of the vectors {v_k+1, ..., v_n} will span the range of T. Notice that there are n - k vectors in this set and their images will be linearly independent.

 

I have given you enough help on this to get you started. I may have given you more than you really need, so if I have you can either finish on your own, ask for more help, or wait for another tutor to post a solution or hint.

 

If you are unaware of the theorem quoted above, and need to work from basic definitions, this may not be of much help to you. Let me know.