Existence of Linear Operators with Matching Subspaces in Vector Spaces

[Dick]

L(v1) = 1
L(v2) = 2
...
For all i<=k: L(vi) = i

I think you mean the right thing. But you certainly aren't saying the right thing. 1 and 2 aren't vectors, are they?

 

[LosTacos]

Sorry. It would be L(v1) = v1, L(v2) = v2,... L(vi)=vi

 

[Dick]

Sorry. It would be L(v1) = v1, L(v2) = v2,... L(vi)=vi

Ok, so L(vi)=vi for i<=k and L(vi)=v1 for i>k? Does that work? Is L(C)=B? Explain why? For extra points tell me some other ways you could have defined L such that L(C)=B.

 

[LosTacos]

Yes. Since {v1, v2, ... , vk} is a basis for B, and {v1, v2, ... , vk, vk+1, vn} is a basis for C,
for i<=k, L(vi) = vi which is apart of both Basis of B and Basis of C because v1 < vi < vk. Then, for any i > k, L(vi) = L(v1) = v1 which is apart of Basis B and Basis C.

 

[LosTacos]

I could of defined L(C)=B such that if i>k, then L(vi) = ker(L) = B = 0

 

[Dick]

I could of defined L(C)=B such that if i>k, then L(vi) = ker(L) = B = 0

Yes, I THINK you've essentially got it, but you don't really express yourself very well, so I'm guessing. The point is that both ways you've defined L, L(C)=span(v1,v2,...,vk). Which defines B. There are many other ways to define L as well, agree?