Recent content by Servarus

So the proof would basically go as follows: 1) Assume U_i \neq U_{i+1}. 2) Then do a proof by contradiction and show that U_i must equal U_{i+1} because U_i is the lowest dimensional subspace.

Oh wow, I'm extremely sorry. Been up for over 24 hours doing homework and obviously not thinking correctly. Now I realized that I do need to make k the smallest dimension of any of the subspaces, and then be able to prove that k=k+1=k+2=...=n=... Also, I'm thinking I would show this by...

Ok, I believe I figured it out after a couple hours. First I must assume that Uk is not zero dimensional. Then I say that another subspace must be zero dimensional since V is finite dimensional. Then prove that Uk is that subspace. Let me know if I am doing this correctly. Thanks in advance.

Homework Statement Let V be a finite dimensional subspace. Let V\supseteqU1\supseteqU2\supseteq...\supseteqUk. Show that there exists k such that Uk=Uk+1=...=Un=...Homework Equations We were also told to assume none of the subspaces are zero dimensional, and to think about how the dimensions...

Alright, I finally understand. Thank you both so much for all the help. It has been quite invaluable.

I apologize, yes I did mean there exists a linear injective map from V to U. And what I am really trying to figure out is how you prove that there does exist a linear injective map when there are no matrices or sets that have to do with either of vector space.

Hey guys, new to the forum but hoping you can help. How do you prove that vector spaces V and U have a linear injective map given V is finite dimensional. I got the linear part but cannot really figure out the injectivity part, although I am thinking that it has to do with the kernel...