Linear Algebra subspace troubles

[Servarus]

Homework Statement

Let V be a finite dimensional subspace. Let V$$\supseteq$$U₁$$\supseteq$$U₂$$\supseteq$$...$$\supseteq$$U_k. Show that there exists k such that U_k=U_k+1=...=U_n=...

Homework Equations

We were also told to assume none of the subspaces are zero dimensional, and to think about how the dimensions can change throughout.

The Attempt at a Solution

I know that all the U_i's are closed under vector addition, but I really don't know what to do with the information. I really don't know where to start. Any and all help will be appreciated.

Thanks guys.

 

[Servarus]

Ok, I believe I figured it out after a couple hours.
First I must assume that U_k is not zero dimensional. Then I say that another subspace must be zero dimensional since V is finite dimensional. Then prove that U_k is that subspace.

Let me know if I am doing this correctly. Thanks in advance.

 

[HallsofIvy]

Ok, I believe I figured it out after a couple hours.
First I must assume that U_k is not zero dimensional. Then I say that another subspace must be zero dimensional since V is finite dimensional. Then prove that U_k is that subspace.

Let me know if I am doing this correctly. Thanks in advance.

Do you realize you just said "assume U_i is not zero dimensional" and then said "prove that U_k is that (zero dimensional) subspace"? So you are going to prove something you are assuming is false!

In any case, you said in your first post that "We were also told to assume none of the subspaces are zero dimensional" so why are you focusing on zero dimensional subspaces?

It is NOT necessary, and you were told to assume it didn't happen, that NONE of the given subspaces is 0 dimensional

For example, suppose V= R³, U₁= {(x, y, 0)}, U₂= {(x, 0, 0)} and, if k> 1, U_k= {(x, 0, 0)}. There are no zero dimensional subspaces in that sequence.

What you can do, since the dimensions of all subspaces must be positive, is let "k" be the smallest dimension of any of the subspaces. Any set of positive integers contains a smallest integer.

 

[Servarus]

What you can do, since the dimensions of all subspaces must be positive, is let "k" be the smallest dimension of any of the subspaces. Any set of positive integers contains a smallest integer.

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 showing that the span{U_k}=span{U_k+1}=... But I'm a little stumped on how I would go about doing this.

 

[HallsofIvy]

If [itex]U_i[itex]is NOT equal to $U_{i+1}$, then, since $U_{i+1}\subset U_i$, $U_{i+1}$ must has lower dimension than $U_i$. What is the smallest possible reduction in dimension? What is the maximum number of times that can happen?[/itex][/itex]

 

[Servarus]

If $U_i$ is NOT equal to $U_{i+1}$, then, since $U_{i+1}\subset U_i$, $U_{i+1}$ must has lower dimension than $U_i$. What is the smallest possible reduction in dimension? What is the maximum number of times that can happen?

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.