Linear Algebra: Subspaces proof

[tylerc1991]

Homework Statement

Suppose $U$ and $W$ are subspaces of $V$, and $U \, \bigcup \, W$ is a subspace of V. Show that $U \subseteq W$.

The Attempt at a Solution

I have been working on this one for a bit and have not made any headway. I wish I could post anything, even a start to this one but I can't seem to see which direction to head in. If anyone could give me some direction I would be grateful!

 

[Shackleford]

Homework Statement

Suppose $U$ and $W$ are subspaces of $V$, and $U \, \bigcup \, W$ is a subspace of V. Show that $U \subseteq W$.

The Attempt at a Solution

I have been working on this one for a bit and have not made any headway. I wish I could post anything, even a start to this one but I can't seem to see which direction to head in. If anyone could give me some direction I would be grateful!

The professor did this in class today. He said it's a bit of a sophisticated proof. He used the proof by contradiction method.

Since this is an iff statement, one direction was fairly trivial, but the other required the contradiction method. He assumed the conclusion failed which violated our assumptions.

 

[tylerc1991]

The professor did this in class today. He said it's a bit of a sophisticated proof. He used the proof by contradiction method.

Since this is an iff statement, one direction was fairly trivial, but the other required the contradiction method. He assumed the conclusion failed which violated our assumptions.

Exactly, the original question was an iff statement. I was posting the part that I was having trouble with. Oddly enough, proof by contradiction was the road I was going down but couldn't get anywhere with. Was there any special trick in particular that made this proof so sophisticated?

 

[Shackleford]

Exactly, the original question was an iff statement. I was posting the part that I was having trouble with. Oddly enough, proof by contradiction was the road I was going down but couldn't get anywhere with. Was there any special trick in particular that made this proof so sophisticated?

It's in the notes. This is actually a homework problem for me too. I kind of stopped paying attention about mid-way through the proof in class. I kind of followed along, but I need to go through it myself. I'm still pretty rusty on linear algebra. I took the sophomore-level class three years ago.

 

[tylerc1991]

It's in the notes. This is actually a homework problem for me too. I kind of stopped paying attention about mid-way through the proof in class. I kind of followed along, but I need to go through it myself. I'm still pretty rusty on linear algebra. I took the sophomore-level class three years ago.

I think we are in very similar circumstances. I took an elementary linear algebra class last semester and essentially all I learned how to do was row-reduce a matrix. Now I am studying the more abstract concepts on my own. I am assuming that you guys are using the book 'Linear Algebra Done Right', which is where I got this problem from. I will keep working on this, now that I know the direction, but if you are working it for homework I would seriously appreciate it if you could let me know what specific steps helped you along. Thanks!

 

[Dick]

The conclusion should really be if U union W is a subspace, then either U is contained in W OR W is contained in U. If U is not contained in W and W is not contained in U, then pick a nonzero element u of U and a nonzero element w of W such that u is not in W and w is not in U. Can u+w be in the union of U and W?

 

[tylerc1991]

The conclusion should really be if U union W is a subspace, then either U is contained in W OR W is contained in U. If U is not contained in W and W is not contained in U, then pick a nonzero element u of U and a nonzero element w of W such that u is not in W and w is not in U. Can u+w be in the union of U and W?

Ugh, stupid mistake on my part. I thought that since the conclusion was 'one subspace is contained in the other', I thought it would be enough to show that one subspace was inside of the other, but not necessarily vice versa (since these were abstract subspaces). Thank you very much for your help!