# Linear Algebra: Subspaces proof

• tylerc1991
In summary, the conversation discusses a proof by contradiction for a homework problem involving subspaces in linear algebra. The conclusion is an "if and only if" statement, and the proof requires showing that either one subspace is contained in the other or vice versa. The conversation also mentions the book "Linear Algebra Done Right" as the source of the problem. The final solution involves picking nonzero elements from each subspace and considering their sum in the union of the subspaces.
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!

tylerc1991 said:

## 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.

Shackleford said:
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?

tylerc1991 said:
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.

Shackleford said:
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!

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?

Last edited:
Dick said:
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!

## 1. What is a subspace in linear algebra?

A subspace in linear algebra is a subset of a vector space that satisfies three properties: closure under vector addition, closure under scalar multiplication, and contains the zero vector.

## 2. How do you prove that a subset is a subspace?

To prove that a subset is a subspace, we need to show that it satisfies the three properties of a subspace: closure under vector addition, closure under scalar multiplication, and contains the zero vector. This can be done by showing that for any two vectors in the subset, their sum and scalar multiple are also in the subset, and that the zero vector is also in the subset.

## 3. Can a subspace contain only one vector?

Yes, a subspace can contain only one vector. This is because a subspace can be defined as a subset of a vector space that satisfies the three properties, and a single vector can satisfy these properties.

## 4. What is the difference between a subspace and a span in linear algebra?

A span is the set of all possible linear combinations of a given set of vectors, while a subspace is a subset of a vector space that satisfies three specific properties. In other words, a span is a set of vectors, while a subspace is a type of set that follows certain rules.

## 5. How do you prove that a set of vectors spans a subspace?

To prove that a set of vectors spans a subspace, we need to show that every vector in the subspace can be expressed as a linear combination of the given set of vectors. This can be done by setting up a system of equations and solving for the coefficients of the linear combination.

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