# Sum of Two Subspaces

1. Jul 6, 2009

This is an example that I am a little confused by:

$$U={(x,0,0)\in\mathbf{F}^3:x\in\mathbf{F}}\text{ and }W={(0,y,0)\in\mathbf{F}^3:y\in\mathbf{F}}$$

Then
$$U+W={(x,y,0):x,y\in\mathbf{F}$$

Okay, I get that. Now it says that U is defined the same as above but now let
$$W={(y,y,0)\in\mathbf{F}^3:y\in\mathbf{F}}$$

Then the sum of U and W is the same as given above. Why is that? What is happening to that y that is in the "x" position?

Perhaps I am confusing the definition of the sum of two lists with the sum of two subspaces.

Last edited: Jul 6, 2009
2. Jul 6, 2009

### Dick

If (a,b,0) is any point there is an x and y that solve it in the first case and there is also an x and y that solve it in the second case. Isn't this so? x and y are ANYTHING, right? Who cares what happened to the y?

3. Jul 6, 2009

Sorry. I guess I am getting lost in all of these definitions.

I guess I don't really understand what it means to 'Add' two subspaces...

See I have it in my head that I should be getting something like (x+y,y,0) ...but I know that is not right.

It says in my book that the "sum of two subspaces is the set of all possible sums of the elements of the two subspaces."

I guess I really don't understand what that means.

4. Jul 7, 2009

### Office_Shredder

Staff Emeritus
So if you have W in the first case:

U+W is the set of all vectors of the form u+w where u is in U and w in W. SO u+w = (x,0,0) + (0,y,0) = (x,y,0) where x and y are arbitrary.

For the second case, u=(x',0,0) for any x', and w = (y',y',0) for any y'. Then u+w = (x'+y',y',0). Now, we want to show that (x,y,0) is in U+W for arbitrary x and y. Given y, set y'=y and now we need that x'+y' =x i.e. x' = x-y. Hence we see we can find x' and y' such that u+w = (x,y,0) as required.

Basically, the point is that the set of all vectors of the form (x+y,y,0) is
1) two dimensional
2) contained in the set of all vectors of the form (x,y,0)

So they must be the same subspace

5. Jul 7, 2009