# Sum of sets of vectors (or whatever it's called)

• July 3rd 2010, 03:29 PM
jayshizwiz
Sum of sets of vectors (or whatever it's called)
S+T = {s+t | s $\in$ S, t $\in$ T}

Can someone please explain this property better?

Let's say S = {(1,0),(0,1)} and T = {(2,1),(1,3)}

What would be S+T?

I would just assume that S+T would be each vector from S plus each vector from T:

S+T = {(3,1),(2,3),(2,2),(1,4)}

But obviously I'm wrong since S+T is a subspace and what I have is clearly not a subspace. How do you get the 0 vector from S+T?

Thanksss.

***Correction - S+T is a subspace only if S and T are both subspaces, and that would explain the zero vector thing... So is my calculation correct?? And does that mean that in this particular example S+T is not a subspace??
• July 3rd 2010, 03:47 PM
HallsofIvy
Quote:

Originally Posted by jayshizwiz
S+T = {s+t | s $\in$ S, t $\in$ T}

Can someone please explain this property better?

Let's say S = {(1,0),(0,1)} and T = {(2,1),(1,3)}

What would be S+T?

I would just assume that S+T would be each vector from S plus each vector from T:

S+T = {(3,1),(2,3),(2,2),(1,4)}

Yes, that is correct.

Quote:

But obviously I'm wrong since S+T is a subspace and what I have is clearly not a subspace. How do you get the 0 vector from S+T?
Why do you conclude that S+ T is a subspace?

Quote:

Thanksss.

***Correction - S+T is a subspace only if S and T are both subspaces, and that would explain the zero vector thing... So is my calculation correct?? And does that mean that in this particular example S+T is not a subspace??
Yes, S and T are not subspaces so S+ T is not necessarily a subspace. And in this case, it isn't.
• July 3rd 2010, 09:20 PM
jayshizwiz
Quote:

Why do you conclude that S+ T is a subspace?
That was simply my original thought. After I posted this, I edited my post with the ***Correction part

Quote:

S+T = {s+t | s S, t T}
So what exactly is this defintion called in English? What is S+T?

One last thought:

Quote:

Let's say S = {(1,0),(0,1)} and T = {(2,1),(1,3)}
Can I say that (1,1) is also in S+T since it is the sum of the two vector of S? Or, you must choose 1 vector from S and 1 vector from T?
• July 3rd 2010, 09:24 PM
roninpro
This is called an internal direct sum.

Direct sum - Wikipedia, the free encyclopedia
• July 3rd 2010, 09:32 PM
jayshizwiz
Quote:

This is called an internal direct sum.
It is not exactly/always a direct sum:

U = (a, b, c | a+b+c=0)

W = (a, b, c | a=c)

In this case U+W = R^3

but (0,0,0) = (0,0,0) + (0,0,0)
but also (0,0,0) = (1,-2,1) + (-1,2,-1)

So Maybe in general S+T is simply called an internal sum