# Closed under addition and closed under scalar

1. Mar 4, 2014

### negation

Could someone explain to me about what closed under addition and closed under scalar multiplication means? I have a patchy idea of what it is but how does it relates to A = {(x,y) | x^2 + y^2 <= 1}?
What does A stands for? What does the language implies?

Edit: My interpretation: Let's suppose there exists a field k with R^n where n = 2) and A is a subset of the field k.
An element is closed under addition iff an element, uA, and, vA such that
u^2+v^2 = <=1.
If u^2+v^2 <=1, then, u and v is a subset of A.

Last edited: Mar 4, 2014
2. Mar 4, 2014

### LCKurtz

In other words, you are considering $R^2$ as a vector space over the reals. Addition means $(u,v)+(x,y) = (u+x,v+y)$ and scalar multiplication means $t(x,y) = (tx,ty)$, the usual vector operations in $R^2$.

You have a very confused interpretation. You don't talk about whether an element is closed under addition, you talk about whether the subset $A$ is closed under addition. That is, if $(a,b)$ and $(c,d)$ are in $A$, is $(a,b)+(c,d)$ always in $A$? If so, $A$ is closed under addition. Similarly, if $(x,y)\in A$ is $t(x,y)\in A$ or not.

3. Mar 4, 2014

### negation

How does x^2 + y^2 relates the the closed addition and scalar multiplication?

4. Mar 4, 2014

### LCKurtz

That's how you tell if $(x,y)\in A$. So if you add two elements in $A$ is the result in $A$?

5. Mar 4, 2014

### negation

So x^2 + y^2 <=1 is a condition. If vector u and vector v both of which are elements of S where S is a set of vectors and if u+v <= 1 then it is closed under addition.
And if it fulfils 2 other necessary conditions- namely, closed under scalar multiplication and where 0 vector is an element of S then S is a subspace.

6. Mar 4, 2014

### Staff: Mentor

Yes, this is the condition by which you can determine whether a vector <x, y> is a member of set A (the name of the set from post #1). S is not mentioned in post #1.
No, that isn't it. If u and v are members of A, and if u + v is also in A, then A is closed under vector addition.
The condition, as you call it, is how you can tell whether a given vector is in set A.
Just to be clear, assuming that u is in A, and t is a scalar, if tu is also in A, then A is closed under scalar multiplication.

7. Mar 4, 2014

### negation

Is it the same thing to say A closed under addition, closed under scalar multiplication and has zero vector and A is the subspace of a vector space?

8. Mar 4, 2014

### Staff: Mentor

Not quite. You need some ifs and a then.

For a set A, a subset of a vector space, if
1. A is closed under vector addition, and
2. A is closed under scalar multiplication, and
3. The zero vector is an element of A,

Then A is a subspace of the vector space. (Whatever vector space we're working with.)