Help me Understand Closed Under Addition and Closed Under Multiplication

[nicknaq]

Help me Understand "Closed Under Addition" and "Closed Under Multiplication"

Linear Algebra...matrices...etc

Examples would be great.

Thanks.

 

[Mark44]

A set is "closed under addition" if the sum of any two members of the set also belongs to the set. For example, the set of even integers. Take any two even integers and add them together. The result is an even integer.

A set is "closed under (scalar) multiplication" if the product of any member and a scalar is also in the set. In other words, if x is in S and a is any scalar then ax will be in the set if the set is closed under scalar multiplication. For example, the set of 2 x 2 diagonal matrices is closed under scalar multiplication.

 

[micromass]

Not sure what you want here, but I'll give it a shot.
In the following, we will consider following subsets of \mathbb{R}^2:

A=\{(x,y)~\vert~y=0\}~\text{and}~B=\{(x,y)~\vert~x+y=1\}

Typical elements of A are (1,0), (2,0),... The element (1,1) is an element not in A.
Typical elements of B are (1,0), (1/2,1/2),... The element (1,1) is an element not in B.

Now A and B carry an addition (i.e. (x,y)+(x',y')=(x+y,x'+y')). Saying that A is closed under addition just means that whenever you take two elements in A, the sum of those elements is again in A. Let's check if this is the case: two elements in A have the form (x,0) and (x',0). The sum of those elements is (x+x',0), and this is again in A. Thus A is closed under addition.

But B is not closed under addition. For example, the element (1,0) is in B. But (1,0)+(1,0)=(2,0) and this sum is not in B. Thus the sum of two elements in B is not necessarily in B. Thus B is not closed under addition.

Closed under multiplication just means that, whenever you take \alpha\in \mathbb{R} and (x,0) in A, we have that \alpha (x,0) is back in A. And this is indeed the case. Thus A is closed under multiplication.
B, on the other hand, is not closed under multiplication, can you see why?