For the first one: .
Closure under vector addition:
Let .
Grouping coefficients,
.
Thus S is closed under vector addition.
Scalar multiplication can be showed similarly. (Hint, define an and show that ).
I have to find which of these subsets are subspaces of
I know that for a subset to be a subspace of a given set, it can't be an empty set, it must be closed under addition and closed under scalar multiplication.
I can see that is going to be a subspace of because if is real, then a real number plus a real number will yield a real number. Also, a real number multiplied by a real number will yield a real number.
But I don't know how to prove it, nor how to analyse any of the others algebraically.
Thanks in advance,
You should also state that the zero vector of R^3 is in your subspace.
This is usually one of the first checks on makes if a subset is a subspace.
Because if the zero vector is not there, then you donīt even have to test for vector addition and scalar multiplication.
Indeed, . But, we know that any vector x in S has its first component: any real, the second component: twice the first component, and the third component: three times the first. In other words, any vector because it satisfies the conditions of membership of S.
We can see that is also such a vector. To make this more clear, let and substitute it:
. Clearly, is also in S.
Because is any arbitrary real rumber, and is any arbitrary vector in S, we have shown that S is closed under scalar multiplication.
Hope that makes sense.
Ok, thanks so far.
For the second one
I can see that if , and then the the equation is valid, therefore it is not an empty set.
if I took and
Then
This still has to satisfy the condition:
so I can say that
So is this subset of only a subspace under certain values of ?
If so how do I define this?
Thanks
Edit:
If I consider and to be on the plane , then any sum of and will also lie on the plane. Therefore closed under vector addition. I know this isn't sufficient, but I think I may be on the right track..
Edit #2: back to and
From this, and therefore, which is not allowed as
Which means that so therefore it is not closed under vector addition!
Please tell me that this one is right.
Thanks,
Let .
The quick way to prove that it is not a subspace is by showing that . I.e. .
But just to show that it isn't closed under vector addition:
Let be any vectors in S. Therefore we know that and likewise .
Then, .
Expanding and grouping terms, we get:
, and as such S is not closed under addition.