Identifying Subspaces and Subspace Addition

[TranscendArcu]

Homework Statement

http://img21.imageshack.us/img21/4580/screenshot20120117at218.png

The Attempt at a Solution

a) Suppose we have two arbitrary vectors of E, call them X,Y. Let X = (2x,x) where x is in R and let Y = (2y,y) where y is in R. If we add X and Y we have (2x,x) + (2y,y) = (2(x + y),x+y) = Z. Addition within the components of Z is just a binary operator. This suggests that if we enter two real numbers for x,y we will get just one real number back. Let z = x +y where x,y,z is an element of R. Since Z is the sum of X and Y, we can write Z = (2z,z) where z is in R, which shows that E is closed under vector addition.

Let r be in R. We will test is rX is still in E. We may say that rX = r(2x,x) = (2rx,rx) = Q. We can restate the argument that multiplication within the elements of Q is again a binary operation. We can say that rx = q, where x,r,q are all in R. This shows that E is closed under scalar multiplication.

b) I won't retype this one out. It basically follows my argument in a) and I concluded that B was also a vector subspace. Unless the former is flawed, I don't think there's a problem.

c) Graphing E,B is R2 seemed pretty easy. Both yielded lines that intersected only at the origin. Thus, I concluded that the only element in both E,B is the zero vector.

d) Consider a scalar m in R such that q does not equal zero and let (2m,m), (m,m) to be two distinct vectors in E U B. Suppose we write, (2m,m) + (m,m) = (3m,2m) = M. This vector is neither in E nor B. Thus, since E U B is not closed under vector addition, it is not a vector space.

e) I wasn't really confident about this last part. I wrote that E + B is just all vectors of the form (2a,a) + (b,b) = (2a + b,a + b). This seems too straightforward, so I think I've done this last part wrong.

 

[lanedance]

Homework Statement

http://img21.imageshack.us/img21/4580/screenshot20120117at218.png

The Attempt at a Solution

a) Suppose we have two arbitrary vectors of E, call them X,Y. Let X = (2x,x) where x is in R and let Y = (2y,y) where y is in R. If we add X and Y we have (2x,x) + (2y,y) = (2(x + y),x+y) = Z. Addition within the components of Z is just a binary operator. This suggests that if we enter two real numbers for x,y we will get just one real number back. Let z = x +y where x,y,z is an element of R. Since Z is the sum of X and Y, we can write Z = (2z,z) where z is in R, which shows that E is closed under vector addition.

Let r be in R. We will test is rX is still in E. We may say that rX = r(2x,x) = (2rx,rx) = Q. We can restate the argument that multiplication within the elements of Q is again a binary operation. We can say that rx = q, where x,r,q are all in R. This shows that E is closed under scalar multiplication.

b) I won't retype this one out. It basically follows my argument in a) and I concluded that B was also a vector subspace. Unless the former is flawed, I don't think there's a problem.

c) Graphing E,B is R2 seemed pretty easy. Both yielded lines that intersected only at the origin. Thus, I concluded that the only element in both E,B is the zero vector.

d) Consider a scalar m in R such that q does not equal zero and let (2m,m), (m,m) to be two distinct vectors in E U B. Suppose we write, (2m,m) + (m,m) = (3m,2m) = M. This vector is neither in E nor B. Thus, since E U B is not closed under vector addition, it is not a vector space.

e) I wasn't really confident about this last part. I wrote that E + B is just all vectors of the form (2a,a) + (b,b) = (2a + b,a + b). This seems too straightforward, so I think I've done this last part wrong.

all looks good to me