Can someone please help me? i have asked a coulple times now if someone could ex
ID: 2963470 • Letter: C
Question
Can someone please help me? i have asked a coulple times now if someone could explain to me about a subspace in linear algebra.
I need help figuring out when/if sets is a subspace....... i need someone to explain it to me like i was a 5 year old, cuz i still can't figure it out! can someone help me out? but like i was kind of slow?
Here is an example: True or False, The set W = (a-3b,b-a,a,b) : a,b in R) is a subspace?
And another example would be : Is it a subspace or not? W is the set of all vectors in R3 such that x1 + x2 + x3=1
keep in mind that i go to class, i have notes, and i have asked this question before i have seen all the usal definitions.......and i STILL dont understand the material (a personal touch might help) .......help me!!!!!!!!!!!
Explanation / Answer
Okay, if W is a nonempty subset of a vector space then to show it is a subspace you need to check two things:
(1) 0 in W
(2) u+w in W or all u and w in W.
(3) k*w in W for all w in W and k in R.
So lets do that.
First example W = { (a-3b, b-a, a, b) | a,b in R } is a subset in the vector space R^4.
(1) We need to check (0,0,0,0) in W. This is obviously true by taking a=b=0.
(2) Suppose u= (a-3b, b-a, a, b) and w= (c-3d, d-c, c, d) are in W. Then
u+w = (a-3b + c-3d, b-a+d-c, a+c, b+d) = ((a+c) - 3(b+d), (b+d) - (a+c), (a+c), (b+d))
and this is in W because (a+c) in R and (b+d) in R (and so u+w takes the form of an element in the set W).
(3) Finally, suppose k in R, then k*(a-3b, b-a, a, b) = (k(a-3b), k(b-a), ka, kb) = (ka - 3(kb), ka-kb, ka, kb) and this is in W because ka, kb are in R.
So the first example is a subspace.
The second example is W = {(x1,x2,x3) | x1+x2+x3 = 1} a subset in the vector space R^3
(1) Firstly, 0 is not in W, that is, (0,0,0) is not in W because 0+0+0 is not equal to 1.
Since this first condition fails we know that in this example, W is not a subspace. You can also see that the other conditions will also fail:
(2) Suppose (x1, x2, x3) and (y1, y2, y3) are in W. Then (x1, x2, x3) + (y1, y2, y3) = (x1+y1, x2+y2, x3+y3). But x1+y1+x2+y2+x3+y3 = (x1 + x2 + x3) + (y1 + y2 + y3) = 1 + 1 = 2 and this is not equal to 1.
(3) Suppose (x1, x2, x3) is in W and k is in R, then k(x1, x2, x3) = (kx1, kx2, kx3) but kx1 + kx2 + kx3 = k(x1 + x2 + x3) = k and this is not equal to 1 unless k=1.
So even though we only needed to show that condition (1) fails, we have shown that all 3 conditions have failed.