Alan G. answered • 03/17/16
Tutor
5
(4)
Successful at helping students improve in math!
The correct answers are (a) and (c).
My reasoning is as follows. Any subspace of R3 must contain the same zero vector and have the same vector operations as R3 (vector addition and scalar multiplication). Notice that (b) and (d) do NOT contain the zero vector (0,0,0), so could not possibly be subspaces of this vector space.
As for (a), it is a two-dimensional plane passing through the origin (the xy-plane), so is a subspace. And (c) is also a plane passing through the origin (it has the equation y = x + z, do you see how?), so it is also a subspace.
If you did not yet know that subspaces of R3 include: the origin (0-dimensional), all lines passing through the origin (1-dimensional), all planes passing through the origin (2-dimensional), and the space itself (3-dimensional), you can still verify that (a) and (c) are subspaces using the Subspace Test. This means you only must verify closure under addition and scalar multiplication to verify it as a subspace.
Please post a reply if you need more details or help.
Alan G.
I will compromise and show you one of the correct proofs.
Recall the Subspace Test states that in order to prove that a subset W of a vector space V is a subspace, you must show:
0) W is not empty, that is, it contains a vector. (This is usually routine.)
1) W is closed under vector addition: if u and v are in W, then u + v are in W.
2) W is closed under scalar multiplication: if c ∈ R and u ∈ W, then cu ∈ W. (R is the real numbers)
Here is how to prove that (a) is a subspace of R3.
First, call the set of vectors P. That is, P = {(a,b,0) | a ∈ R and b ∈ R}.
0) P is non empty. Proof: Since the zero vector (0,0,0) ∈ P, this is established.
1) P is closed under addition. Proof: Suppose u and v are in P. Then u = (a,b,0) for some real numbers a and b. Also, v = (a′,b′,0) for some reals numbers a′ and b′. Then, u + v = (a,b,0) + (a′,b′,0) = (a+a′, b+b′, 0+0) = (a+a′, b+b′, 0). Remember that the real numbers are closed under addition, so a+b ∈ R and a′+b′ ∈ R. Thus, u + v ∈ P since it has the form of a vector in P. This shows P is closed under addition.
2) P is closed under scalar multiplication. Proof: Suppose c ∈ R and u ∈ P. The object is to show that cu ∈ P. Let u = (a,b,0). Then, according to the definition of scalar multiplication in P, cu = c(a,b,0) = (ca, cb, c0) = (ca, cb, 0). Again, multiplication of numbers is close in R, so ca and cb ∈ R. This shows that cu ∈ P, and hence that P is closed under scalar multiplication.
Thus, P is a subspace of R3 by the Subspace Test.
Note 1: I usually provide an excess of detail and explanation when I have taught proofs in Linear Algebra because the object of a mathematical proof is to show someone who probably knows less than you do why a theorem is true. Most students are impatient with this approach, but it serves to demonstrate (prove) that a proof is a serious undertaking and should be written clearly and completely, despite the hardship it may cause the author. This is the basic reason why mathematics works!
Note 2: To prove that the origin, lines and planes passing through the origin, and the space itself are subspaces of R3, and that these are ALL of the subspaces of R3, requires a little bit more theory which your teacher or professor will surely give you in your class soon. It reduces to the idea of dimension of a vector space and this is a relatively simple but important concept.
Report
03/18/16
Rola M.
03/18/16