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How do i determine if U is a subspace of R3

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine whether U is a subspace of R3
    U= [0 s t|s and t in R]


    2. Relevant equations
    My textbook, which is vague in its explinations, says the following
    "a set of U vectors is called a subspace of Rn if it satisfies the following properties
    1. The zero vector 0 is in U
    2. If X and Y are in U, then X+Y is also in U
    3. If X is in U then aX is in U for every real number a


    3. The attempt at a solution
    ive been trying to figure out how to do this, but i dont understand the three rules and how to apply them to the question
     
  2. jcsd
  3. Dec 3, 2008 #2

    Mark44

    Staff: Mentor

    What does "0 s t" mean?
    The three rules are very clear.
    Does the set U have the zero vector in it?
    If you add together any two vectors in U, is the sum in U?
    If you multiply any vector in U by a scalar, is the new vector still in U?

    Before you can verify these rules, though, you need to understand exactly how U is defined. I have no idea what this set looks like from your description.
     
  4. Dec 3, 2008 #3
    sorry that the vector [0 s t]^T
     
  5. Dec 3, 2008 #4

    Mark44

    Staff: Mentor

    I don't know what that means either. I still don't know what 0 s t means.
     
  6. Dec 3, 2008 #5
    They are the entries in a 3x1 vector U. [tex] U_{11} = 0, U_{21} = s, U_{31} = t [/tex] and T represents the transpose to put it in vector notation.



    [0 0 0]^T is in U since s=t=0 is in R.

    Suppose X and Y are in U then
    X = [0 a b ]^t
    Y= [0 c d]^t

    a,b,c,d are in r

    Then add the two vectors together to form a new vector:

    [0 (a+c) (b+d)]

    Since a,b,c,d are all in R, the sum of any real numbers is still a real number so X+Y is in U

    Try out the same sort of method with aX where a is a real number
     
  7. Dec 3, 2008 #6
    Err whoops, U is a set of vectors, not a single vector.
     
  8. Dec 3, 2008 #7
    ok so i choose a=2
    X= 2[0 a b]
    = [0 2a 2b]
    correct?
    under what circumstances would this last principle make the vector not be in the subspace?
     
  9. Dec 4, 2008 #8
    That's correct.

    You should make your language very precise. The last principle states that U is only a subspace if and only if the last principle is true; it doesn't mention the last principle failing sometimes so that we get a vector not in the subspace. Below, I have traced out my interpretation of subspaces in Rn.

    It may be easier to understand what the principles are saying once you look at the problem geometrically. Suppose our set U is a subspace that contains vectors in Rn. Picture each of those vectors having their tail at the origin i.e. emanating from the origin. The heads of all the vectors in U trace out a shape in Rn, and we will use the 3 principles to understand what this shape "looks" like.

    Of course, we could imagine that some arbitrary set U of vectors could contain a finite number of vectors, such that the heads of those vectors look just like several unconnected points in Rn; then U would not have any "nice" shape. Or we could imagine U as containing an infinite number of vectors, but in disconnected pieces. But if U is a subspace of vectors in Rn, then the principles guarantee that it has a nice and connected shape. Now, let U not be some arbitrary set but a subspace.

    First, what does principle 3 say? It says that if U contains a vector v, it must also contain any vector that is v scaled, namely av. Intuitively, if U is a subspace of Rn and contains v, it must contain an infinite number of other vectors in the same direction of v. The head of those vectors positioned at the origin trace out the entire line in the direction of v running through the origin. When we say that U is a line, this is what we mean.

    In particular, U must contain [itex]0 \cdot \vec{v}[/itex] for all [itex]\vec{v} \in U[/itex], so it must contain the 0 vector. Principle 1 is listed explicitly but is primarily a consequence of principle 3.

    We will use this geometric interpretation of principle 3 to help understand what principle 2 does to the shape of U. Principle 2 says that any two different vectors in U must add to another vector in U. Let's call these two vectors u and v. Suppose that they point in different directions i.e. u is not a multiple of v. We already know that principle 3 tells us U contains the two lines in the u and v directions through the origin. By appropriately adding scaled vectors along those two lines, it turns out that it is possible to end up with a vector sum that has its head on any point "between" those two lines in the plane containing them. Therefore, whenever we have two vectors in U in different directions, the entire plane containing those two vectors is in U.

    Let us suppose that we have a vector in a third direction that is not in the plane containing u and v. Then it turns out that with those 3 vectors and judicious use of principle 3 to scale vectors along the line in those 3 directions and principle 2 to add the vectors together, every point in 3 dimensional space is reachable. Therefore, U is all of space. In fact, in order for U to contain those 3 vectors and still be completely connected in an intuitively vague way that I have not rigorously defined, its size must be at least of all of 3-D space. This is some sort of geometric way to look at the notion of subspace.

    Here is a specific example. If we are in ordinary 3-dimensional space, R3, and U is a subspace, then there are only a few possibilities for the shape of U. The smallest possible U can be is [itex]\{\vec{0}\}[/itex]. In this case, U is 0-dimensional and just a single point: the origin. Remember that before I said principle 1 can be derived from principle 3. This is true for all subspaces except [itex]\{\vec{0}\}[/itex], because there are no other vectors we can scale as needed in principle 3. I hope you see that this is sort of a technicality required in part to model the geometric picture.

    What about if there are only vectors in one direction in U? We know that then U is a line through the origin; it is a subspace R1 embedded in R3. If there is just one other vector in U that points in a different direction, we are guaranteed that U contains the entire plane between them: principle 3 guarantees U must be at least 2 lines through the origin, principle 2 then guarantees U must contain all points between those 2 lines in the plane. Then, U will be a subspace R2 through the origin embedded in R3. Finally, if U contains 3 vectors that are not all coplanar, then U must be all of R3. Take a look at your problem; what type of subspace do you think you have?

    So that was the geometric picture. Why do we want to describe sets that look like the geometry I've just described? Now we take a look at the algebraic properties that come from such a shape. Euclidean space Rn obeys the three properties described; it is closed under vector addition and scalar multiplication. With this closure, a lot of interesting results can be derived that have applications in subjects such as linear algebra, so subspaces are an essential concept to understand when you study more mathematics.


    So how does this all apply to your question? How do you use the 3 properties to check of a set U is a subspace? Simply choose arbitrary vectors in U when you need them: to do this, use variables to stand in place for whether variable numbers would go. Add them or scale them, and see if the final result fits the defining property of U. If so, then U is a subspace.
     
    Last edited: Dec 4, 2008
  10. Dec 4, 2008 #9
    I'd be careful with that:

    3. If X is in U then aX is in U for every real number a

    You only showed it was true for 1 value of a. And you didn't state why it was true for a=2 (Because any real number multiplied by other real number, yields a real number). I know these things are obvious, but that is the reason that aX is still an element of U.
     
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