1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find the following subspaces

  1. Aug 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Find subspaces [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex] of [tex]\mathbb{R}^3[/tex] so that [tex]A \cap B \cap C \ne \{\vec{0}\}[/tex] and [tex](A + B) \cap C \ne A \cap C + B \cap C[/tex].
    You can specify a subspace by the form [tex]A = span\{\vec{e}_1, \vec{e}_2\}[/tex].

    2. Relevant equations
    [tex]A + B[/tex] is the set of all vectors in [tex]\mathbb{R}^3[/tex] of the form [tex]\vec{a} + \vec{b}[/tex], where [tex]\vec{a}[/tex] is in [tex]A[/tex] and [tex]\vec{b}[/tex] is in [tex]B[/tex].

    3. The attempt at a solution
    I’ve been trying to guess. I’ve tried [tex]A[/tex] as a line, [tex]B[/tex] as a plane containing [tex]A[/tex], and [tex]C[/tex] as [tex]\mathbb{R}^3[/tex] itself. I’ve tried making all three subspaces planes that intersect in a line. I’ve tried making [tex]A[/tex] and [tex]B[/tex] planes that intersect in a line and C as [tex]\mathbb{R}^3[/tex] itself. I’ve tried making [tex]A[/tex] equal to [tex]B[/tex] and [tex]C[/tex] another plane intersecting them in a line.
    Finding [tex]A \cap B \cap C \ne \{\vec{0}\}[/tex] is not my problem. I’m having trouble satisfying the second specification: [tex](A + B) \cap C \ne A \cap C + B \cap C[/tex].
    Last edited: Aug 12, 2007
  2. jcsd
  3. Aug 12, 2007 #2
    if A and B are subspaces then so is A+B. Can you find a basis of A+B knowing a basis of A and B? How does this help?

    We also know that the intersection of subspaces is a subspace.
    Last edited: Aug 12, 2007
  4. Aug 12, 2007 #3
    I can't find the bases without knowing how many dimensions each subspace is. How do I even know what dimension to make the subspaces A, B, and C?
  5. Aug 12, 2007 #4
    Check my awkward method below,

    Imagine stuff like Cartesian coordinate systems, then we have X, Y,Z axis(orthongol to each other).The R^3 can be expressed by points in OXYZ, with XZ plane at the bottom.

    XY plane and bottom make a line x. YZ plane and bottom make a line z.[WHY I CANNOT COPY YOUR EQUATIONS AND JUST PASTE THEM HERE?]

    According to definition of XY+YZ, the set has another POINT OTHER THAN x line and z line , but that point, e.g. point (1,0,1)=(1,0,0)+(0,0,1) ,lie on the bottom plane.

    Or using set language

    A={(x,y,z);x=0}, B={(x,y,z);z=0}, C={(x,y,x);y=0}

    A and C make {(x,y,z);x=0,y=0}, B and C make {(x,y,z);z=0,y=0};

    (A+B) and C includes a point {(1,0,1)},This compltes your problem.
    Last edited: Aug 12, 2007
  6. Aug 12, 2007 #5
    In this situation the subspaces [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex] do not satisfy the first requirement: [tex]A \cap B \cap C \neq \{\vec{0}\}[/tex]. The only vector in common among the 3 planes is the zero vector. But the problem is asking for subspaces [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex], that intersect in more than just the case of the zero vector.
  7. Aug 12, 2007 #6

    I had thought your questions are seperate (since you said meeting the first one is not your problem).

    But it is handy as well. Shift the bottom to one unit.See below

    A={(x,y,z);x=0}, B={(x,y,z);z=0}, C={(x,y,z);y=1}

    A and C make {(x,y,z);x=0,y=1}, B and C make {(x,y,z);z=0,y=1};

    (A+B) and C includes a point (1,1,1),This compltes your problem
    Last edited: Aug 12, 2007
  8. Aug 12, 2007 #7
    The point (1, 1, 1) is not the same as the vector <1, 1, 1>. If the planes only meet at one particular point, then their intersection is just the zero vector. So the intersection of these 3 planes is still just the zero vector. I need the 3 planes to intersect in a line segment, not just one point.
  9. Aug 12, 2007 #8
    Where is your statement from? I haven't touched mathematics for many years, but I believe your understanding is completely wrong.

    A={(x,y,z);x=0}, B={(x,y,z);z=0}, C={(x,y,z);y=1}

    A AND B AND C={(0,1,0)} , IS THIS A ZERO VECTOR IN R^3? You may be confused by the point(working as a vector) with the vector with magnitude and direction stuff? That can be called a located vector (by serge lang), which can be taken as an ordered pair of points.In vector space (subspace), points (if you think of it as an element in your R^3) ARE vectors. Also, don't get confused by the vector used in tensor analysis. Vector space is a set. It is easier to use some set language to help your analysis. But vector space can be an element in another set. What really is an element, or a set, or a vector space in your analysis of the problem, is your decision depending on what kind of problems ;and once you decide (e.g. a point is a vector), then the analysis must be consistent. Since all your statements are sets in R^3, then A,B,C ARE SETS including many elements (each element belongs to R^3, and each element can be expressed by an ordered triple or a point with the three coordinates). That's why I write them as written.Noting I embedded a coordinate system to help our analysis. Under such a coordinate system, the problem can be easily described. Now, I only need to find one case that DISobey the corresponding assertion.

    A and C make {(x,y,z);x=0,y=1}, B and C make {(x,y,z);z=0,y=1};

    (make sure you know what A+B is about according to its definition, also make sure THE ELEMENTS IN A+B IS STILL POINTS OR an ordered triple WITH THREE COORDINATES )
    (A+B) and C includes a point (1,1,1), (this is just an alternative to equation problem, but this can be obtained without resorting to equations, only need to make a big guess as your case is simple). This point will be neither in {(x,y,z);x=0,y=1} nor in {(x,y,z);z=0,y=1} (this is obvious). This compltes your problem
  10. Aug 12, 2007 #9
    This is what I eventually came up with and I got confirmation from a member of another math forum.

    Let [tex]A = span\{\left[\begin{array}{c}1\\0\\0\end{array}\right] \left[\begin{array}{c}0\\0\\1\end{array}\right]\}[/tex].
    Let [tex]B = span\{\left[\begin{array}{c}0\\1\\0\end{array}\right] \left[\begin{array}{c}0\\0\\1\end{array}\right]\}[/tex].
    Let [tex]C = span\{\left[\begin{array}{c}1\\1\\0\end{array}\right] \left[\begin{array}{c}0\\0\\1\end{array}\right]\}[/tex].

    Then [tex]A \cap B \cap C = span\{\left[\begin{array}{c}0\\0\\1\end{array}\right]\}[/tex],
    [tex](A+B) \cap C = span\{\left[\begin{array}{c}1\\1\\0\end{array}\right] \left[\begin{array}{c}0\\0\\1\end{array}\right]\}[/tex],
    and [tex]A \cap C + B \cap C = span\{\left[\begin{array}{c}0\\0\\1\end{array}\right]\}[/tex].

    [tex]\therefore (A + B) \cap C \ne A \cap C + B \cap C[/tex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook