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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].
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