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Basis of the intersection of two spaces

  1. Feb 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider two vector spaces ##A=span\{(1,1,0),(0,2,0)\}## and ##B=\{(x,y,z)\in\mathbb{R}^3 s.t. x-y=0\}##. Find a basis of ##A\cap B##.

    I get the solution but I also inferred it without all the calculations. Is my reasoning correct

    2. Relevant equations

    linear dependence definition.

    3. The attempt at a solution

    First of all I look for a basis of ##B##, which is ##(1,1,0),(0,0,1)##.
    Now, the complete method for finding the basis of ##A\cap B## is the following one. I set
    \begin{equation}
    a_{1}(1,1,0)+a_{2}(0,2,0)=b_{1}(1,1,0)+b_{2}(0,0,1)
    \end{equation}
    or, equivalently (the coefficients are unknown so the sign is not that important as I can name ##-b_{i}=b_{i}##),
    \begin{equation}
    a_{1}(1,1,0)+a_{2}(0,2,0)+b_{1}(1,1,0)+b_{2}(0,0,1)=(0,0,0)
    \end{equation}
    By solving the linear system I get:
    \begin{equation}
    \begin{cases} a_{1}+0+b_{1}+0=0\\
    a_{1}+2a_{2}+b_{1}+0=0 \\ 0+0+0+b_{2}=0 \end{cases}
    \Leftrightarrow \begin{cases} a_{1}=-b_{1}\\
    a_{1}+2a_{2}+b_{1}=0 \\ b_{2}=0 \end{cases}
    \end{equation}
    ##a_{1}=-b_{1},a_{2}=b_{2}=0##. Thus, the generic vector ##\textbf{v}\in A\cap B## can be written as the linear combination of the basis of ##A## with ##a_{1}## and ##a_{2}## as coefficents.
    \begin{equation}
    \textbf{v}=-b_{1}(1,1,0)+0\cdot(0,2,0)=-b_{1}(1,1,0)
    \end{equation}
    The conclusion is that the searched basis is ##(1,1,0)##.

    Now, I know this is the correct procedure but is it correct to say that, since I know a basis of ##A## and a basis of ##B##, the basis of ##A\cap B## are the common vectors to both basis, if there is at least one?
    This way I could have immediately said that ##\mathcal{B}_A:(1,1,0),(0,2,0), \mathcal{B}_B:(1,1,0),(0,0,1)\Rightarrow \mathcal{B}_{A\cap B}:(1,1,0)##.
     
  2. jcsd
  3. Feb 6, 2017 #2

    RUber

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    Homework Helper

    I think the answer is yes...but you have to be careful. If you are able to write a basis vector for one space in terms of basis vectors from another, then clearly, those vectors will be in the intersection.
    However, your space A is simply the xy-plane. It could have been written with other basis vectors that would not be so simple as [1,1,0] to pick out. Consider [1,0,0] and [0,1,0].
    If you have two 2-D spaces, as above, your intersection is either all of the 2D space (if A = B) , spanned by a single vector, or empty (parallel planes).
     
  4. Feb 6, 2017 #3

    PeroK

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    Science Advisor
    Homework Helper
    Gold Member

    You could get the dimension of ##A \cap B## from considering the dimensions of the subspaces. If the second basis vector for ##B## is not in ##A##, then the intersection must be the 1D subspace spanned by the common basis vector.

    Geometrically, two planes through the origin, if not equal, intersect in a line through the origin.
     
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