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Linear Algebra: intersection of subspaces

  1. Mar 7, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm working on a problem that involves looking at the dimension of the intersection of two subspaces of a vector space.


    2. Relevant equations
    [itex]M \subset V [/itex]
    [itex]N \subset V [/itex]
    dim(M [itex]\cap[/itex] N)
    [itex][\vec{v}]_{B_M}[/itex] is the coordinate representation of a vector v with respect to the basis for M



    3. The attempt at a solution
    I reformulated M [itex]\cap[/itex] N in a bunch of different ways that would be too long to copy down here, but I finally came to this (which may or may not be useful to me in my larger problem but I'm wondering if it is valid itself):

    [itex]\vec{v}[/itex] is itself, so it must have the same dimension in both M and N, and since the bases are ordered, for each [itex]\vec{b}_{Mi}[/itex] in [itex]B_M[/itex] for which the corresponding scalar is not zero in the linear combination of elements of [itex]B_M[/itex] equal to [itex]\vec{v}[/itex], and each [itex]\vec{b}_{Nj}[/itex] in [itex]B_N[/itex] for which the corresponding scalar is not zero in the linear combination of elements of [itex]B_N[/itex] equal to [itex]\vec{v}[/itex], if i=j then [itex]\vec{b}_{Mi}[/itex] and [itex]\vec{b}_{Nj}[/itex] are dependent

    and [itex][\vec{v}]_{B_M}[/itex] has zeros in the same places as [itex][\vec{v}]_{B_N}[/itex]

    but there is a major problem here with the fact that we may have dimM ≠ dimN
     
  2. jcsd
  3. Mar 7, 2012 #2
    first imaging a basis for the intersection {b1,b2,...,bp}, then expand it to a basis of M, {b1,b2,...,bp,m1,m2,...,mq},
    also expand it to a basis for N, {b1,...,bp,n1,n2,...,nr}, dimension of the intersection is p, dim(M)=p+q, dim(N)=p+r.
     
  4. Mar 8, 2012 #3
    beautiful
     
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