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Homework Help: Linear Map Question

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that V and W are finite dimensional and that U is a subspace of V. Prove that there exists [tex]T \in L(V,W)[/tex] such that null T = U if and only if [tex]dim U \geq dim V - dim W.[/tex]


    2. Relevant equations
    thm: If [tex]T \in L(V,W)[/tex], then range T is a subspace of W.

    thm: If V is a finite dimensional vector space and [tex]T \in L(V,W)[/tex] then range T is a finite-dimensional subspace of W and dim V = dim null T + dim range T.


    3. The attempt at a solution
    forward direction: by thm, range T is a subspace of W implies that
    [tex]dim range T \leq dim range W[/tex].

    by thm, dim V = dim null T + dim range T
    dim V = dim U + dim range T (since U = null T)
    dim V - dim range T = dim U

    [tex]dim V - dim W \leq dim U[/tex] since [tex]dim range T \leq dim range W[/tex].

    i think the forward direction is good. comments?

    backward direction:
    we have [tex]dim V - dim W \leq dim U[/tex]. Let [tex](u_{1},...,u_{n})[/tex] be a basis for U. extend this to a basis for V: [tex](u_{1},...,u_{n},u_{n+1},...u_{m})[/tex]. then dim U = n, and dim V = m. Then any [tex]v \in V[/tex] can be written as [tex]a_{1}u_{1}+...+a_{m}u_{m}[/tex].

    I think i'm in the right direction but i'm confused as to what to do. since we have dim V - dim W is less than dim U, i want to say that dim W is greater than or equal to m, but i don't know how to define T so that null T = U. If i make all of the T(u_i} in the basis 0, then null T = U, but how does that relate to the relation of [tex]dim V - dim W \leq dim U[/tex]?

    thanks.
     
  2. jcsd
  3. Sep 26, 2008 #2

    HallsofIvy

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    Science Advisor

    Yes, thatis correct. I had to stop and think about it for a moment. Since range T is a subset of W, dim Range T[itex]\le[/itex] dim W. Therefore, -dim Range T[itex]\ge[/itex] -dim W, therefore dim V- dim W[itex]\le[/itex]dimV- dim Range T= dim U.

    Think about what you want to prove: if dim V- dim W[itex]\le[/itex] dim U, then there exists a linear transformation T such that null T= U. dim V- dim W[itex]\le[/itex] dim U means dim V- dim U[itex]\le[/itex] dim W. Choose a basis for U, extend it to a basis for V. Now, define T so that Tu= 0 for any basis vector of U. Tv, for v a basis vector for v not in the basis for U, can be anything in W.
     
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