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Ideas on Linear Algebra proof

  1. Jan 29, 2009 #1
    I came across a question saying
    Let S: U -> V and T:V -> W be linear maps, where U V and W are vectors spaces over the same field K
    Show that Rank(TS) =< Rank(T)

    This is my attempt
    the im(TS) is a subspace of W
    and so is the im(T)

    am I missing out something ?
  2. jcsd
  3. Jan 29, 2009 #2
    This is my second attempot oon it
    im(TS) is a subspace of im(T)
    and im(T) is a subspace of W
    therefore rank(TS) =< rank (T)
    is it correct ?

    but isn't im(T) = W ?
  4. Jan 30, 2009 #3
    [tex]Im(T) = W[/tex] when [tex]T[/tex] is onto.

    Your second attempt is the right direction, you might want to explain more of the details.
  5. Feb 1, 2009 #4
    You were pretty much done, but you did some extra stuff that wasn't necessary.

    Assuming the subspaces involved are finite-dimensional, [itex]\operatorname{rank} TS \le \operatorname{rank} T[/itex] means, by definition, [itex]\dim(\operatorname{im} TS) \le \dim(\operatorname{im} T)[/itex]. This happens exactly when [itex]\operatorname{im} TS \subseteq \operatorname{im} T[/itex], which you know is true.
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