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Math Help - Is this statement about the rank of a linear map true or false?

  1. #1
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    Is this statement about the rank of a linear map true or false?

    Is this statement true or false

    if false a counterexample is needed

    if true then an explanation

    If T : U \rightarrow V is a linear map, then Rank(T) \le(dim(U) + dim(V ))/2
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  2. #2
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    Well, you know that

    \text{rank}(T)\le\dim(U) and

    \text{rank}(T)\le\dim(V), right? Can you see where to go from here?
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  3. #3
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    hey, I think I will use inequalities, I have an idea using the fact that

    dimU = rank(T) + nullity(T) and the inequalities you provided?

    but I do not quite understand yet even from the link why exactly

    rank (T) \leq dim (U)

    rank (T) \leq dim (V)
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  4. #4
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    Quote Originally Posted by maximus101 View Post
    hey, I think I will use inequalities, I have an idea using the fact that

    dimU = rank(T) + nullity(T) and the inequalities you provided?

    but I do not quite understand yet even from the link why exactly

    rank (T) \leq dim (U)

    rank (T) \leq dim (V)


    Add sidewise both given inequalities...

    Tonio
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  5. #5
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    Quote Originally Posted by maximus101 View Post
    hey, I think I will use inequalities, I have an idea using the fact that

    dimU = rank(T) + nullity(T) and the inequalities you provided?

    but I do not quite understand yet even from the link why exactly

    rank (T) \leq dim (U)
    Any linear transformation from U to V maps a basis for U onto a basis for T(U).
    If T is "one to one", then it maps the set of basis vectors of U onto an independent set and so maps U onto a subspace of V having exactly the same dimension as U. If not, then it maps U onto a subset of V having smaller dimension. "rank T" is the dimension of T(U).

    rank (T) \leq dim (V)
    "rank T" is the dimension of T(U) as above and, of course, T(U) is a subspace of V.
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