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Math Help - Diemsion of Hom

  1. #1
    Super Member Showcase_22's Avatar
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    Diemsion of Hom

    I started by making e_1, e_2,........,e_n a basis of U and f_1,f_2,.....,f_m a basis of V.

    Transforming the basis of U by T gives T(e_1),T(e_2),......,T(e_n). These vectors may not be linearly independent so sifting them gives:

    T(e_1),T(e_2),......,T(e_r) where r \leq n. Let T(e_i)=f_i which gives me the basis of V.

    Is this the correct way to start? I really can't see another way of doing this question.

    EDIT: Sorry for mistyping "Dimension" as the title.
    Last edited by Showcase_22; February 8th 2009 at 03:11 AM.
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    I started by making e_1, e_2,........,e_n a basis of U and f_1,f_2,.....,f_m a basis of V.
    Prove that the lin. transfs T_{ij} such that T_{ij} (e_i) = f_j(actually they are just maps according to definition, but you can extend it linearly), form a basis of \text{Hom (U,V)}
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  3. #3
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    Let U and V be vector spaces of dimensions n and m over K, and let Hom_K(U,V) be the vector space over K of all linear maps from U to V.

    Find the dimension and describe a basis of Hom_K(U,V).
    Let T \in \hom_K (U,V) and (u_1,...,u_n) be a basis for U and (v_1,...,v_m) be a basis for V. This means that T(u_j) = \Sigma_i a_{ij}v_i. Let M(T) be the matrix (a_{ij}) i.e. M(T) is the matrix associated with T. We see that M(T_1+T_2) = M(T_1)+M(T_2) and M(kT) = kM(T). Therefore, F: \hom_K(U,V)\to M_{m\times n}(K) defined by F(T) = M(T) is a linear transformation between the vector space \hom_K(U,V) and M_{m\times n}(K) - the m\times n matrices over K. This linear transformation is one-to-one because if F(T) = 0 it means M(T) = 0 and so T(u_j) = 0 for each 1\leq j \leq n. All elements in U are linear combinations of u_j and so if T(u_j)=0 then it means T is the zero linear transformation. Thus, \ker (T) = \{ \bold{0}\} and that implies T is one-to-one. It remains to show that F is onto. Thus, F provides a vector space isomorphism between \hom_K(U,V) and M_{m\times n}(K). Now, M_{m\times n}(K) has dimension mn with the basis \{ \bold{e}_{ij}\} where \bold{e}_{ij} is a matrix having 1 in ij location and 0's everywhere else. Thus, \text{dim}( \hom_k(U,V) ) = mn.
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  4. #4
    Super Member Showcase_22's Avatar
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    Is it true that bijectivity implies that the dimensions are the same?

    What's the best way to prove that it's surjective? We're not really sure how to start.

    Do both Hom_K(U,V) and M_{m \times n} have the same basis?
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  5. #5
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    Quote Originally Posted by Showcase_22 View Post
    Is it true that bijectivity implies that the dimensions are the same?
    If U and V are isomorphic (bijective linear transformation) then they have the same dimension.
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  6. #6
    Super Member Showcase_22's Avatar
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    Can you argue that F is surjective because every linear map can be represented as a matrix, since Hom_K(U,V) is the set of all these linear maps then a row in the matrix M _{m \times n}(K) can be mapped to the linear map it represents.


    Hence it must be surjective.
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