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Thread: Diemsion of Hom

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

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

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

    $\displaystyle T(e_1),T(e_2),......,T(e_r)$ where $\displaystyle r \leq n$. Let $\displaystyle T(e_i)=f_i$ 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; Feb 8th 2009 at 02:11 AM.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by Showcase_22 View Post
    I started by making $\displaystyle e_1, e_2,........,e_n$ a basis of U and $\displaystyle f_1,f_2,.....,f_m$ a basis of V.
    Prove that the lin. transfs $\displaystyle T_{ij}$ such that $\displaystyle T_{ij} (e_i) = f_j$(actually they are just maps according to definition, but you can extend it linearly), form a basis of $\displaystyle \text{Hom (U,V)}$
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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 $\displaystyle Hom_K(U,V)$ be the vector space over K of all linear maps from U to V.

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


    Hence it must be surjective.
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