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Thread: Linear Algebra help!

  1. #1
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    Linear Algebra help!

    Hi I'm new here and I have a bit of trouble understanding a problem that was given to us:

    Let V and W be finite-dimensional vector spaces and let T:V->W be an isomorphism. Let Vo be a subspace of V. Prove that 1) T(Vo) is a subspace of W, and 2) dim (Vo)=dim(T(Vo))

    I would appreciate any help. thank you
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by x3non25 View Post
    Hi I'm new here and I have a bit of trouble understanding a problem that was given to us:

    Let V and W be finite-dimensional vector spaces and let T:V->W be an isomorphism. Let Vo be a subspace of V. Prove that 1) T(Vo) is a subspace of W, and 2) dim (Vo)=dim(T(Vo))

    I would appreciate any help. thank you

    the first one is easy..
    i)
    clearly, $\displaystyle T(V_0) \subseteq W$ by taking an arbitrary element of $\displaystyle V_0 \subseteq V$ and apply the mapping.

    ii)
    suppose $\displaystyle u,w \in T(V_0)$, then there exist $\displaystyle v_1, v_2 \in V_0$ such that $\displaystyle u=T(v_1)$ and $\displaystyle w=T(v_2)$.
    now, $\displaystyle u+w = T(v_1)+T(v_2) = T(v_1 + v_2)$ since $\displaystyle T$ is an isomorphism. and since $\displaystyle v_1, v_2 \in V_0$, so does $\displaystyle v_1 + v_2$. thus $\displaystyle u+v = T(v_1 + v_2) \in T(V_0)$

    iii)
    Let $\displaystyle a \in F$ and $\displaystyle v \in T(V_0)$. Then $\displaystyle \exists v_0 \in V_0$ such that $\displaystyle v = T(v_0)$
    now, $\displaystyle av = aT(v_0) = T(av_0)$ and just like the reasoning above, $\displaystyle av = T(av_0) \in T(V_0)$.

    for the second one..
    i)
    let dim$\displaystyle (V_0) = n$. take $\displaystyle \left\{ v_1, v_2, ..., v_n \right\}$ be a basis for $\displaystyle V_0$. Let $\displaystyle w \in T(V_0)$. then, there exists a $\displaystyle v \in V_0$ such that $\displaystyle w = T(v)$, say $\displaystyle v = a_1v_1 + a_2v_2 + ... + a_nv_n$..

    $\displaystyle w = T(v) = T(a_1v_1 + a_2v_2 + ... + a_nv_n) = T(a_1v_1) + T(a_2v_2) + ... T(a_nv_n)$
    $\displaystyle =a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n)$

    thus $\displaystyle \left\{ T(v_1), T(v_2), ..., T(v_n)\right\}$ is a spanning set for $\displaystyle T(V_0)$

    ii)
    suppose $\displaystyle a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) = 0$ (the zero vector for $\displaystyle T(V_0)$).
    then,
    $\displaystyle a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) = T(a_1v_1 + a_2v_2 + ... + a_nv_n) = 0$
    $\displaystyle \Longleftrightarrow a_1v_1 + a_2v_2 + ... + a_nv_n = 0$ (the zero vector for $\displaystyle V_0$).. since $\displaystyle \left\{ v_1, v_2, ..., v_n \right\}$ be a basis for $\displaystyle V_0$,
    $\displaystyle a_1v_1 + a_2v_2 + ... + a_nv_n = 0 \Longleftrightarrow a_i = 0 \, \, \forall \, i = 1,...,n$..

    this implies that $\displaystyle \left\{ T(v_1), T(v_2), ..., T(v_n)\right\}$ is a basis for $\displaystyle T(V_0)$. Therefore dim$\displaystyle T(V_0) = n$.. QED
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