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Math Help - Dimension of Vectorspaces

  1. #1
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    Question Dimension of Vectorspaces

    Hi

    how to show that the dimension of a subvector space ist smaller or at least equal to the dimension of the main vectorspace?

    I know that the dimension is the length of the basis but how to use this information here?

    thank you
    greetings
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  2. #2
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    Quote Originally Posted by Herbststurm View Post
    Hi

    how to show that the dimension of a subvector space ist smaller or at least equal to the dimension of the main vectorspace?

    I know that the dimension is the length of the basis but how to use this information here?

    thank you
    greetings
    If V is a subspace of W then say \text{dim}(V) > \text{dim}(W) then it means we can find a basis set for V, \{v_1,...,v_n\}. But these vectors are linearly independent. The problem is that a set of linearly independent vectors cannot exceede the dimension and we see that v_1,...,v_n exceede the dimension of W. This is a contradiction.
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  3. #3
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    Sorry, I don't understand your post

    How could you write that the dimension of the subspace is bigger than the main space?

    Ad why do you say the base is linear dependent. I thought a base is always a linear indipendent creation system?

    greetings
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  4. #4
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    Quote Originally Posted by Herbststurm View Post
    Sorry, I don't understand your post

    How could you write that the dimension of the subspace is bigger than the main space?

    Ad why do you say the base is linear dependent. I thought a base is always a linear indipendent creation system?

    greetings
    Say \text{dim}(W) = 3 and V\subseteq W as a subspace has \text{dim}(V) = 4. Then this means V has a basis of four elements, say, \{ v_1,v_2,v_3,v_4\}. This is a linearly independent set in V. Since v_i \in V \implies v_i \in W because V\subseteq W. Therefore, \{ v_1,v_2,v_3,v_4\} is a linearly independent set in W. But it is impossible to have a linearly independent set so that the number of elements in the set exceede the dimension of the space. Here the number of elements are 4 and the dimension of W is 3 and we see that 4>3. This is a contradiction. Thus, the basis for V must have fewer or equal to than 3 elements.
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