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Math Help - Dimension and basis

  1. #1
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    Dimension and basis

    Let M be a subspace of the vector space \mathbb{R}_2[t] generated by p_1(T)=t^2+t+1 and p_2(T)=1-t^2, and N be a subspace generated by q_1(T)=t^2+2t+3 and q_2(T)=t^2-t+1. Show the dimension of the following subspaces:  M+N, M \cap N, and give a basis for each.
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    Re: Dimension and basis

    Quote Originally Posted by noricka View Post
    Let M be a subspace of the vector space \mathbb{R}_2[t] generated by p_1(T)=t^2+t+1 and p_2(T)=1-t^2, and N be a subspace generated by q_1(T)=t^2+2t+3 and q_2(T)=t^2-t+1. Show the dimension of the following subspaces:  M+N, M \cap N, and give a basis for each.
    So, what are your thoughts?
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    Re: Dimension and basis

    Quote Originally Posted by Drexel28 View Post
    So, what are your thoughts?
    I have tried the following: if I take the linear combination of p_1 p_2 q_1 q_2, I get (a+b+c+d)t^2 + (a+2c-d)t +(a+b+3c+d). And a basis of this polynomial is \{1,t,t^2\}, which means the dimension of M+N is 3.

    And if M and N are finite dimension subspaces then dim(M+N)=dim M + dim N- dim(M \cap N). But what is the dimension of M and N?
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  4. #4
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    Re: Dimension and basis

    how many basis elements do M and N have?
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    Re: Dimension and basis

    Both of them have 3?
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  6. #6
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    Re: Dimension and basis

    how can a set generated by 2 elements have 3 basis vectors???

    to be fair, p1 and p2 are only listed as generating elements, its not explicitly stated whether or not {p1,p2} forms a basis.

    but, by the very definition of "generate" they are elements of a span-set. are they linearly independent? (if you decide they are, how can you be sure?

    and what does this mean dim(M) is?)
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    Re: Dimension and basis

    p1 and p2 are linearly independent, so their dimension is 2?
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    Re: Dimension and basis

    elements do not have dimension, vector spaces (and their subspaces) have dimension.

    the dimension of a vector space, V, is defined to be the number of elements in ANY basis.

    if S is a subset of V, then dim(span(S)) ≤ |S|.

    if S is a linearly independent set, then dim(span(S)) = |S|.
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  9. #9
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    Re: Dimension and basis

    So the dimension of the subspace M generated by p1, and p2 is 2, then. And that's the same with N as well.
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  10. #10
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    Re: Dimension and basis

    Was my idea of the dimension of M+N right? And what about the bases?
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