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Math Help - [x,y]=0 is a subspace

  1. #1
    Member Mollier's Avatar
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    [x,y]=0 is a subspace

    Hi,

    problem:

    Prove that if y is a linear functional on an n-dimensional vector space V, then the set of all those vectors x for which [x,y]=0 is a subspace of V; what is the dimension of that subspace?

    attempt:

    Let X be the set of vectors x in V for which [x,y]=0 where y is a linear functional. Let x_1\;and\;x_2 be any two vectors in X and let \alpha\;and\;\beta be arbitrary scalars. Then,
     [\alpha x_1+\beta x_2,y]=\alpha[x_1,y]+\beta[x_2,y]=0 and so X is a subspace of V.

    Don't know the dimension, maybe n-1

    Thanks!
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  2. #2
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    Quote Originally Posted by Mollier View Post
    Hi,

    problem:

    Prove that if y is a linear functional on an n-dimensional vector space V, then the set of all those vectors x for which [x,y]=0 is a subspace of V; what is the dimension of that subspace?

    attempt:

    Let X be the set of vectors x in V for which [x,y]=0 where y is a linear functional. Let x_1\;and\;x_2 be any two vectors in X and let \alpha\;and\;\beta be arbitrary scalars. Then,
     [\alpha x_1+\beta x_2,y]=\alpha[x_1,y]+\beta[x_2,y]=0 and so X is a subspace of V.

    Don't know the dimension, maybe n-1

    Thanks!

    Apparently you guys use the notation y(x)=[x,y] ...weird. Anyway, use the dimensions theorem since, after all, alinear function is just a linear transformation from

    a vector space V_\mathbb{F}\rightarrow \mathbb{F} , \mathbb{F}= a field.

    About the dimension: if the functional is the zero func. then___, but if it is now the zero func. then it automatically is ONTO, so___

    Dimensions theorem: if T:V\rightarrow W is a linear transfromation between vectors spaces over the same field, and if \dim V=n<\infty , we get

    \dim(ker\, T)+\dim(Im\, T)=n

    Tonio
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by tonio View Post
    Apparently you guys use the notation y(x)=[x,y] ...weird.
    Halmos uses y(x)=[x,y] in his books. I was just giving it a try

    Quote Originally Posted by tonio View Post
    About the dimension: if the functional is the zero func. then___, but if it is now the zero func. then it automatically is ONTO, so___
    I let \{x_1,\cdots,x_n\} be a basis of V. Any vector in the subspace, say N, can be written as
    x=\xi_1x_1+\cdots+\xi_nx_n.
    y(x)=0 \rightarrow \xi_1y(x_1)+\cdots+\xi_ny(x_n)=0

    I know for a earlier theorem that if V is an n-dimensional vector space with a basis \{x_1,\cdots,x_n\}, then there is a uniquely determined basis \{y_1,\cdots,y_n\} (dual basis) with the property that y_j(x_i)=\delta_{ij}

    So in my y(x), there is some y(x_k) that is equal to 1 and if y(x)=0, then \xi_k=0.

    So I end up with dim(N)=n-1..

    Quote Originally Posted by tonio View Post
    Dimensions theorem: if T:V\rightarrow W is a linear transfromation between vectors spaces over the same field, and if \dim V=n<\infty , we get

    \dim(ker\, T)+\dim(Im\, T)=n
    I will have to read up on this a bit.

    Thanks.
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