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Thread: [x,y]=0 is a subspace

  1. #1
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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 $\displaystyle x_1\;and\;x_2$ be any two vectors in X and let $\displaystyle \alpha\;and\;\beta$ be arbitrary scalars. Then,
    $\displaystyle [\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 $\displaystyle x_1\;and\;x_2$ be any two vectors in X and let $\displaystyle \alpha\;and\;\beta$ be arbitrary scalars. Then,
    $\displaystyle [\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 $\displaystyle y(x)=[x,y]$ ...weird. Anyway, use the dimensions theorem since, after all, alinear function is just a linear transformation from

    a vector space $\displaystyle V_\mathbb{F}\rightarrow \mathbb{F}$ , $\displaystyle \mathbb{F}=$ 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 $\displaystyle T:V\rightarrow W$ is a linear transfromation between vectors spaces over the same field, and if $\displaystyle \dim V=n<\infty$ , we get

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

    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 $\displaystyle y(x)=[x,y]$ ...weird.
    Halmos uses $\displaystyle y(x)=[x,y]$ 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 $\displaystyle \{x_1,\cdots,x_n\}$ be a basis of $\displaystyle V$. Any vector in the subspace, say $\displaystyle N$, can be written as
    $\displaystyle x=\xi_1x_1+\cdots+\xi_nx_n$.
    $\displaystyle y(x)=0 \rightarrow \xi_1y(x_1)+\cdots+\xi_ny(x_n)=0$

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

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

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

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

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

    Thanks.
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