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Math Help - Dimension of the Null Space of Linear Functional

  1. #1
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    Dimension of the Null Space of Linear Functional

    I need help with showing this:

    If f is a linear functional on an n-dimensional vector space X, what dimension can the Null Space N(f) have?

    Answer:

    dim (N(f)) = n or n-1.
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  2. #2
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    Quote Originally Posted by frater_cp View Post
    I need help with showing this:

    If f is a linear functional on an n-dimensional vector space X, what dimension can the Null Space N(f) have?

    Answer:

    dim (N(f)) = n or n-1.
    If N(f) = X then its dimension is n. Otherwise, there must be some element x_0\in X which is not in the null space of f, and by taking a scalar multiple of it, we may assume that f(x_0)=1. If x is any element of X then x = f(x)x_0 + (x-f(x)x_0), and x-f(x)x_0\in N(f). So every element of X is a sum of something in the 1-dimensional space spanned by x_0 and something in N(f). You should be able to deduce from this that dim(N(f)) = n1.
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  3. #3
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    Full of Gratitude

    Hi Opalg, thank you so much for your reply.
    Unfortunately I dont follow the whole argument yet.
    Can you help me a bit more please?

    I follow first part, then...

    So every element of X is a sum of something in the 1-dimensional space spanned by and something in N(f). You should be able to deduce from this that dim(N(f)) = n1.....

    I intuitively can see that dim(span{x_0})=1 and N(f) must then be of dimension n-1 since dimX=n. Is there a mathematical way to show this using n-1 linear independent vectors?
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  4. #4
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    Let \{x_1,x_2,\ldots,x_k\} be a basis for N(f) (so that k is the dimension of n(f), which we don't yet know). Then the calculation in my previous comment shows that \{x_0,x_1,x_2,\ldots,x_k\} is a basis for the whole of X, and therefore contains n vectors. So k+1=n, or k=n1.
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  5. #5
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    Gratitude

    excellent excellent excellently done!!
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