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Homework Help: Linear functionals

  1. Sep 14, 2006 #1
    Here is the problem I have been asked to solve:

    Assume that m < n and l1, l2, . . . , lm are linear functionals on an n-dimensional vector space X.
    (a) Prove there exists a non-zero vector x in X such that the scalar product < x, lj >= 0 for 1 <= j <= m. What does this say about the solution of systems of linear equations?
    (b) Under what conditions on the scalars b1, . . . , bm is it true that there exists a vector x in X such that the scalar product < x, lj >= bj for 1 <= j <= m? What does this say about the solution of systems of linear equations?

    I am having trouble understanding the concept of a linear functional and how it relates to the vector space X.
    Last edited: Sep 14, 2006
  2. jcsd
  3. Sep 14, 2006 #2


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    x is a vector in X, and lj is a linear functional on X, so what is < x, lj >? Do you mean lj(x)? Linear functionals are linear functions which map vectors to scalars. By linearity, they are uniquely determined by their behaviour on the basis vectors of the vector space. Let f be a linear functional on a vector space V, and let v1, ..., vn be basis vectors for V. If x is any vector in X, then there exist scalars x1, ..., xn such that

    x = x1v1 + ... + xnvn


    = f(x1v1 + ... + xnvn)
    = f(x1v1) + ... + f(xnvn)
    = x1f(v1) + ... + xnf(vn)
    = (x1, ..., xn).(f(v1), ..., f(vn))

    If l1, ..., lm are linear functionals, consider the row vectors (l1(v1), ..., l1(vn)), ..., (lm(v1), ..., lm(vn)). Do you see how the question is reduced to the following:

    If m < n, and {li(vj) | i = 1, ..., m; j = 1, ..., n} is any set of scalars, do there exists scalars x1, ..., xn such that for each i, it holds that:

    (x1, ..., xn).(li(v1), ..., li(vn))


    [tex](\mbox{Span}\{(l_i(v_1),\, \dots ,\, l_i (v_n))\ :\ 1 \leq i \leq m\})^{\perp}[/tex]
  4. Sep 14, 2006 #3
    < x, lj > is the scalar/dot product
  5. Sep 14, 2006 #4


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    I know that. Suppose v is a vector and x is toenail. What is <v,x>?
  6. Sep 14, 2006 #5
    Yeah, I still have no idea what I'm proving for this problem, nor do I understand the final question, though I imagine if I could do the proof, the question would make sense. A hint would be appreciated.
  7. Sep 15, 2006 #6


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    The dot product takes two vectors of the same space and gives you a scalar. A linear functional on V is not a vector in V. There is a strong relation between the value of a linear functional at a vector and the dot product of that vector with a particular other vector, but that's still a little off in the distance for us. First, you need to get the basics.

    7 is a real number. So is [itex]-\pi[/itex]. It makes sense to talk about, say, 7 x [itex]-\pi[/itex]. You can multiply numbers. Can you multiply a number by a function? What is 7 x sine? It doesn't really make sense (okay, you can make sense of it, but try to understand the basic idea). Similarly, v and x are both vectors, say. Then it makes sense to take their dot product, v.x If f is a linear functional, then what sense does it make to take v.f? You can multiply two numbers together, but you can't multiply a number by a toenail, it's just absurd. Likewise, you can take the dot product of a vector with another vector, but you can't take the dot product of a vector with a functional, it's just absurd. It makes as much sense as taken the dot product of a vector with a toenail.

    If V is a vector space over a field F, then a linear functional is a function: f : V -> F that is linear. So a linear functional (I'll just call it a functional) is a function, which is just something that maps elements of one set to elements of another set. In addition, it is a special kind of function, it is a linear function. That means f(v+w) = f(v)+f(w), and f(av) = af(v), where v and w are vectors in V, and a is a scalar in F. Do you understand this so far?

    Suppose you have an n-dimensional vector space V (over a field F), and m (with m < n) linear functionals f1, ..., fm. You want to prove that there is a non-zero x such that fi(x) = 0 for 1 < i < m.

    Let me give you a related problem. Let V be an n-dimensional vector space. Let {v1, ..., vm} be any set of m vectors from V, with m < n. Prove that there exists a non-zero x such that <x,vi> = 0 for 1 < i < m.
  8. Sep 15, 2006 #7

    matt grime

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    Duals always seem to cause problem's, conceptually, and I don't know why.

    If V and W are vector spaces you are really happy with what the space of linear maps between V and W is. If we are the kind of person who likes bases then it is the set of dim(W) by dim(V) matrices. Well, all we've done is take a general case you're happy with and looked at one particular example, when W is one dimensional. Surely that is even easier to understand than the general case?
  9. Sep 21, 2011 #8
    I have the same question as wurthskidder23 and ive read the reply by AKG. Can anybody give some further hint as how to solve the question. Im not sure how to prove that x is non-zero?
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