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Math Help - Simple question about linear functionals

  1. #1
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    Simple question about linear functionals

    Hi everyone. Paul Halmos asks a seemingly-innocuous question in Finite Dimensional Vector Spaces: let V be a vector space over a field F and let y and z be linear functionals from V to F. Suppose, furthermore that y(x) = 0 whenever z(x) = 0, for all x in V. Prove that y = az for some scalar a.

    Here's where I am so far: if z = 0 then the problem is trivial, so without loss of generality z(x0) is not zero for some vector x0 in V. Then the only possibility for a is a = y(x0)/z(x0). Now I need to prove the identity for all x in V:
    y(x) = y(x0)z(x)/z(x0)

    But I don't see how to do this. I have absolutely no background with linear functionals, and only know undergraduate-level linear algebra. I'd appreciate a tip.
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  2. #2
    Super Member girdav's Avatar
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    Re: Simple question about linear functionals

    Write x=x-\frac{z(x)}{z(x_0)}x_0+\frac{z(x)}{z(x_0)}x_0 and compute y\left(x-\frac{z(x)}{z(x_0)}x_0\right).
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  3. #3
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    Re: Simple question about linear functionals

    I'm having trouble even with the hint. I would love to compute y\left(x-\frac{z(x)}{z(x_0)}x_0\right). Indeed, this quantity needs to be zero, and then the result immediately follows. But showing that this result is equal to zero is really a trivial modification of the problem:

    y\left(x-\frac{z(x)}{z(x_0)}x_0\right)=y(x)-y\left(\frac{z(x)}{z(x_0)}x_0\right)
    =y(x)-\frac{y(x_0)}{z(x_0)}z(x)
    =(*)

    is precisely the quantity I need to show is zero. I don't see any way to "compute" this further, so that's kind of my problem. I could give it a try:

    (*)=\frac{1}{z(x_0)}\left(y(x)z(x_0)-y(x_0)z(x)\right)
    =\frac{1}{z(x_0)}z\left(y(x)x_0-y(x_0)x\right)

    Basically I'm just trying to use linearity here. I am not aware of any other possible manipulations (short of adding and subtracting a quantity which I just don't see). Could I please have a bit more help?
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  4. #4
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    Re: Simple question about linear functionals

    Ah hah. I see what I needed to do. I do not need to evaluate y(...). I need to show that z(...) = 0, and then by hypotheses, y(...) = 0. But showing that z(...) = 0 is very easy. That solves the problem. Thank you. I was being silly - I should have looked for a way to use the given hypotheses.

    Thanks again! And by the way, a great hint that didn't give away too much of the fun!
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