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Math Help - linearly dependent in Q, linearly independent in R

  1. #1
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    linearly dependent in Q, linearly independent in R

    I need help with this one.

    Find \lambda \in \mathbb{R} such that \{(\lambda,1,0),(1,\lambda,1),(0,1,\lambda)\} <br />
is linearly dependent in \mathbb{Q} and linearly independent in \mathbb{R}.

    Here's what I did:
    \alpha (\lambda, 1, 0)+\beta(1, \lambda, 1)+\gamma (0, 1, \lambda)=0
    Solving this gets me:
    \alpha \lambda+\beta=0
    \alpha + \beta \lambda + \gamma=0 and
    \beta + \gamma \lambda=0.

    This gets me:
    \lambda(\gamma - \alpha)=0 which means that either \lambda=0 or \gamma - \alpha=0.

    The first case gives me \alpha, \beta, \gamma=0, which means that the set is linearly independent both in \mathbb{R} and \mathbb{Q} so that is obviously not the \lambda I'm looking for.

    Now, the other case, \gamma-\alpha=0. This means that \alpha=\gamma. The three equations I had are now two equations:
    \alpha \lambda + \beta=0 and
    2\alpha + \beta \lambda=0.

    Solving that gets me:
    \alpha (2-\lambda^2)=0.

    Now, if \lambda=\pm \sqrt{2} \in \mathbb{R}, I get that \alpha, \beta, \gamma are not necessarily zero, but if \lambda \in \mathbb{Q}, the equation \alpha (2-\lambda^2)=0 implies that \alpha=0, and so \beta=0 and \gamma=0.
    Obviously, \lambda=\pm \sqrt{2} is the one we're looking for.


    Is this correct?
    Is this the only possible \lambda?

    Thank you!
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by harriette View Post
    I need help with this one.

    Find \lambda \in \mathbb{R} such that \{(\lambda,1,0),(1,\lambda,1),(0,1,\lambda)\} <br />
is linearly dependent in \mathbb{Q} and linearly independent in \mathbb{R}.

    Here's what I did:
    \alpha (\lambda, 1, 0)+\beta(1, \lambda, 1)+\gamma (0, 1, \lambda)=0
    Solving this gets me:
    \alpha \lambda+\beta=0
    \alpha + \beta \lambda + \gamma=0 and
    \beta + \gamma \lambda=0.

    This gets me:
    \lambda(\gamma - \alpha)=0 which means that either \lambda=0 or \gamma - \alpha=0.

    The first case gives me \alpha, \beta, \gamma=0, which means that the set is linearly independent both in \mathbb{R} and \mathbb{Q} so that is obviously not the \lambda I'm looking for.

    Now, the other case, \gamma-\alpha=0. This means that \alpha=\gamma. The three equations I had are now two equations:
    \alpha \lambda + \beta=0 and
    2\alpha + \beta \lambda=0.

    Solving that gets me:
    \alpha (2-\lambda^2)=0.

    Now, if \lambda=\pm \sqrt{2} \in \mathbb{R}, I get that \alpha, \beta, \gamma are not necessarily zero, but if \lambda \in \mathbb{Q}, the equation \alpha (2-\lambda^2)=0 implies that \alpha=0, and so \beta=0 and \gamma=0.
    Obviously, \lambda=\pm \sqrt{2} is the one we're looking for.


    Is this correct?
    Is this the only possible \lambda?

    Thank you!
    Hmm...surely if \lambda = 0 then you just have \beta=0 and that \alpha + \gamma=0, not that \alpha=0=\gamma...however, you are right - you can discard this case as it is linearly dependent in both the reals and the rationals.

    And yes, \pm\sqrt{2} are your only two solutions, and your working looks correct. Further, b=-a\lambda. So clearly one of a or b is in \mathbb{R} (as \lambda \not\in \mathbb{Q}). And so the set is linearly independent over \mathbb{Q} but linearly dependent over \mathbb{R} (which is the opposite of what you said, but I think you just wrote them the wrong way round...or I have written them the wrong way round...but I'm pretty sure I haven't!)
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    Hmm...surely if \lambda = 0 then you just have \beta=0 and that \alpha + \gamma=0, not that \alpha=0=\gamma...however, you are right - you can discard this case as it is linearly dependent in both the reals and the rationals.
    Yes, I see this now!
    But how do I conclude that the set is lineraly dependent? Is it because \alpha and \gamma can be anything, as long as it's \alpha=\gamma? So they are not necessarily 0?

    (which is the opposite of what you said, but I think you just wrote them the wrong way round...or I have written them the wrong way round...but I'm pretty sure I haven't!)
    I mixed it up, of course.
    Thank you!!!
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by harriette View Post
    Yes, I see this now!
    But how do I conclude that the set is lineraly dependent? Is it because \alpha and \gamma can be anything, as long as it's \alpha=\gamma? So they are not necessarily 0?
    Watch - you missed a minus sign! It is linearly dependent as for any \alpha, \gamma such that \alpha=-\gamma you get zero. So, as always with a counter-example, plug in a number. So \alpha=1 and \gamma=-1 work.
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  5. #5
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    Thank you once again!
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