linearly dependent in Q, linearly independent in R

**harriette** · April 12th 2011, 01:03 PM

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!

**Swlabr** · April 12th 2011, 01:22 PM

Originally Posted by harriette

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!)

**harriette** · April 12th 2011, 01:30 PM

Originally Posted by Swlabr

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!!!

**Swlabr** · April 12th 2011, 01:42 PM

Originally Posted by harriette

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.

**harriette** · April 12th 2011, 01:44 PM

Thank you once again!

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